{"id":965,"date":"2016-02-15T22:16:24","date_gmt":"2016-02-15T22:16:24","guid":{"rendered":"https:\/\/courses.candelalearning.com\/nrocarithmetic\/?post_type=chapter&#038;p=965"},"modified":"2024-12-03T22:43:59","modified_gmt":"2024-12-03T22:43:59","slug":"5-2-1-factor-trinomials","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/beginalgebra\/chapter\/5-2-1-factor-trinomials\/","title":{"raw":"More Factoring Methods","rendered":"More Factoring Methods"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Factor by Grouping\r\n<ul>\r\n \t<li>Identify patterns that result from multiplying two binomials and how they affect factoring by grouping<\/li>\r\n \t<li>Factor a four term polynomial by grouping terms<\/li>\r\n<\/ul>\r\n<\/li>\r\n \t<li>Methods for Factoring Trinomials\r\n<ul>\r\n \t<li>Apply an algorithm to rewrite a trinomial as a four term polynomial<\/li>\r\n \t<li>Use factoring by grouping to factor a trinomial<\/li>\r\n \t<li>Use a shortcut to factor trinomials of the form [latex]x^2+bx+c[\/latex]<\/li>\r\n \t<li>Factor trinomials of the form [latex]ax^2+bx+c[\/latex]<\/li>\r\n \t<li>Recognize where to place negative signs when factoring a trinomial<\/li>\r\n \t<li>Recognize when a polynomial is a difference of squares, and how it would factor as the product of two binomials<\/li>\r\n<\/ul>\r\n<\/li>\r\n<\/ul>\r\n<\/div>\r\nWhen we learned to multiply two binomials, we found that the result, before combining like terms, was a four term polynomial, as in this example:\u00a0[latex]\\left(x+4\\right)\\left(x+2\\right)=x^{2}+2x+4x+8[\/latex].\r\n\r\nWe can apply what we have learned about factoring out a common monomial \u00a0to return a four term polynomial to the product of two binomials. Why would we even want to do this?\r\n\r\n[caption id=\"attachment_4825\" align=\"aligncenter\" width=\"391\"]<img class=\" wp-image-4825\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/117\/2016\/06\/12195953\/Screen-Shot-2016-06-12-at-12.59.11-PM-300x138.png\" alt=\"Thought bubble with the words .....and i should care why?\" width=\"391\" height=\"180\" \/> Why Should I Care?[\/caption]\r\n\r\nBecause it is an important step in learning techniques for factoring trinomials, such as the one you get when you simplify the product of the two binomials from above:\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}\\left(x+4\\right)\\left(x+2\\right)\\\\=x^{2}+2x+4x+8\\\\=x^2+6x+8\\end{array}[\/latex]<\/p>\r\nAdditionally, factoring by grouping is a technique that allows us to factor a polynomial whose terms don't all share a GCF. In the following example, we will introduce you to the technique. Remember, one of the main reasons to factor is because it will help solve polynomial equations.\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\nFactor [latex]a^2+3a+5a+15[\/latex]\r\n[reveal-answer q=\"437455\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"437455\"]\r\n\r\nThere isn't a common factor between all four terms, so we will group the terms into pairs that will enable us to find a GCF for them. \u00a0For example, we wouldn't want to group [latex]a^2\\text{ and }15[\/latex] because they don't share a common factor.\r\n<p style=\"text-align: center;\">[latex]\\left(a^2+3a\\right)+\\left(5a+15\\right)[\/latex]<\/p>\r\nFind the GCF of the first pair of terms.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}\\,\\,\\,\\,a^2=a\\cdot{a}\\\\\\,\\,\\,\\,3a=3\\cdot{a}\\\\\\text{GCF}=a\\end{array}[\/latex]<\/p>\r\nFactor the GCF, <i>a<\/i>, out of the first group.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}\\left(a\\cdot{a}+a\\cdot{3}\\right)+\\left(5a+15\\right)\\\\a\\left(a+3\\right)+\\left(5a+15\\right)\\end{array}[\/latex]<\/p>\r\nFind the GCF of the second pair of terms.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}5a=5\\cdot{a}\\\\15=5\\cdot3\\\\\\text{GCF}=5\\,\\,\\,\\,\\,\\,\\,\\end{array}[\/latex]<\/p>\r\nFactor 5 out of the second group.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}a\\left(a+3\\right)+\\left(5\\cdot{a}+5\\cdot3\\right)\\\\a\\left(a+3\\right)+5\\left(a+3\\right)\\end{array}[\/latex]<\/p>\r\nNotice that the two terms have a common factor [latex]\\left(a+3\\right)[\/latex].\r\n<p style=\"text-align: center;\">[latex]a\\left(a+3\\right)+5\\left(a+3\\right)[\/latex]<\/p>\r\nFactor out the common factor [latex]\\left(a+3\\right)[\/latex] from the two terms.\r\n<p style=\"text-align: center;\">[latex]\\left(a+3\\right)\\left(a+5\\right)[\/latex]<\/p>\r\nNote how the a and 5 become a binomial sum, and the other factor. \u00a0This is probably the most confusing part of factoring by grouping.\r\n<h4>Answer<\/h4>\r\n[latex]a^2+3a+5a+15=\\left(a+3\\right)\\left(a+5\\right)[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nNotice that when you factor a two term polynomial, the result is a monomial times a polynomial. But the factored form of a four-term polynomial is the product of two binomials. As we noted before, this is an important middle step in learning how to factor a three term polynomial.\r\n\r\nThis process is called the <i>grouping technique<\/i>. Broken down into individual steps, here's how to do it (you can also follow this process in the example below).\r\n<ul>\r\n \t<li>Group the terms of the polynomial into pairs that share a GCF.<\/li>\r\n \t<li>Find the greatest common factor and then use the distributive property to pull out the GCF<\/li>\r\n \t<li>Look for the common binomial\u00a0between the factored terms<\/li>\r\n \t<li>Factor the common binomial\u00a0out of the groups, the other factors will make the other binomial<\/li>\r\n<\/ul>\r\nLet\u2019s try factoring a few more four-term polynomials. Note how there is a now a constant in front of the [latex]x^2[\/latex] term. We will just consider this another factor when we are finding the GCF.\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\nFactor [latex]2x^{2}+4x+5x+10[\/latex].\r\n[reveal-answer q=\"313122\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"313122\"]Group terms of the polynomial into pairs.\r\n<p style=\"text-align: center;\">[latex]\\left(2x^{2}+4x\\right)+\\left(5x+10\\right)[\/latex]<\/p>\r\nFactor out the like factor, <i>2x<\/i>, from the first group.\r\n<p style=\"text-align: center;\">[latex]2x\\left(x+2\\right)+\\left(5x+10\\right)[\/latex]<\/p>\r\nFactor out the like factor, 5, from the second group.\r\n<p style=\"text-align: center;\">[latex]2x\\left(x+2\\right)+5\\left(x+2\\right)[\/latex]<\/p>\r\nLook for common factors between the factored forms of the paired terms. Here, the common factor is [latex](x+2)[\/latex].\r\n\r\nFactor out the common factor, [latex]\\left(x+2\\right)[\/latex], from both terms.\r\n<p style=\"text-align: center;\">[latex]\\left(x+2\\right)\\left(2x+5\\right)[\/latex]<\/p>\r\nThe polynomial is now factored.\r\n<h4>Answer<\/h4>\r\n[latex]\\left(x+2\\right)\\left(2x+5\\right)[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nAnother example follows that contains subtraction. Note how we choose a positive GCF\u00a0from each group of terms, and the negative signs stay.\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\nFactor [latex]2x^{2}\u20133x+8x\u201312[\/latex].\r\n[reveal-answer q=\"715080\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"715080\"]Group terms into pairs.\r\n<p style=\"text-align: center;\">[latex](2x^{2}\u20133x)+(8x\u201312)[\/latex]<\/p>\r\nFactor the common factor, <i>x<\/i>, out of the first group and the common factor, 4, out of the second group.\r\n<p style=\"text-align: center;\">[latex]x\\left(2x\u20133\\right)+4\\left(2x\u20133\\right)[\/latex]<\/p>\r\nFactor out the common factor, [latex]\\left(2x\u20133\\right)[\/latex], from both terms.\r\n<p style=\"text-align: center;\">[latex]\\left(x+4\\right)\\left(2x\u20133\\right)[\/latex]<\/p>\r\n\r\n<h4>Answer<\/h4>\r\n[latex]\\left(x+4\\right)\\left(2x-3\\right)[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nThe video that follows provides another example of factoring by grouping.\r\n\r\nhttps:\/\/youtu.be\/RR5nj7RFSiU\r\n\r\nIn the next example, we will have a GCF that is negative. \u00a0It is important to pay attention to what happens to the resulting binomial when the GCF is negative.\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\nFactor [latex]3x^{2}+3x\u20132x\u20132[\/latex].\r\n[reveal-answer q=\"744005\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"744005\"]Group terms into pairs.\r\n<p style=\"text-align: center;\">[latex]\\left(3x^{2}+3x\\right)+\\left(-2x-2\\right)[\/latex]<\/p>\r\nFactor the common factor 3<i>x<\/i> out of first group.\r\n<p style=\"text-align: center;\">[latex]3x\\left(x+1\\right)+\\left(-2x-2\\right)[\/latex]<\/p>\r\nFactor the common factor [latex]\u22122[\/latex] out of the second group. Notice what happens to the signs within the parentheses once [latex]\u22122[\/latex] is factored out.\r\n<p style=\"text-align: center;\">[latex]3x\\left(x+1\\right)-2\\left(x+1\\right)[\/latex]<\/p>\r\nFactor out the common factor, [latex]\\left(x+1\\right)[\/latex], from both terms.\r\n<p style=\"text-align: center;\">[latex]\\left(x+1\\right)\\left(3x-2\\right)[\/latex]<\/p>\r\n\r\n<h4>Answer<\/h4>\r\n[latex]\\left(x+1\\right)\\left(3x-2\\right)[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nIn the following video we present another example of factoring by grouping when one of the GCF is negative.\r\n\r\nhttps:\/\/youtu.be\/0dvGmDGVC5U\r\n\r\nSometimes, you will encounter polynomials that, despite your best efforts, cannot be factored into the product of two binomials.\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\nFactor [latex]7x^{2}\u201321x+5x\u20135[\/latex].\r\n[reveal-answer q=\"262926\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"262926\"]Group terms into pairs.\r\n<p style=\"text-align: center;\">[latex]\\left(7x^{2}\u201321x\\right)+\\left(5x\u20135\\right)[\/latex]<\/p>\r\nFactor the common factor 7<i>x<\/i> out of the first group.\r\n<p style=\"text-align: center;\">[latex]7x\\left(x-3\\right)+\\left(5x-5\\right)[\/latex]<\/p>\r\nFactor the common factor 5 out of the second group.\r\n<p style=\"text-align: center;\">[latex]7x\\left(x-3\\right)+5\\left(x-1\\right)[\/latex]<\/p>\r\nThe two groups [latex]7x\\left(x\u20133\\right)[\/latex] and [latex]5\\left(x\u20131\\right)[\/latex] do not have any common factors, so this polynomial cannot be factored any further.\r\n<p style=\"text-align: center;\">[latex]7x\\left(x\u20133\\right)+5\\left(x\u20131\\right)[\/latex]<\/p>\r\n\r\n<h4>Answer<\/h4>\r\nCannot be factored\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nIn the example above, each pair can be factored, but then there is no common factor between the pairs!\r\n<h2>Factor Trinomials Part I<\/h2>\r\nIn the last section we introduced the technique of factoring by grouping as a means to be able to factor a trinomial. Now we will actually get to the work of starting with a three term polynomial, and rewriting it as a four term polynomial so it can be factored.\r\n\r\nWe will start with factoring trinomials of the form [latex]x^{2}+bx+c[\/latex] that don't have a coefficient in front of the [latex]x^2[\/latex] term.\r\n\r\nRemember that when\u00a0[latex]\\left(x+2\\right)[\/latex] and [latex]\\left(x+5\\right)[\/latex], are multiplied, the result is a four term polynomial and then it is\u00a0simplified into a trinomial:\r\n<p style=\"text-align: center;\">\u00a0[latex]\\left(x+2\\right)\\left(x+5\\right)=x^2+5x+2x+10=x^2+7x+10[\/latex]<\/p>\r\nFactoring is the reverse of multiplying, so let\u2019s go in reverse and factor the trinomial [latex]x^{2}+7x+10[\/latex]. The individual terms [latex]x^{2}[\/latex], [latex]7x[\/latex], and 10 share no common factors. If we\u00a0rewrite the middle term as the sum of the two terms [latex]7x=5x+2x[\/latex] then we can use the grouping technique:\r\n<p style=\"text-align: center;\">[latex](x^{2}+5x)+(2x+10)[\/latex]<\/p>\r\n<p style=\"text-align: center;\">Factor each pair:\u00a0[latex]\\begin{array}{l}x\\underbrace{\\left(x+5\\right)}+2\\underbrace{\\left(x+5\\right)}\\\\\\text{common binomial factor}\\end{array}[\/latex]<\/p>\r\n<p style=\"text-align: center;\">Then pull\u00a0out the common binomial factor: [latex]\\left(x+5\\right)\\left(x+2\\right)[\/latex]<\/p>\r\nWhat would have happened if we had\u00a0rewritten [latex]7x[\/latex] as [latex]6x+x[\/latex]?\r\n<p style=\"text-align: center;\">[latex](x^{2}+6x)+(x+10)[\/latex]<\/p>\r\n<p style=\"text-align: center;\">Factor each pair:\u00a0[latex]x\\left(x+6\\right)+1\\left(x+10\\right)[\/latex]<\/p>\r\nThen we don't have a\u00a0common factor of [latex]\\left(x+5\\right)[\/latex] like we did before. There is a method to the madness of choosing how to rewrite the middle terms so that you will end up with a common binomial factor.\r\n\r\n[caption id=\"attachment_4833\" align=\"aligncenter\" width=\"300\"]<img class=\"size-medium wp-image-4833\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/117\/2016\/06\/12211615\/Screen-Shot-2016-06-12-at-2.15.41-PM-300x281.png\" alt=\"Shakespeare quote: &quot;Though this be madness, yet there is method in it.&quot;\" width=\"300\" height=\"281\" \/> Method to the Madness[\/caption]\r\n\r\nThe following is a summary of the method, then we will show some examples of how to use it.\r\n<div class=\"textbox shaded\">\r\n<h3>Factoring Trinomials in the form\u00a0[latex]x^{2}+bx+c[\/latex]<\/h3>\r\nTo factor a trinomial in the form [latex]x^{2}+bx+c[\/latex], find two integers, <i>r<\/i> and <i>s<\/i>, whose product is <i>c <\/i>and whose sum is <i>b<\/i>.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}r\\cdot{s}=c\\\\\\text{ and }\\\\r+s=b\\end{array}[\/latex]<\/p>\r\nRewrite the trinomial as [latex]x^{2}+rx+sx+c[\/latex]\u00a0and then use grouping and the distributive property to factor the polynomial. The resulting factors will be [latex]\\left(x+r\\right)[\/latex] and [latex]\\left(x+s\\right)[\/latex].\r\n\r\n<\/div>\r\nFor example, to factor [latex]x^{2}+7x+10[\/latex], you are looking for two numbers whose sum is 7 (the coefficient of the middle term) and whose product is 10 (the last term).\r\n\r\nLook at factor pairs of 10:1 and 10, 2, and 5. Do either of these pairs have a sum of 7? Yes, 2 and 5. So you can rewrite\u00a0[latex]7x[\/latex] as [latex]2x+5x[\/latex], and continue factoring as in the example above. Note that you can also rewrite [latex]7x[\/latex] as [latex]5x+2x[\/latex]. Both will work.\r\n\r\nLet\u2019s factor the trinomial [latex]x^{2}+5x+6[\/latex]. In this polynomial, the <i>b<\/i> part of the middle term is 5 and the <i>c<\/i> term is 6. A chart will help us organize possibilities. On the left, list all possible factors of the <i>c<\/i> term, 6; on the right you'll find the sums.\r\n<table style=\"width: 20%;\">\r\n<thead>\r\n<tr>\r\n<th>Factors whose product is 6<\/th>\r\n<th>Sum of the factors<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr>\r\n<td>[latex]1\\cdot6=6[\/latex]<\/td>\r\n<td>[latex]1+6=7[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]2\\cdot3=6[\/latex]<\/td>\r\n<td>[latex]2+3=5[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nThere are only two possible factor combinations, 1 and 6, and 2 and 3. You can see that\u00a0[latex]2+3=5[\/latex]. So [latex]2x+3x=5x[\/latex], giving us the correct middle term.\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\nFactor\u00a0[latex]x^{2}+5x+6[\/latex].\r\n\r\n[reveal-answer q=\"141663\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"141663\"]Use values from the chart above. Replace\u00a0[latex]5x[\/latex] with [latex]2x+3x[\/latex].\r\n<p style=\"text-align: center;\">[latex]x^{2}+2x+3x+6[\/latex]<\/p>\r\nGroup the pairs of terms.\r\n<p style=\"text-align: center;\">[latex]\\left(x^{2}+2x\\right)+\\left(3x+6\\right)[\/latex]<\/p>\r\nFactor <i>x<\/i> out of the first pair of terms\r\n<p style=\"text-align: center;\">[latex]x\\left(x+2\\right)+\\left(3x+6\\right)[\/latex]<\/p>\r\nFactor 3 out of the second pair of terms.\r\n<p style=\"text-align: center;\">[latex]x\\left(x+2\\right)+3\\left(x+2\\right)[\/latex]<\/p>\r\nFactor out [latex]\\left(x+2\\right)[\/latex].\r\n<p style=\"text-align: center;\">[latex]\\left(x+2\\right)\\left(x+3\\right)[\/latex]<\/p>\r\n\r\n<h4>Answer<\/h4>\r\n[latex]\\left(x+2\\right)\\left(x+3\\right)[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nNote that if you wrote [latex]x^{2}+5x+6[\/latex] as [latex]x^{2}+3x+2x+6[\/latex] and grouped the pairs as [latex]\\left(x^{2}+3x\\right)+\\left(2x+6\\right)[\/latex]; then factored, [latex]x\\left(x+3\\right)+2\\left(x+3\\right)[\/latex], and factored out [latex]x+3[\/latex], the answer would be [latex]\\left(x+3\\right)\\left(x+2\\right)[\/latex]. Since multiplication is commutative, the order of the factors does not matter. So this answer is correct as well; they are equivalent answers.\r\n\r\nIn the following video, we present another example of how to use grouping to factor a quadratic polynomial.\r\n\r\nhttps:\/\/youtu.be\/_Rtp7nSxf6c\r\n\r\nFinally, let\u2019s take a look at the trinomial [latex]x^{2}+x\u201312[\/latex]. In this trinomial, the <i>c<\/i> term is [latex]\u221212[\/latex]. So look at all of the combinations of factors whose product is [latex]\u221212[\/latex]. Then see which of these combinations will give you the correct middle term, where <i>b<\/i> is 1.\r\n<table style=\"width: 30%;\">\r\n<thead>\r\n<tr>\r\n<th>Factors whose product is [latex]\u221212[\/latex]<\/th>\r\n<th>Sum of the factors<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr>\r\n<td>[latex]1\\cdot\u221212=\u221212[\/latex]<\/td>\r\n<td>[latex]1+\u221212=\u221211[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]2\\cdot\u22126=\u221212[\/latex]<\/td>\r\n<td>[latex]2+\u22126=\u22124[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]3\\cdot\u22124=\u221212[\/latex]<\/td>\r\n<td>[latex]3+\u22124=\u22121[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]4\\cdot\u22123=\u221212[\/latex]<\/td>\r\n<td>[latex]4+\u22123=1[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]6\\cdot\u22122=\u221212[\/latex]<\/td>\r\n<td>[latex]6+\u22122=4[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]12\\cdot\u22121=\u221212[\/latex]<\/td>\r\n<td>[latex]12+\u22121=11[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nThere is only one combination where the product is [latex]\u221212[\/latex] and the sum is 1, and that is when [latex]r=4[\/latex], and [latex]s=\u22123[\/latex]. Let\u2019s use these to factor our original trinomial.\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\nFactor\u00a0[latex]x^{2}+x\u201312[\/latex].\r\n\r\n[reveal-answer q=\"205737\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"205737\"]Rewrite the trinomial using the values from the chart above. Use values [latex]r=4[\/latex] and [latex]s=\u22123[\/latex].\r\n<p style=\"text-align: center;\">[latex]x^{2}+4x+\u22123x\u201312[\/latex]<\/p>\r\nGroup pairs of terms.\r\n<p style=\"text-align: center;\">[latex]\\left(x^{2}+4x\\right)+\\left(\u22123x\u201312\\right)[\/latex]<\/p>\r\nFactor <em>x<\/em> out of the first group.\r\n<p style=\"text-align: center;\">[latex]x\\left(x+4\\right)+\\left(-3x-12\\right)[\/latex]<\/p>\r\nFactor \u22123 out of the second group.\r\n<p style=\"text-align: center;\">[latex]x\\left(x+4\\right)\u20133\\left(x+4\\right)[\/latex]<\/p>\r\nFactor out [latex]\\left(x+4\\right)[\/latex].\r\n<p style=\"text-align: center;\">[latex]\\left(x+4\\right)\\left(x-3\\right)[\/latex]<\/p>\r\n\r\n<h4>Answer<\/h4>\r\n[latex]\\left(x+4\\right)\\left(x-3\\right)[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nIn the above example, you could also rewrite [latex]x^{2}+x-12[\/latex] as [latex]x^{2}\u2013 3x+4x\u201312[\/latex] first. Then factor [latex]x\\left(x \u2013 3\\right)+4\\left(x\u20133\\right)[\/latex], and factor out [latex]\\left(x\u20133\\right)[\/latex] getting [latex]\\left(x\u20133\\right)\\left(x+4\\right)[\/latex]. Since multiplication is commutative, this is the same answer.\r\n<h3>Factoring Tips<\/h3>\r\nFactoring trinomials is a matter of practice and patience. Sometimes, the appropriate number combinations will just pop out and seem so obvious! Other times, despite trying many possibilities, the correct combinations are hard to find. And, there are times when the trinomial cannot be factored.\r\n\r\nWhile there is no foolproof way to find the right combination on the first guess, there are some tips that can ease the way.\r\n<div class=\"textbox shaded\">\r\n<h3>Tips for Finding Values that Work<\/h3>\r\nWhen factoring a trinomial in the form [latex]x^{2}+bx+c[\/latex], consider the following tips.\r\n\r\nLook at the <i>c<\/i> term first.\r\n<ul>\r\n \t<li>If the <i>c<\/i> term is a positive number, then the factors of <i>c<\/i> will both be positive or both be negative. In other words, <i>r<\/i> and <i>s<\/i> will have the same sign.<\/li>\r\n \t<li>If the <i>c<\/i> term is a negative number, then one factor of <i>c<\/i> will be positive, and one factor of <i>c<\/i> will be negative. Either <i>r<\/i> or <i>s<\/i> will be negative, but not both.<\/li>\r\n<\/ul>\r\nLook at the <i>b<\/i> term second.\r\n<ul>\r\n \t<li>If the <i>c<\/i> term is positive and the <i>b<\/i> term is positive, then both <i>r<\/i> and <i>s<\/i> are positive.<\/li>\r\n \t<li>If the <i>c<\/i> term is positive and the <i>b<\/i> term is negative, then both <i>r <\/i>and <i>s<\/i> are negative.<\/li>\r\n \t<li>If the <i>c<\/i> term is negative and the <i>b<\/i> term is positive, then the factor that is positive will have the greater absolute value. That is, if [latex]|r|&gt;|s|[\/latex], then <i>r<\/i> is positive and <i>s <\/i>is negative.<\/li>\r\n \t<li>If the <i>c<\/i> term is negative and the <i>b<\/i> term is negative, then the factor that is negative will have the greater absolute value. That is, if [latex]|r|&gt;|s|[\/latex],<i> <\/i>then <i>r<\/i> is negative and <i>s <\/i>is positive.<\/li>\r\n<\/ul>\r\n<\/div>\r\nAfter you have factored a number of trinomials in the form [latex]x^{2}+bx+c[\/latex], you may notice that the numbers you identify for <i>r<\/i> and <i>s<\/i> end up being included in the factored form of the trinomial. Have a look at the following chart, which reviews the three problems you have seen so far.\r\n<table>\r\n<tbody>\r\n<tr>\r\n<th>Trinomial<\/th>\r\n<th>[latex]x^{2}+7x+10[\/latex]<\/th>\r\n<th>[latex]x^{2}+5x+6[\/latex]<\/th>\r\n<th>[latex]x^{2}+x-12[\/latex]<\/th>\r\n<\/tr>\r\n<tr>\r\n<th><i>r<\/i> and <i>s<\/i> values<\/th>\r\n<td>[latex]r=+5,s=+2[\/latex]<\/td>\r\n<td>[latex]r=+2,s=+3[\/latex]<\/td>\r\n<td>[latex]r=+4,s=\u20133[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<th>Factored form<\/th>\r\n<td>[latex]\\left(x+5\\right)\\left(x+2\\right)[\/latex]<\/td>\r\n<td>[latex]\\left(x+2\\right)\\left(x+3\\right)[\/latex]<\/td>\r\n<td>[latex](x+4)(x\u20133)[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<h2>The Shortcut<\/h2>\r\nNotice that in each of the examples above, the <i>r<\/i> and <i>s<\/i> values are repeated in the factored form of the trinomial.\u00a0So what does this mean? It means that in trinomials of the form [latex]x^{2}+bx+c[\/latex] (where the coefficient in front of [latex]x^{2}[\/latex]\u00a0is 1), if you can identify the correct <i>r<\/i> and <i>s<\/i> values, you can effectively skip the grouping steps and go right to the factored form. For those of you that like shortcuts, let's look at some examples where we use this idea.\r\n\r\n[caption id=\"attachment_4884\" align=\"aligncenter\" width=\"735\"]<img class=\" wp-image-4884\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/117\/2016\/06\/14174839\/Screen-Shot-2016-06-14-at-10.43.58-AM-300x167.png\" alt=\"Picture of a sidewalk leading to a parking lot. There is a path through the grass to teh right of the sidewalk through the trees that has been made by people walking on the grass. The shortcut to the parking lot is the preferred way.\" width=\"735\" height=\"409\" \/> Shortcut This Way[\/caption]\r\n\r\nIn the next two examples, we will show how you can skip the step of factoring by grouping and move directly to the factored form of a product of two binomials with the r and s values that you find. The idea is that you can build factors for a trinomial in this form: [latex]x^2+bx+c[\/latex] by finding r and s, then placing them in two binomial factors like this:\r\n<p style=\"text-align: center;\">[latex]\\left(x+r\\right)\\left(x+s\\right)\\text{ OR }\\left(x+s\\right)\\left(x+r\\right)[\/latex]<\/p>\r\n\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nFactor: [latex]y^2+6y-27[\/latex]\r\n[reveal-answer q=\"601131\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"601131\"]\r\n\r\nFind r and s:\r\n<table style=\"width: 20%;\">\r\n<thead>\r\n<tr>\r\n<th>Factors whose product is -27<\/th>\r\n<th>Sum of the factors<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr>\r\n<td>[latex]1\\cdot{-27}=-27[\/latex]<\/td>\r\n<td>[latex]1-27=-26[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]3\\cdot{-9}=-27[\/latex]<\/td>\r\n<td>[latex]3-9=-6[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]-3\\cdot{9}=-27[\/latex]<\/td>\r\n<td>\u00a0[latex]-3+9=6[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nInstead of rewriting the middle term, we will use the values of r and s that give the product and sum that we need.\r\n\r\nIn this case:\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}r=-3\\\\s=9\\end{array}[\/latex]<\/p>\r\nIt helps to start by writing two empty sets of parentheses:\r\n<p style=\"text-align: center;\">[latex]\\left(\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\right)\\left(\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\right)[\/latex]<\/p>\r\n<p style=\"text-align: left;\">The squared term is y, so we will place a y in each set of parentheses:<\/p>\r\n<p style=\"text-align: center;\">[latex]\\left(y\\,\\,\\,\\,\\,\\,\\,\\,\\right)\\left(y\\,\\,\\,\\,\\,\\,\\,\\,\\right)[\/latex]<\/p>\r\n<p style=\"text-align: left;\">Now we can fill in the rest of each binomial with the values we found for r and s.<\/p>\r\n<p style=\"text-align: center;\">[latex]\\left(y-3\\right)\\left(y+9\\right)[\/latex]<\/p>\r\n<p style=\"text-align: left;\">Note how we kept the sign on each of the values. \u00a0The nice thing about factoring is you can check your work. Multiply the binomials together to see if you did it correctly.<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}\\left(y-3\\right)\\left(y+9\\right)\\\\=y^2+9y-3y-27\\\\=y^2+6y-27\\end{array}[\/latex]<\/p>\r\n\r\n<h4 style=\"text-align: left;\">Answer<\/h4>\r\n[latex]\\left(y-3\\right)\\left(y+9\\right)[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nWe will show one more example so you can gain more experience.\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nFactor: [latex]-m^2+16m-48[\/latex]\r\n[reveal-answer q=\"402116\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"402116\"]\r\n\r\nThere is a negative in front of the squared term, so we will factor out a negative one from the whole trinomial first. Remember, this boils down to changing the sign of all the terms:\r\n<p style=\"text-align: center;\">[latex]-m^2+16m-48=-1\\left(m^2-16m+48\\right)[\/latex]<\/p>\r\n<p style=\"text-align: left;\">Now we can factor [latex]\\left(m^2-16m+48\\right)[\/latex] by finding r and s. Note that b is negative, and c\u00a0is positive so we are probably looking for two negative numbers:<\/p>\r\n\r\n<table style=\"width: 20%;\">\r\n<thead>\r\n<tr>\r\n<th>Factors whose product is 48<\/th>\r\n<th>Sum of the factors<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr>\r\n<td>[latex]-1\\cdot{-48}=48[\/latex]<\/td>\r\n<td>[latex]-1-48=-49[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]-2\\cdot{-12}=48[\/latex]<\/td>\r\n<td>[latex]-2-12=-14[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]-3\\cdot{-16}=-48[\/latex]<\/td>\r\n<td>[latex]-3-16=-19[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]-4\\cdot{-12}=-48[\/latex]<\/td>\r\n<td>[latex]-4-12=-16[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<p style=\"text-align: left;\">There are more factors whose product is 48, but we have found the ones that sum to -16, so we can stop.<\/p>\r\n<p style=\"text-align: left;\">[latex]\\begin{array}{l}r=-4\\\\s=-12\\end{array}[\/latex]<\/p>\r\n<p style=\"text-align: left;\">Now we can fill in each binomial with the values we found for r and s, make sure to use the correct variable!<\/p>\r\n<p style=\"text-align: center;\">[latex]\\left(m-4\\right)\\left(m-12\\right)[\/latex]<\/p>\r\n<p style=\"text-align: left;\">We are not done yet, remember that we factored out a negative sign in the first step. We need to remember to include that.<\/p>\r\n<p style=\"text-align: center;\">\u00a0[latex]-1\\left(m-4\\right)\\left(m-12\\right)[\/latex]<\/p>\r\n\r\n<h4 style=\"text-align: left;\">Answer<\/h4>\r\n[latex]-m^2+16m-48=-1\\left(m-4\\right)\\left(m-12\\right)[\/latex]\r\n<p style=\"text-align: left;\">[\/hidden-answer]<\/p>\r\n\r\n<\/div>\r\nIn the following video, we present two more examples of factoring a trinomial using the shortcut presented here.\r\n\r\nhttps:\/\/youtu.be\/-SVBVVYVNTM\r\n<h2>Factor Trinomials Part II<\/h2>\r\nThe next goal is for you to be comfortable with recognizing where to place negative signs, and whether a trinomial can even be factored. \u00a0Additionally, we will explore\u00a0one special case to look out for at the end of this page.\r\n\r\nNot all trinomials look like [latex]x^{2}+5x+6[\/latex], where the coefficient in front of the [latex]x^{2}[\/latex]\u00a0term is 1. In these cases, your first step should be to look for common factors for the three terms.\r\n<table>\r\n<thead>\r\n<tr>\r\n<th>Trinomial<\/th>\r\n<th>Factor out Common Factor<\/th>\r\n<th>Factored<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr>\r\n<td>[latex]2x^{2}+10x+12[\/latex]<\/td>\r\n<td>[latex]2(x^{2}+5x+6)[\/latex]<\/td>\r\n<td>[latex]2\\left(x+2\\right)\\left(x+3\\right)[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]\u22125a^{2}\u221215a\u221210[\/latex]<\/td>\r\n<td>[latex]\u22125(a^{2}+3a+2)[\/latex]<\/td>\r\n<td>[latex]\u22125\\left(a+2\\right)\\left(a+1\\right)[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]c^{3}\u20138c^{2}+15c[\/latex]<\/td>\r\n<td>[latex]c\\left(c^{2}\u20138c+15\\right)[\/latex]<\/td>\r\n<td>[latex]c\\left(c\u20135\\right)\\left(c\u20133\\right)[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]y^{4}\u20139y^{3}\u201310y^{2}[\/latex]<\/td>\r\n<td>[latex]y^{2}\\left(y^{2}\u20139y\u201310\\right)[\/latex]<\/td>\r\n<td>[latex]y^{2}\\left(y\u201310\\right)\\left(y+1\\right)[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nNotice that once you have identified and pulled out the common factor, you can factor the remaining trinomial as usual. This process is shown below.\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\nFactor\u00a0[latex]3x^{3}\u20133x^{2}\u201390x[\/latex].\r\n\r\n[reveal-answer q=\"298928\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"298928\"]Since 3 is a common factor for the three terms, factor out the 3.\r\n<p style=\"text-align: center;\">[latex]3\\left(x^{3}\u2013x^{2}\u201330x\\right)[\/latex]<\/p>\r\n<i>x<\/i> is also a common factor, so factor out <i>x<\/i>.\r\n<p style=\"text-align: center;\">[latex]3x\\left(x^{2}\u2013x\u201330\\right)[\/latex]<\/p>\r\nNow you can factor the trinomial\u00a0[latex]x^{2}\u2013x\u201330[\/latex]. To find <i>r<\/i> and <i>s<\/i>, identify two numbers whose product is [latex]\u221230[\/latex] and whose sum is [latex]\u22121[\/latex].\r\n\r\nThe pair of factors is [latex]\u22126[\/latex] and [latex]5[\/latex]. So replace [latex]\u2013x[\/latex] with [latex]\u22126x+5x[\/latex].\r\n<p style=\"text-align: center;\">[latex]3x\\left(x^{2}\u20136x+5x\u201330\\right)[\/latex]<\/p>\r\nUse grouping to consider the terms in pairs.\r\n<p style=\"text-align: center;\">[latex]3x\\left[\\left(x^{2}\u20136x\\right)+\\left(5x\u201330\\right)\\right][\/latex]<\/p>\r\nFactor <i>x<\/i> out of the first group and factor 5 out of the second group.\r\n<p style=\"text-align: center;\">[latex]3x\\left[\\left(x\\left(x\u20136\\right)\\right)+5\\left(x\u20136\\right)\\right][\/latex]<\/p>\r\nThen factor out [latex]x\u20136[\/latex].\r\n<p style=\"text-align: center;\">[latex]3x\\left(x\u20136\\right)\\left(x+5\\right)[\/latex]<\/p>\r\n\r\n<h4>Answer<\/h4>\r\n[latex]3x\\left(x\u20136\\right)\\left(x+5\\right)[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nThe following video contains two more examples of factoring a quadratic trinomial where the first step is to factor out a GCF. We use the shortcut method instead of factoring by grouping.\r\n\r\nhttps:\/\/youtu.be\/pgH77rAtsbs\r\n\r\nThe general form of trinomials with a leading coefficient of <i>a<\/i> is [latex]ax^{2}+bx+c[\/latex]. Sometimes the factor of <i>a<\/i> can be factored as you saw above; this happens when <i>a<\/i> can be factored out of all three terms. The remaining trinomial that still needs factoring will then be simpler, with the leading term only being an [latex]x^{2}[\/latex]\u00a0term, instead of an [latex]ax^{2}[\/latex]\u00a0term.\r\n\r\nHowever, if the coefficients of all three terms of a trinomial don\u2019t have a common factor, then you will need to factor the trinomial with a coefficient of something other than 1.\r\n<div class=\"textbox shaded\">\r\n<h3>Factoring Trinomials in the form [latex]ax^{2}+bx+c[\/latex]<\/h3>\r\nTo factor a trinomial in the form [latex]ax^{2}+bx+c[\/latex], find two integers, <i>r<\/i> and <i>s<\/i>, whose sum is <i>b<\/i> and whose product is <i>ac.<\/i>\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}r\\cdot{s}=a\\cdot{c}\\\\r+s=b\\end{array}[\/latex]<\/p>\r\nRewrite the trinomial as [latex]ax^{2}+rx+sx+c[\/latex] and then use grouping and the distributive property to factor the polynomial.\r\n\r\n<\/div>\r\nThis is almost the same as factoring trinomials in the form [latex]x^{2}+bx+c[\/latex],<i> <\/i>as in this form\u00a0[latex]a=1[\/latex].<i> <\/i>Now you are looking for two factors whose product is [latex]a\\cdot{c}[\/latex],\u00a0and whose sum is <i>b<\/i>.\r\n\r\nLet\u2019s see how this strategy works by factoring [latex]6z^{2}+11z+4[\/latex].\r\n\r\nIn this trinomial, [latex]a=6[\/latex], [latex]b=11[\/latex], and [latex]c=4[\/latex]. According to the strategy, you need to find two factors, <i>r<\/i> and <i>s<\/i>, whose sum is [latex]b=11[\/latex]\u00a0and whose product is [latex]a\\cdot{c}=6\\cdot4=24[\/latex]. You can make a chart to organize the possible factor combinations. (Notice that this chart only has positive numbers. Since <i>ac<\/i> is positive and <i>b<\/i> is positive, you can be certain that the two factors you're looking for are also positive numbers.)\r\n<table style=\"width: 20%;\">\r\n<thead>\r\n<tr>\r\n<th>Factors whose product is 24<\/th>\r\n<th>Sum of the factors<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr>\r\n<td>[latex]1\\cdot24=24[\/latex]<\/td>\r\n<td>[latex]1+24=25[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]2\\cdot12=24[\/latex]<\/td>\r\n<td>[latex]2+12=14[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]3\\cdot8=24[\/latex]<\/td>\r\n<td>[latex]3+8=11[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]4\\cdot6=24[\/latex]<\/td>\r\n<td>[latex]4+6=10[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nThere is only one combination where the product is 24 and the sum is 11, and that is when [latex]r=3[\/latex], and [latex]s=8[\/latex]. Let\u2019s use these values to factor the original trinomial.\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\nFactor [latex]6z^{2}+11z+4[\/latex].\r\n[reveal-answer q=\"796129\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"796129\"]Rewrite the middle term, [latex]11z[\/latex], as [latex]3z + 8z[\/latex] (from the chart above.)\r\n<p style=\"text-align: center;\">[latex]6z^{2}+3z+8z+4[\/latex]<\/p>\r\nGroup pairs. Use grouping to consider the terms in pairs.\r\n<p style=\"text-align: center;\">[latex]\\left(6z^{2}+3z\\right)+\\left(8z+4\\right)[\/latex]<\/p>\r\nFactor 3<i>z<\/i> out of the first group and 4 out of the second group.\r\n<p style=\"text-align: center;\">[latex]3z\\left(2z+1\\right)+4\\left(2z+1\\right)[\/latex]<\/p>\r\nFactor out [latex]\\left(2z+1\\right)[\/latex].\r\n<p style=\"text-align: center;\">[latex]\\left(2z+1\\right)\\left(3z+4\\right)[\/latex]<\/p>\r\n\r\n<h4>Answer<\/h4>\r\n[latex]\\left(2z+1\\right)\\left(3z+4\\right)[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nIn the following video, we present another example of factoring a trinomial using grouping. \u00a0In this example, the middle term, b, is negative. Note how having a negative middle term and a positive c term influence the options for r and s when factoring.\r\n\r\nhttps:\/\/youtu.be\/agDaQ_cZnNc\r\n\r\nBefore going any further, it is worth mentioning that not all trinomials can be factored using integer pairs. Take the trinomial [latex]2z^{2}+35z+7[\/latex], for instance. Can you think of two integers whose sum is [latex]b=35[\/latex] and whose product is [latex]a\\cdot{c}=2\\cdot7=14[\/latex]? There are none! This type of trinomial, which cannot be factored using integers, is called a prime trinomial.\r\n\r\nIn some situations, <i>a<\/i> is negative, as in [latex]\u22124h^{2}+11h+3[\/latex]. It often makes sense to factor out [latex]\u22121[\/latex] as the first step in factoring, as doing so will change the sign of [latex]ax^{2}[\/latex]\u00a0from negative to positive, making the remaining trinomial easier to factor.\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\nFactor\u00a0[latex]\u22124h^{2}+11h+3[\/latex]\r\n[reveal-answer q=\"471034\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"471034\"]Factor [latex]\u22121[\/latex] out of the trinomial. Notice that the signs of all three terms have changed.\r\n<p style=\"text-align: center;\">[latex]\u22121\\left(4h^{2}\u201311h\u20133\\right)[\/latex]<\/p>\r\nTo factor the trinomial, you need to figure out how to rewrite [latex]\u221211h[\/latex].\u00a0The product of [latex]rs=4\\cdot\u22123=\u221212[\/latex], and the sum of [latex]rs=\u221211[\/latex].\r\n<table style=\"width: 20%;\">\r\n<tbody>\r\n<tr>\r\n<td>[latex]r\\cdot{s}=\u221212[\/latex]<\/td>\r\n<td>[latex]r+s=\u221211[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]\u221212\\cdot1=\u221212[\/latex]<\/td>\r\n<td>[latex]\u221212+1=\u221211[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]\u22126\\cdot2=\u221212[\/latex]<\/td>\r\n<td>[latex]\u22126+2=\u22124[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]\u22124\\cdot3=\u221212[\/latex]<\/td>\r\n<td>[latex]\u22124+3=\u22121[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nRewrite the middle term\u00a0[latex]\u221211h[\/latex] as [latex]\u221212h+1h[\/latex].\r\n<p style=\"text-align: center;\">[latex]\u22121\\left(4h^{2}\u201312h+1h\u20133\\right)[\/latex]<\/p>\r\nGroup terms.\r\n<p style=\"text-align: center;\">[latex]\u22121\\left[\\left(4h^{2}\u201312h\\right)+\\left(1h\u20133\\right)\\right][\/latex]<\/p>\r\nFactor out 4<i>h<\/i> from the first pair. The second group cannot be factored further, but you can write it as [latex]+1\\left(h\u20133\\right)[\/latex]\u00a0since [latex]+1\\left(h\u20133\\right)=\\left(h\u20133\\right)[\/latex]. This helps with factoring in the next step.\r\n<p style=\"text-align: center;\">[latex]\u22121\\left[4h\\left(h\u20133\\right)+1\\left(h\u20133\\right)\\right][\/latex]<\/p>\r\nFactor out a common factor of [latex]\\left(h\u20133\\right)[\/latex]. Notice you are left with [latex]\\left(h\u20133\\right)\\left(4h+1\\right)[\/latex]; the [latex]+1[\/latex] comes from the term [latex]+1\\left(h\u20133\\right)[\/latex]\u00a0in the previous step.\r\n<p style=\"text-align: center;\">[latex]\u22121\\left[\\left(h\u20133\\right)\\left(4h+1\\right)\\right][\/latex]<\/p>\r\n\r\n<h4>Answer<\/h4>\r\n[latex]\u22121\\left(h\u20133\\right)\\left(4h+1\\right)[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nNote that the answer above can also be written as [latex]\\left(\u2212h+3\\right)\\left(4h+1\\right)[\/latex] or [latex]\\left(h\u20133\\right)\\left(\u22124h\u20131\\right)[\/latex] if you multiply [latex]\u22121[\/latex]\u00a0times one of the other factors.\r\n\r\nIn the following video we present another example of factoring a trinomial in the form [latex]-ax^2+bx+c[\/latex] using the grouping technique.\r\n\r\nhttps:\/\/youtu.be\/zDAMjdBfkDs\r\n<h2>Difference of Squares<\/h2>\r\nWe would be remiss if we failed to introduce one more type of polynomial that can be factored. This polynomial can be factored into two binomials but has only two terms. \u00a0Let's start from the product of two binomials to see the pattern.\r\n\r\nGiven the product of two binomials: [latex]\\left(x-2\\right)\\left(x+2\\right)[\/latex], if we multiply them together, we lose the middle term that we are used to seeing as a result.\r\n<p style=\"text-align: left;\">Multiply:<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}\\left(x-2\\right)\\left(x+2\\right)\\\\\\text{}\\\\=x^2-2x+2x-2^2\\\\\\text{}\\\\=x^2-2^2\\\\\\text{}\\\\=x^2-4\\end{array}[\/latex]<\/p>\r\n\u00a0The polynomial [latex]x^2-4[\/latex] is called a difference of squares because teach term can be written as something squared. \u00a0A difference of squares will always factor in the following way:\r\n<div class=\"textbox shaded\">\r\n<h3>Factor a Difference of Squares<\/h3>\r\nGiven [latex]a^2-b^2[\/latex], it's factored form will be [latex]\\left(a+b\\right)\\left(a-b\\right)[\/latex]\r\n\r\n<\/div>\r\nLet\u2019s factor [latex]x^{2}\u20134[\/latex]\u00a0by writing it as a trinomial, [latex]x^{2}+0x\u20134[\/latex]. \u00a0This\u00a0is similar in format to the\u00a0trinomials we have been factoring so far, so let\u2019s\u00a0use the same method.\r\n<p style=\"text-align: center;\">Find the factors of [latex]a\\cdot{c}[\/latex]\u00a0whose sum is <i>b, <\/i>in this case, 0<i>:<\/i><\/p>\r\n\r\n<table style=\"width: 20%;\">\r\n<thead>\r\n<tr>\r\n<th>Factors of [latex]\u22124[\/latex]<\/th>\r\n<th>Sum of the factors<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr>\r\n<td>[latex]1\\cdot-4=\u22124[\/latex]<\/td>\r\n<td>[latex]1-4=\u22123[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]2\\cdot\u22122=\u22124[\/latex]<\/td>\r\n<td>[latex]2-2=0[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]-1\\cdot4=\u22124[\/latex]<\/td>\r\n<td>[latex]-1+4=3[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n2, and -2 have a sum of 0. You can use these to factor [latex]x^{2}\u20134[\/latex].\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\nFactor [latex]x^{2}\u20134[\/latex].\r\n[reveal-answer q=\"23133\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"23133\"]Rewrite\u00a0[latex]0x[\/latex] as [latex]\u22122x+2x[\/latex].\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}x^{2}+0x-4\\\\x^{2}-2x+2x-4\\end{array}[\/latex]<\/p>\r\nGroup pairs.\r\n<p style=\"text-align: center;\">[latex]\\left(x^{2}\u20132x\\right)+\\left(2x\u20134\\right)[\/latex]<\/p>\r\nFactor <i>x<\/i> out of the first group. Factor 2 out of the second group.\r\n<p style=\"text-align: center;\">[latex]x\\left(x\u20132\\right)+2\\left(x\u20132\\right)[\/latex]<\/p>\r\nFactor out [latex]\\left(x\u20132\\right)[\/latex].\r\n<p style=\"text-align: center;\">[latex]\\left(x\u20132\\right)\\left(x+2\\right)[\/latex]<\/p>\r\n\r\n<h4>Answer<\/h4>\r\n[latex]\\left(3x\u20132\\right)\\left(3x+2\\right)[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nSince order doesn't matter with multiplication, the answer can also be written as [latex]\\left(x+2\\right)\\left(x\u20132\\right)[\/latex].\r\n\r\nYou can check the answer by multiplying [latex]\\left(x\u20132\\right)\\left(x+2\\right)=x^{2}+2x\u20132x\u20134=x^{2}\u20134[\/latex].\r\n\r\nThe following video show two more examples of factoring a difference of squares.\r\n\r\nhttps:\/\/youtu.be\/Li9IBp5HrFA\r\n<h2>Summary<\/h2>\r\nWhen a trinomial is in the form of [latex]ax^{2}+bx+c[\/latex], where <i>a<\/i> is a coefficient other than 1, look first for common factors for all three terms. Factor out the common factor first, then factor the remaining simpler trinomial. If the remaining trinomial is still of the form [latex]ax^{2}+bx+c[\/latex], find two integers, <i>r<\/i> and <i>s<\/i>, whose sum is <i>b<\/i> and whose product is <i>ac.<\/i> Then rewrite the trinomial as [latex]ax^{2}+rx+sx+c[\/latex]\u00a0and use grouping and the distributive property to factor the polynomial.\r\n\r\nWhen [latex]ax^{2}[\/latex] is negative, you can factor [latex]\u22121[\/latex] out of the whole trinomial before continuing.\r\n\r\nA difference of squares [latex]a^2-b^2[\/latex] will factor in this way: [latex]\\left(a+b\\right)\\left(a-b\\right)[\/latex].\r\n<h2>Summary<\/h2>\r\nTrinomials in the form [latex]x^{2}+bx+c[\/latex] can be factored by finding two integers, <i>r<\/i> and <i>s<\/i>, whose sum is <i>b<\/i> and whose product is <i>c.<\/i> Rewrite the trinomial as [latex]x^{2}+rx+sx+c[\/latex]\u00a0and then use grouping and the distributive property to factor the polynomial.\r\n\r\n&nbsp;","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Factor by Grouping\n<ul>\n<li>Identify patterns that result from multiplying two binomials and how they affect factoring by grouping<\/li>\n<li>Factor a four term polynomial by grouping terms<\/li>\n<\/ul>\n<\/li>\n<li>Methods for Factoring Trinomials\n<ul>\n<li>Apply an algorithm to rewrite a trinomial as a four term polynomial<\/li>\n<li>Use factoring by grouping to factor a trinomial<\/li>\n<li>Use a shortcut to factor trinomials of the form [latex]x^2+bx+c[\/latex]<\/li>\n<li>Factor trinomials of the form [latex]ax^2+bx+c[\/latex]<\/li>\n<li>Recognize where to place negative signs when factoring a trinomial<\/li>\n<li>Recognize when a polynomial is a difference of squares, and how it would factor as the product of two binomials<\/li>\n<\/ul>\n<\/li>\n<\/ul>\n<\/div>\n<p>When we learned to multiply two binomials, we found that the result, before combining like terms, was a four term polynomial, as in this example:\u00a0[latex]\\left(x+4\\right)\\left(x+2\\right)=x^{2}+2x+4x+8[\/latex].<\/p>\n<p>We can apply what we have learned about factoring out a common monomial \u00a0to return a four term polynomial to the product of two binomials. Why would we even want to do this?<\/p>\n<div id=\"attachment_4825\" style=\"width: 401px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-4825\" class=\"wp-image-4825\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/117\/2016\/06\/12195953\/Screen-Shot-2016-06-12-at-12.59.11-PM-300x138.png\" alt=\"Thought bubble with the words .....and i should care why?\" width=\"391\" height=\"180\" \/><\/p>\n<p id=\"caption-attachment-4825\" class=\"wp-caption-text\">Why Should I Care?<\/p>\n<\/div>\n<p>Because it is an important step in learning techniques for factoring trinomials, such as the one you get when you simplify the product of the two binomials from above:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}\\left(x+4\\right)\\left(x+2\\right)\\\\=x^{2}+2x+4x+8\\\\=x^2+6x+8\\end{array}[\/latex]<\/p>\n<p>Additionally, factoring by grouping is a technique that allows us to factor a polynomial whose terms don&#8217;t all share a GCF. In the following example, we will introduce you to the technique. Remember, one of the main reasons to factor is because it will help solve polynomial equations.<\/p>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>Factor [latex]a^2+3a+5a+15[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q437455\">Show Solution<\/span><\/p>\n<div id=\"q437455\" class=\"hidden-answer\" style=\"display: none\">\n<p>There isn&#8217;t a common factor between all four terms, so we will group the terms into pairs that will enable us to find a GCF for them. \u00a0For example, we wouldn&#8217;t want to group [latex]a^2\\text{ and }15[\/latex] because they don&#8217;t share a common factor.<\/p>\n<p style=\"text-align: center;\">[latex]\\left(a^2+3a\\right)+\\left(5a+15\\right)[\/latex]<\/p>\n<p>Find the GCF of the first pair of terms.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}\\,\\,\\,\\,a^2=a\\cdot{a}\\\\\\,\\,\\,\\,3a=3\\cdot{a}\\\\\\text{GCF}=a\\end{array}[\/latex]<\/p>\n<p>Factor the GCF, <i>a<\/i>, out of the first group.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}\\left(a\\cdot{a}+a\\cdot{3}\\right)+\\left(5a+15\\right)\\\\a\\left(a+3\\right)+\\left(5a+15\\right)\\end{array}[\/latex]<\/p>\n<p>Find the GCF of the second pair of terms.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}5a=5\\cdot{a}\\\\15=5\\cdot3\\\\\\text{GCF}=5\\,\\,\\,\\,\\,\\,\\,\\end{array}[\/latex]<\/p>\n<p>Factor 5 out of the second group.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}a\\left(a+3\\right)+\\left(5\\cdot{a}+5\\cdot3\\right)\\\\a\\left(a+3\\right)+5\\left(a+3\\right)\\end{array}[\/latex]<\/p>\n<p>Notice that the two terms have a common factor [latex]\\left(a+3\\right)[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]a\\left(a+3\\right)+5\\left(a+3\\right)[\/latex]<\/p>\n<p>Factor out the common factor [latex]\\left(a+3\\right)[\/latex] from the two terms.<\/p>\n<p style=\"text-align: center;\">[latex]\\left(a+3\\right)\\left(a+5\\right)[\/latex]<\/p>\n<p>Note how the a and 5 become a binomial sum, and the other factor. \u00a0This is probably the most confusing part of factoring by grouping.<\/p>\n<h4>Answer<\/h4>\n<p>[latex]a^2+3a+5a+15=\\left(a+3\\right)\\left(a+5\\right)[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>Notice that when you factor a two term polynomial, the result is a monomial times a polynomial. But the factored form of a four-term polynomial is the product of two binomials. As we noted before, this is an important middle step in learning how to factor a three term polynomial.<\/p>\n<p>This process is called the <i>grouping technique<\/i>. Broken down into individual steps, here&#8217;s how to do it (you can also follow this process in the example below).<\/p>\n<ul>\n<li>Group the terms of the polynomial into pairs that share a GCF.<\/li>\n<li>Find the greatest common factor and then use the distributive property to pull out the GCF<\/li>\n<li>Look for the common binomial\u00a0between the factored terms<\/li>\n<li>Factor the common binomial\u00a0out of the groups, the other factors will make the other binomial<\/li>\n<\/ul>\n<p>Let\u2019s try factoring a few more four-term polynomials. Note how there is a now a constant in front of the [latex]x^2[\/latex] term. We will just consider this another factor when we are finding the GCF.<\/p>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>Factor [latex]2x^{2}+4x+5x+10[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q313122\">Show Solution<\/span><\/p>\n<div id=\"q313122\" class=\"hidden-answer\" style=\"display: none\">Group terms of the polynomial into pairs.<\/p>\n<p style=\"text-align: center;\">[latex]\\left(2x^{2}+4x\\right)+\\left(5x+10\\right)[\/latex]<\/p>\n<p>Factor out the like factor, <i>2x<\/i>, from the first group.<\/p>\n<p style=\"text-align: center;\">[latex]2x\\left(x+2\\right)+\\left(5x+10\\right)[\/latex]<\/p>\n<p>Factor out the like factor, 5, from the second group.<\/p>\n<p style=\"text-align: center;\">[latex]2x\\left(x+2\\right)+5\\left(x+2\\right)[\/latex]<\/p>\n<p>Look for common factors between the factored forms of the paired terms. Here, the common factor is [latex](x+2)[\/latex].<\/p>\n<p>Factor out the common factor, [latex]\\left(x+2\\right)[\/latex], from both terms.<\/p>\n<p style=\"text-align: center;\">[latex]\\left(x+2\\right)\\left(2x+5\\right)[\/latex]<\/p>\n<p>The polynomial is now factored.<\/p>\n<h4>Answer<\/h4>\n<p>[latex]\\left(x+2\\right)\\left(2x+5\\right)[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>Another example follows that contains subtraction. Note how we choose a positive GCF\u00a0from each group of terms, and the negative signs stay.<\/p>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>Factor [latex]2x^{2}\u20133x+8x\u201312[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q715080\">Show Solution<\/span><\/p>\n<div id=\"q715080\" class=\"hidden-answer\" style=\"display: none\">Group terms into pairs.<\/p>\n<p style=\"text-align: center;\">[latex](2x^{2}\u20133x)+(8x\u201312)[\/latex]<\/p>\n<p>Factor the common factor, <i>x<\/i>, out of the first group and the common factor, 4, out of the second group.<\/p>\n<p style=\"text-align: center;\">[latex]x\\left(2x\u20133\\right)+4\\left(2x\u20133\\right)[\/latex]<\/p>\n<p>Factor out the common factor, [latex]\\left(2x\u20133\\right)[\/latex], from both terms.<\/p>\n<p style=\"text-align: center;\">[latex]\\left(x+4\\right)\\left(2x\u20133\\right)[\/latex]<\/p>\n<h4>Answer<\/h4>\n<p>[latex]\\left(x+4\\right)\\left(2x-3\\right)[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>The video that follows provides another example of factoring by grouping.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-1\" title=\"Ex 2: Intro to Factor By Grouping Technique\" width=\"500\" height=\"375\" src=\"https:\/\/www.youtube.com\/embed\/RR5nj7RFSiU?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<p>In the next example, we will have a GCF that is negative. \u00a0It is important to pay attention to what happens to the resulting binomial when the GCF is negative.<\/p>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>Factor [latex]3x^{2}+3x\u20132x\u20132[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q744005\">Show Solution<\/span><\/p>\n<div id=\"q744005\" class=\"hidden-answer\" style=\"display: none\">Group terms into pairs.<\/p>\n<p style=\"text-align: center;\">[latex]\\left(3x^{2}+3x\\right)+\\left(-2x-2\\right)[\/latex]<\/p>\n<p>Factor the common factor 3<i>x<\/i> out of first group.<\/p>\n<p style=\"text-align: center;\">[latex]3x\\left(x+1\\right)+\\left(-2x-2\\right)[\/latex]<\/p>\n<p>Factor the common factor [latex]\u22122[\/latex] out of the second group. Notice what happens to the signs within the parentheses once [latex]\u22122[\/latex] is factored out.<\/p>\n<p style=\"text-align: center;\">[latex]3x\\left(x+1\\right)-2\\left(x+1\\right)[\/latex]<\/p>\n<p>Factor out the common factor, [latex]\\left(x+1\\right)[\/latex], from both terms.<\/p>\n<p style=\"text-align: center;\">[latex]\\left(x+1\\right)\\left(3x-2\\right)[\/latex]<\/p>\n<h4>Answer<\/h4>\n<p>[latex]\\left(x+1\\right)\\left(3x-2\\right)[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>In the following video we present another example of factoring by grouping when one of the GCF is negative.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-2\" title=\"Ex 1: Intro to Factor By Grouping Technique\" width=\"500\" height=\"375\" src=\"https:\/\/www.youtube.com\/embed\/0dvGmDGVC5U?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<p>Sometimes, you will encounter polynomials that, despite your best efforts, cannot be factored into the product of two binomials.<\/p>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>Factor [latex]7x^{2}\u201321x+5x\u20135[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q262926\">Show Solution<\/span><\/p>\n<div id=\"q262926\" class=\"hidden-answer\" style=\"display: none\">Group terms into pairs.<\/p>\n<p style=\"text-align: center;\">[latex]\\left(7x^{2}\u201321x\\right)+\\left(5x\u20135\\right)[\/latex]<\/p>\n<p>Factor the common factor 7<i>x<\/i> out of the first group.<\/p>\n<p style=\"text-align: center;\">[latex]7x\\left(x-3\\right)+\\left(5x-5\\right)[\/latex]<\/p>\n<p>Factor the common factor 5 out of the second group.<\/p>\n<p style=\"text-align: center;\">[latex]7x\\left(x-3\\right)+5\\left(x-1\\right)[\/latex]<\/p>\n<p>The two groups [latex]7x\\left(x\u20133\\right)[\/latex] and [latex]5\\left(x\u20131\\right)[\/latex] do not have any common factors, so this polynomial cannot be factored any further.<\/p>\n<p style=\"text-align: center;\">[latex]7x\\left(x\u20133\\right)+5\\left(x\u20131\\right)[\/latex]<\/p>\n<h4>Answer<\/h4>\n<p>Cannot be factored<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>In the example above, each pair can be factored, but then there is no common factor between the pairs!<\/p>\n<h2>Factor Trinomials Part I<\/h2>\n<p>In the last section we introduced the technique of factoring by grouping as a means to be able to factor a trinomial. Now we will actually get to the work of starting with a three term polynomial, and rewriting it as a four term polynomial so it can be factored.<\/p>\n<p>We will start with factoring trinomials of the form [latex]x^{2}+bx+c[\/latex] that don&#8217;t have a coefficient in front of the [latex]x^2[\/latex] term.<\/p>\n<p>Remember that when\u00a0[latex]\\left(x+2\\right)[\/latex] and [latex]\\left(x+5\\right)[\/latex], are multiplied, the result is a four term polynomial and then it is\u00a0simplified into a trinomial:<\/p>\n<p style=\"text-align: center;\">\u00a0[latex]\\left(x+2\\right)\\left(x+5\\right)=x^2+5x+2x+10=x^2+7x+10[\/latex]<\/p>\n<p>Factoring is the reverse of multiplying, so let\u2019s go in reverse and factor the trinomial [latex]x^{2}+7x+10[\/latex]. The individual terms [latex]x^{2}[\/latex], [latex]7x[\/latex], and 10 share no common factors. If we\u00a0rewrite the middle term as the sum of the two terms [latex]7x=5x+2x[\/latex] then we can use the grouping technique:<\/p>\n<p style=\"text-align: center;\">[latex](x^{2}+5x)+(2x+10)[\/latex]<\/p>\n<p style=\"text-align: center;\">Factor each pair:\u00a0[latex]\\begin{array}{l}x\\underbrace{\\left(x+5\\right)}+2\\underbrace{\\left(x+5\\right)}\\\\\\text{common binomial factor}\\end{array}[\/latex]<\/p>\n<p style=\"text-align: center;\">Then pull\u00a0out the common binomial factor: [latex]\\left(x+5\\right)\\left(x+2\\right)[\/latex]<\/p>\n<p>What would have happened if we had\u00a0rewritten [latex]7x[\/latex] as [latex]6x+x[\/latex]?<\/p>\n<p style=\"text-align: center;\">[latex](x^{2}+6x)+(x+10)[\/latex]<\/p>\n<p style=\"text-align: center;\">Factor each pair:\u00a0[latex]x\\left(x+6\\right)+1\\left(x+10\\right)[\/latex]<\/p>\n<p>Then we don&#8217;t have a\u00a0common factor of [latex]\\left(x+5\\right)[\/latex] like we did before. There is a method to the madness of choosing how to rewrite the middle terms so that you will end up with a common binomial factor.<\/p>\n<div id=\"attachment_4833\" style=\"width: 310px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-4833\" class=\"size-medium wp-image-4833\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/117\/2016\/06\/12211615\/Screen-Shot-2016-06-12-at-2.15.41-PM-300x281.png\" alt=\"Shakespeare quote: &quot;Though this be madness, yet there is method in it.&quot;\" width=\"300\" height=\"281\" \/><\/p>\n<p id=\"caption-attachment-4833\" class=\"wp-caption-text\">Method to the Madness<\/p>\n<\/div>\n<p>The following is a summary of the method, then we will show some examples of how to use it.<\/p>\n<div class=\"textbox shaded\">\n<h3>Factoring Trinomials in the form\u00a0[latex]x^{2}+bx+c[\/latex]<\/h3>\n<p>To factor a trinomial in the form [latex]x^{2}+bx+c[\/latex], find two integers, <i>r<\/i> and <i>s<\/i>, whose product is <i>c <\/i>and whose sum is <i>b<\/i>.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}r\\cdot{s}=c\\\\\\text{ and }\\\\r+s=b\\end{array}[\/latex]<\/p>\n<p>Rewrite the trinomial as [latex]x^{2}+rx+sx+c[\/latex]\u00a0and then use grouping and the distributive property to factor the polynomial. The resulting factors will be [latex]\\left(x+r\\right)[\/latex] and [latex]\\left(x+s\\right)[\/latex].<\/p>\n<\/div>\n<p>For example, to factor [latex]x^{2}+7x+10[\/latex], you are looking for two numbers whose sum is 7 (the coefficient of the middle term) and whose product is 10 (the last term).<\/p>\n<p>Look at factor pairs of 10:1 and 10, 2, and 5. Do either of these pairs have a sum of 7? Yes, 2 and 5. So you can rewrite\u00a0[latex]7x[\/latex] as [latex]2x+5x[\/latex], and continue factoring as in the example above. Note that you can also rewrite [latex]7x[\/latex] as [latex]5x+2x[\/latex]. Both will work.<\/p>\n<p>Let\u2019s factor the trinomial [latex]x^{2}+5x+6[\/latex]. In this polynomial, the <i>b<\/i> part of the middle term is 5 and the <i>c<\/i> term is 6. A chart will help us organize possibilities. On the left, list all possible factors of the <i>c<\/i> term, 6; on the right you&#8217;ll find the sums.<\/p>\n<table style=\"width: 20%;\">\n<thead>\n<tr>\n<th>Factors whose product is 6<\/th>\n<th>Sum of the factors<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<td>[latex]1\\cdot6=6[\/latex]<\/td>\n<td>[latex]1+6=7[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]2\\cdot3=6[\/latex]<\/td>\n<td>[latex]2+3=5[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>There are only two possible factor combinations, 1 and 6, and 2 and 3. You can see that\u00a0[latex]2+3=5[\/latex]. So [latex]2x+3x=5x[\/latex], giving us the correct middle term.<\/p>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>Factor\u00a0[latex]x^{2}+5x+6[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q141663\">Show Solution<\/span><\/p>\n<div id=\"q141663\" class=\"hidden-answer\" style=\"display: none\">Use values from the chart above. Replace\u00a0[latex]5x[\/latex] with [latex]2x+3x[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]x^{2}+2x+3x+6[\/latex]<\/p>\n<p>Group the pairs of terms.<\/p>\n<p style=\"text-align: center;\">[latex]\\left(x^{2}+2x\\right)+\\left(3x+6\\right)[\/latex]<\/p>\n<p>Factor <i>x<\/i> out of the first pair of terms<\/p>\n<p style=\"text-align: center;\">[latex]x\\left(x+2\\right)+\\left(3x+6\\right)[\/latex]<\/p>\n<p>Factor 3 out of the second pair of terms.<\/p>\n<p style=\"text-align: center;\">[latex]x\\left(x+2\\right)+3\\left(x+2\\right)[\/latex]<\/p>\n<p>Factor out [latex]\\left(x+2\\right)[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]\\left(x+2\\right)\\left(x+3\\right)[\/latex]<\/p>\n<h4>Answer<\/h4>\n<p>[latex]\\left(x+2\\right)\\left(x+3\\right)[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>Note that if you wrote [latex]x^{2}+5x+6[\/latex] as [latex]x^{2}+3x+2x+6[\/latex] and grouped the pairs as [latex]\\left(x^{2}+3x\\right)+\\left(2x+6\\right)[\/latex]; then factored, [latex]x\\left(x+3\\right)+2\\left(x+3\\right)[\/latex], and factored out [latex]x+3[\/latex], the answer would be [latex]\\left(x+3\\right)\\left(x+2\\right)[\/latex]. Since multiplication is commutative, the order of the factors does not matter. So this answer is correct as well; they are equivalent answers.<\/p>\n<p>In the following video, we present another example of how to use grouping to factor a quadratic polynomial.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-3\" title=\"Ex 1:  Factor a Quadratic Expression Using Grouping When a = 1\" width=\"500\" height=\"375\" src=\"https:\/\/www.youtube.com\/embed\/_Rtp7nSxf6c?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<p>Finally, let\u2019s take a look at the trinomial [latex]x^{2}+x\u201312[\/latex]. In this trinomial, the <i>c<\/i> term is [latex]\u221212[\/latex]. So look at all of the combinations of factors whose product is [latex]\u221212[\/latex]. Then see which of these combinations will give you the correct middle term, where <i>b<\/i> is 1.<\/p>\n<table style=\"width: 30%;\">\n<thead>\n<tr>\n<th>Factors whose product is [latex]\u221212[\/latex]<\/th>\n<th>Sum of the factors<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<td>[latex]1\\cdot\u221212=\u221212[\/latex]<\/td>\n<td>[latex]1+\u221212=\u221211[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]2\\cdot\u22126=\u221212[\/latex]<\/td>\n<td>[latex]2+\u22126=\u22124[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]3\\cdot\u22124=\u221212[\/latex]<\/td>\n<td>[latex]3+\u22124=\u22121[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]4\\cdot\u22123=\u221212[\/latex]<\/td>\n<td>[latex]4+\u22123=1[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]6\\cdot\u22122=\u221212[\/latex]<\/td>\n<td>[latex]6+\u22122=4[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]12\\cdot\u22121=\u221212[\/latex]<\/td>\n<td>[latex]12+\u22121=11[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>There is only one combination where the product is [latex]\u221212[\/latex] and the sum is 1, and that is when [latex]r=4[\/latex], and [latex]s=\u22123[\/latex]. Let\u2019s use these to factor our original trinomial.<\/p>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>Factor\u00a0[latex]x^{2}+x\u201312[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q205737\">Show Solution<\/span><\/p>\n<div id=\"q205737\" class=\"hidden-answer\" style=\"display: none\">Rewrite the trinomial using the values from the chart above. Use values [latex]r=4[\/latex] and [latex]s=\u22123[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]x^{2}+4x+\u22123x\u201312[\/latex]<\/p>\n<p>Group pairs of terms.<\/p>\n<p style=\"text-align: center;\">[latex]\\left(x^{2}+4x\\right)+\\left(\u22123x\u201312\\right)[\/latex]<\/p>\n<p>Factor <em>x<\/em> out of the first group.<\/p>\n<p style=\"text-align: center;\">[latex]x\\left(x+4\\right)+\\left(-3x-12\\right)[\/latex]<\/p>\n<p>Factor \u22123 out of the second group.<\/p>\n<p style=\"text-align: center;\">[latex]x\\left(x+4\\right)\u20133\\left(x+4\\right)[\/latex]<\/p>\n<p>Factor out [latex]\\left(x+4\\right)[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]\\left(x+4\\right)\\left(x-3\\right)[\/latex]<\/p>\n<h4>Answer<\/h4>\n<p>[latex]\\left(x+4\\right)\\left(x-3\\right)[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>In the above example, you could also rewrite [latex]x^{2}+x-12[\/latex] as [latex]x^{2}\u2013 3x+4x\u201312[\/latex] first. Then factor [latex]x\\left(x \u2013 3\\right)+4\\left(x\u20133\\right)[\/latex], and factor out [latex]\\left(x\u20133\\right)[\/latex] getting [latex]\\left(x\u20133\\right)\\left(x+4\\right)[\/latex]. Since multiplication is commutative, this is the same answer.<\/p>\n<h3>Factoring Tips<\/h3>\n<p>Factoring trinomials is a matter of practice and patience. Sometimes, the appropriate number combinations will just pop out and seem so obvious! Other times, despite trying many possibilities, the correct combinations are hard to find. And, there are times when the trinomial cannot be factored.<\/p>\n<p>While there is no foolproof way to find the right combination on the first guess, there are some tips that can ease the way.<\/p>\n<div class=\"textbox shaded\">\n<h3>Tips for Finding Values that Work<\/h3>\n<p>When factoring a trinomial in the form [latex]x^{2}+bx+c[\/latex], consider the following tips.<\/p>\n<p>Look at the <i>c<\/i> term first.<\/p>\n<ul>\n<li>If the <i>c<\/i> term is a positive number, then the factors of <i>c<\/i> will both be positive or both be negative. In other words, <i>r<\/i> and <i>s<\/i> will have the same sign.<\/li>\n<li>If the <i>c<\/i> term is a negative number, then one factor of <i>c<\/i> will be positive, and one factor of <i>c<\/i> will be negative. Either <i>r<\/i> or <i>s<\/i> will be negative, but not both.<\/li>\n<\/ul>\n<p>Look at the <i>b<\/i> term second.<\/p>\n<ul>\n<li>If the <i>c<\/i> term is positive and the <i>b<\/i> term is positive, then both <i>r<\/i> and <i>s<\/i> are positive.<\/li>\n<li>If the <i>c<\/i> term is positive and the <i>b<\/i> term is negative, then both <i>r <\/i>and <i>s<\/i> are negative.<\/li>\n<li>If the <i>c<\/i> term is negative and the <i>b<\/i> term is positive, then the factor that is positive will have the greater absolute value. That is, if [latex]|r|>|s|[\/latex], then <i>r<\/i> is positive and <i>s <\/i>is negative.<\/li>\n<li>If the <i>c<\/i> term is negative and the <i>b<\/i> term is negative, then the factor that is negative will have the greater absolute value. That is, if [latex]|r|>|s|[\/latex],<i> <\/i>then <i>r<\/i> is negative and <i>s <\/i>is positive.<\/li>\n<\/ul>\n<\/div>\n<p>After you have factored a number of trinomials in the form [latex]x^{2}+bx+c[\/latex], you may notice that the numbers you identify for <i>r<\/i> and <i>s<\/i> end up being included in the factored form of the trinomial. Have a look at the following chart, which reviews the three problems you have seen so far.<\/p>\n<table>\n<tbody>\n<tr>\n<th>Trinomial<\/th>\n<th>[latex]x^{2}+7x+10[\/latex]<\/th>\n<th>[latex]x^{2}+5x+6[\/latex]<\/th>\n<th>[latex]x^{2}+x-12[\/latex]<\/th>\n<\/tr>\n<tr>\n<th><i>r<\/i> and <i>s<\/i> values<\/th>\n<td>[latex]r=+5,s=+2[\/latex]<\/td>\n<td>[latex]r=+2,s=+3[\/latex]<\/td>\n<td>[latex]r=+4,s=\u20133[\/latex]<\/td>\n<\/tr>\n<tr>\n<th>Factored form<\/th>\n<td>[latex]\\left(x+5\\right)\\left(x+2\\right)[\/latex]<\/td>\n<td>[latex]\\left(x+2\\right)\\left(x+3\\right)[\/latex]<\/td>\n<td>[latex](x+4)(x\u20133)[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<h2>The Shortcut<\/h2>\n<p>Notice that in each of the examples above, the <i>r<\/i> and <i>s<\/i> values are repeated in the factored form of the trinomial.\u00a0So what does this mean? It means that in trinomials of the form [latex]x^{2}+bx+c[\/latex] (where the coefficient in front of [latex]x^{2}[\/latex]\u00a0is 1), if you can identify the correct <i>r<\/i> and <i>s<\/i> values, you can effectively skip the grouping steps and go right to the factored form. For those of you that like shortcuts, let&#8217;s look at some examples where we use this idea.<\/p>\n<div id=\"attachment_4884\" style=\"width: 745px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-4884\" class=\"wp-image-4884\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/117\/2016\/06\/14174839\/Screen-Shot-2016-06-14-at-10.43.58-AM-300x167.png\" alt=\"Picture of a sidewalk leading to a parking lot. There is a path through the grass to teh right of the sidewalk through the trees that has been made by people walking on the grass. The shortcut to the parking lot is the preferred way.\" width=\"735\" height=\"409\" \/><\/p>\n<p id=\"caption-attachment-4884\" class=\"wp-caption-text\">Shortcut This Way<\/p>\n<\/div>\n<p>In the next two examples, we will show how you can skip the step of factoring by grouping and move directly to the factored form of a product of two binomials with the r and s values that you find. The idea is that you can build factors for a trinomial in this form: [latex]x^2+bx+c[\/latex] by finding r and s, then placing them in two binomial factors like this:<\/p>\n<p style=\"text-align: center;\">[latex]\\left(x+r\\right)\\left(x+s\\right)\\text{ OR }\\left(x+s\\right)\\left(x+r\\right)[\/latex]<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Factor: [latex]y^2+6y-27[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q601131\">Show Solution<\/span><\/p>\n<div id=\"q601131\" class=\"hidden-answer\" style=\"display: none\">\n<p>Find r and s:<\/p>\n<table style=\"width: 20%;\">\n<thead>\n<tr>\n<th>Factors whose product is -27<\/th>\n<th>Sum of the factors<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<td>[latex]1\\cdot{-27}=-27[\/latex]<\/td>\n<td>[latex]1-27=-26[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]3\\cdot{-9}=-27[\/latex]<\/td>\n<td>[latex]3-9=-6[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]-3\\cdot{9}=-27[\/latex]<\/td>\n<td>\u00a0[latex]-3+9=6[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>Instead of rewriting the middle term, we will use the values of r and s that give the product and sum that we need.<\/p>\n<p>In this case:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}r=-3\\\\s=9\\end{array}[\/latex]<\/p>\n<p>It helps to start by writing two empty sets of parentheses:<\/p>\n<p style=\"text-align: center;\">[latex]\\left(\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\right)\\left(\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\right)[\/latex]<\/p>\n<p style=\"text-align: left;\">The squared term is y, so we will place a y in each set of parentheses:<\/p>\n<p style=\"text-align: center;\">[latex]\\left(y\\,\\,\\,\\,\\,\\,\\,\\,\\right)\\left(y\\,\\,\\,\\,\\,\\,\\,\\,\\right)[\/latex]<\/p>\n<p style=\"text-align: left;\">Now we can fill in the rest of each binomial with the values we found for r and s.<\/p>\n<p style=\"text-align: center;\">[latex]\\left(y-3\\right)\\left(y+9\\right)[\/latex]<\/p>\n<p style=\"text-align: left;\">Note how we kept the sign on each of the values. \u00a0The nice thing about factoring is you can check your work. Multiply the binomials together to see if you did it correctly.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}\\left(y-3\\right)\\left(y+9\\right)\\\\=y^2+9y-3y-27\\\\=y^2+6y-27\\end{array}[\/latex]<\/p>\n<h4 style=\"text-align: left;\">Answer<\/h4>\n<p>[latex]\\left(y-3\\right)\\left(y+9\\right)[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>We will show one more example so you can gain more experience.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Factor: [latex]-m^2+16m-48[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q402116\">Show Solution<\/span><\/p>\n<div id=\"q402116\" class=\"hidden-answer\" style=\"display: none\">\n<p>There is a negative in front of the squared term, so we will factor out a negative one from the whole trinomial first. Remember, this boils down to changing the sign of all the terms:<\/p>\n<p style=\"text-align: center;\">[latex]-m^2+16m-48=-1\\left(m^2-16m+48\\right)[\/latex]<\/p>\n<p style=\"text-align: left;\">Now we can factor [latex]\\left(m^2-16m+48\\right)[\/latex] by finding r and s. Note that b is negative, and c\u00a0is positive so we are probably looking for two negative numbers:<\/p>\n<table style=\"width: 20%;\">\n<thead>\n<tr>\n<th>Factors whose product is 48<\/th>\n<th>Sum of the factors<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<td>[latex]-1\\cdot{-48}=48[\/latex]<\/td>\n<td>[latex]-1-48=-49[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]-2\\cdot{-12}=48[\/latex]<\/td>\n<td>[latex]-2-12=-14[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]-3\\cdot{-16}=-48[\/latex]<\/td>\n<td>[latex]-3-16=-19[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]-4\\cdot{-12}=-48[\/latex]<\/td>\n<td>[latex]-4-12=-16[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p style=\"text-align: left;\">There are more factors whose product is 48, but we have found the ones that sum to -16, so we can stop.<\/p>\n<p style=\"text-align: left;\">[latex]\\begin{array}{l}r=-4\\\\s=-12\\end{array}[\/latex]<\/p>\n<p style=\"text-align: left;\">Now we can fill in each binomial with the values we found for r and s, make sure to use the correct variable!<\/p>\n<p style=\"text-align: center;\">[latex]\\left(m-4\\right)\\left(m-12\\right)[\/latex]<\/p>\n<p style=\"text-align: left;\">We are not done yet, remember that we factored out a negative sign in the first step. We need to remember to include that.<\/p>\n<p style=\"text-align: center;\">\u00a0[latex]-1\\left(m-4\\right)\\left(m-12\\right)[\/latex]<\/p>\n<h4 style=\"text-align: left;\">Answer<\/h4>\n<p>[latex]-m^2+16m-48=-1\\left(m-4\\right)\\left(m-12\\right)[\/latex]<\/p>\n<p style=\"text-align: left;\"><\/div>\n<\/div>\n<\/div>\n<p>In the following video, we present two more examples of factoring a trinomial using the shortcut presented here.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-4\" title=\"Factor a Trinomial Using the Shortcut Method - Form x^2+bx+c\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/-SVBVVYVNTM?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<h2>Factor Trinomials Part II<\/h2>\n<p>The next goal is for you to be comfortable with recognizing where to place negative signs, and whether a trinomial can even be factored. \u00a0Additionally, we will explore\u00a0one special case to look out for at the end of this page.<\/p>\n<p>Not all trinomials look like [latex]x^{2}+5x+6[\/latex], where the coefficient in front of the [latex]x^{2}[\/latex]\u00a0term is 1. In these cases, your first step should be to look for common factors for the three terms.<\/p>\n<table>\n<thead>\n<tr>\n<th>Trinomial<\/th>\n<th>Factor out Common Factor<\/th>\n<th>Factored<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<td>[latex]2x^{2}+10x+12[\/latex]<\/td>\n<td>[latex]2(x^{2}+5x+6)[\/latex]<\/td>\n<td>[latex]2\\left(x+2\\right)\\left(x+3\\right)[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]\u22125a^{2}\u221215a\u221210[\/latex]<\/td>\n<td>[latex]\u22125(a^{2}+3a+2)[\/latex]<\/td>\n<td>[latex]\u22125\\left(a+2\\right)\\left(a+1\\right)[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]c^{3}\u20138c^{2}+15c[\/latex]<\/td>\n<td>[latex]c\\left(c^{2}\u20138c+15\\right)[\/latex]<\/td>\n<td>[latex]c\\left(c\u20135\\right)\\left(c\u20133\\right)[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]y^{4}\u20139y^{3}\u201310y^{2}[\/latex]<\/td>\n<td>[latex]y^{2}\\left(y^{2}\u20139y\u201310\\right)[\/latex]<\/td>\n<td>[latex]y^{2}\\left(y\u201310\\right)\\left(y+1\\right)[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>Notice that once you have identified and pulled out the common factor, you can factor the remaining trinomial as usual. This process is shown below.<\/p>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>Factor\u00a0[latex]3x^{3}\u20133x^{2}\u201390x[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q298928\">Show Solution<\/span><\/p>\n<div id=\"q298928\" class=\"hidden-answer\" style=\"display: none\">Since 3 is a common factor for the three terms, factor out the 3.<\/p>\n<p style=\"text-align: center;\">[latex]3\\left(x^{3}\u2013x^{2}\u201330x\\right)[\/latex]<\/p>\n<p><i>x<\/i> is also a common factor, so factor out <i>x<\/i>.<\/p>\n<p style=\"text-align: center;\">[latex]3x\\left(x^{2}\u2013x\u201330\\right)[\/latex]<\/p>\n<p>Now you can factor the trinomial\u00a0[latex]x^{2}\u2013x\u201330[\/latex]. To find <i>r<\/i> and <i>s<\/i>, identify two numbers whose product is [latex]\u221230[\/latex] and whose sum is [latex]\u22121[\/latex].<\/p>\n<p>The pair of factors is [latex]\u22126[\/latex] and [latex]5[\/latex]. So replace [latex]\u2013x[\/latex] with [latex]\u22126x+5x[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]3x\\left(x^{2}\u20136x+5x\u201330\\right)[\/latex]<\/p>\n<p>Use grouping to consider the terms in pairs.<\/p>\n<p style=\"text-align: center;\">[latex]3x\\left[\\left(x^{2}\u20136x\\right)+\\left(5x\u201330\\right)\\right][\/latex]<\/p>\n<p>Factor <i>x<\/i> out of the first group and factor 5 out of the second group.<\/p>\n<p style=\"text-align: center;\">[latex]3x\\left[\\left(x\\left(x\u20136\\right)\\right)+5\\left(x\u20136\\right)\\right][\/latex]<\/p>\n<p>Then factor out [latex]x\u20136[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]3x\\left(x\u20136\\right)\\left(x+5\\right)[\/latex]<\/p>\n<h4>Answer<\/h4>\n<p>[latex]3x\\left(x\u20136\\right)\\left(x+5\\right)[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>The following video contains two more examples of factoring a quadratic trinomial where the first step is to factor out a GCF. We use the shortcut method instead of factoring by grouping.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-5\" title=\"Factor a Trinomial With A Common Factor Using the Shortcut Method - Form x^2+bx+c\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/pgH77rAtsbs?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<p>The general form of trinomials with a leading coefficient of <i>a<\/i> is [latex]ax^{2}+bx+c[\/latex]. Sometimes the factor of <i>a<\/i> can be factored as you saw above; this happens when <i>a<\/i> can be factored out of all three terms. The remaining trinomial that still needs factoring will then be simpler, with the leading term only being an [latex]x^{2}[\/latex]\u00a0term, instead of an [latex]ax^{2}[\/latex]\u00a0term.<\/p>\n<p>However, if the coefficients of all three terms of a trinomial don\u2019t have a common factor, then you will need to factor the trinomial with a coefficient of something other than 1.<\/p>\n<div class=\"textbox shaded\">\n<h3>Factoring Trinomials in the form [latex]ax^{2}+bx+c[\/latex]<\/h3>\n<p>To factor a trinomial in the form [latex]ax^{2}+bx+c[\/latex], find two integers, <i>r<\/i> and <i>s<\/i>, whose sum is <i>b<\/i> and whose product is <i>ac.<\/i><\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}r\\cdot{s}=a\\cdot{c}\\\\r+s=b\\end{array}[\/latex]<\/p>\n<p>Rewrite the trinomial as [latex]ax^{2}+rx+sx+c[\/latex] and then use grouping and the distributive property to factor the polynomial.<\/p>\n<\/div>\n<p>This is almost the same as factoring trinomials in the form [latex]x^{2}+bx+c[\/latex],<i> <\/i>as in this form\u00a0[latex]a=1[\/latex].<i> <\/i>Now you are looking for two factors whose product is [latex]a\\cdot{c}[\/latex],\u00a0and whose sum is <i>b<\/i>.<\/p>\n<p>Let\u2019s see how this strategy works by factoring [latex]6z^{2}+11z+4[\/latex].<\/p>\n<p>In this trinomial, [latex]a=6[\/latex], [latex]b=11[\/latex], and [latex]c=4[\/latex]. According to the strategy, you need to find two factors, <i>r<\/i> and <i>s<\/i>, whose sum is [latex]b=11[\/latex]\u00a0and whose product is [latex]a\\cdot{c}=6\\cdot4=24[\/latex]. You can make a chart to organize the possible factor combinations. (Notice that this chart only has positive numbers. Since <i>ac<\/i> is positive and <i>b<\/i> is positive, you can be certain that the two factors you&#8217;re looking for are also positive numbers.)<\/p>\n<table style=\"width: 20%;\">\n<thead>\n<tr>\n<th>Factors whose product is 24<\/th>\n<th>Sum of the factors<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<td>[latex]1\\cdot24=24[\/latex]<\/td>\n<td>[latex]1+24=25[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]2\\cdot12=24[\/latex]<\/td>\n<td>[latex]2+12=14[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]3\\cdot8=24[\/latex]<\/td>\n<td>[latex]3+8=11[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]4\\cdot6=24[\/latex]<\/td>\n<td>[latex]4+6=10[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>There is only one combination where the product is 24 and the sum is 11, and that is when [latex]r=3[\/latex], and [latex]s=8[\/latex]. Let\u2019s use these values to factor the original trinomial.<\/p>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>Factor [latex]6z^{2}+11z+4[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q796129\">Show Solution<\/span><\/p>\n<div id=\"q796129\" class=\"hidden-answer\" style=\"display: none\">Rewrite the middle term, [latex]11z[\/latex], as [latex]3z + 8z[\/latex] (from the chart above.)<\/p>\n<p style=\"text-align: center;\">[latex]6z^{2}+3z+8z+4[\/latex]<\/p>\n<p>Group pairs. Use grouping to consider the terms in pairs.<\/p>\n<p style=\"text-align: center;\">[latex]\\left(6z^{2}+3z\\right)+\\left(8z+4\\right)[\/latex]<\/p>\n<p>Factor 3<i>z<\/i> out of the first group and 4 out of the second group.<\/p>\n<p style=\"text-align: center;\">[latex]3z\\left(2z+1\\right)+4\\left(2z+1\\right)[\/latex]<\/p>\n<p>Factor out [latex]\\left(2z+1\\right)[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]\\left(2z+1\\right)\\left(3z+4\\right)[\/latex]<\/p>\n<h4>Answer<\/h4>\n<p>[latex]\\left(2z+1\\right)\\left(3z+4\\right)[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>In the following video, we present another example of factoring a trinomial using grouping. \u00a0In this example, the middle term, b, is negative. Note how having a negative middle term and a positive c term influence the options for r and s when factoring.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-6\" title=\"Factor a Trinomial in the Form ax^2+bx+c Using the Grouping Technique\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/agDaQ_cZnNc?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<p>Before going any further, it is worth mentioning that not all trinomials can be factored using integer pairs. Take the trinomial [latex]2z^{2}+35z+7[\/latex], for instance. Can you think of two integers whose sum is [latex]b=35[\/latex] and whose product is [latex]a\\cdot{c}=2\\cdot7=14[\/latex]? There are none! This type of trinomial, which cannot be factored using integers, is called a prime trinomial.<\/p>\n<p>In some situations, <i>a<\/i> is negative, as in [latex]\u22124h^{2}+11h+3[\/latex]. It often makes sense to factor out [latex]\u22121[\/latex] as the first step in factoring, as doing so will change the sign of [latex]ax^{2}[\/latex]\u00a0from negative to positive, making the remaining trinomial easier to factor.<\/p>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>Factor\u00a0[latex]\u22124h^{2}+11h+3[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q471034\">Show Solution<\/span><\/p>\n<div id=\"q471034\" class=\"hidden-answer\" style=\"display: none\">Factor [latex]\u22121[\/latex] out of the trinomial. Notice that the signs of all three terms have changed.<\/p>\n<p style=\"text-align: center;\">[latex]\u22121\\left(4h^{2}\u201311h\u20133\\right)[\/latex]<\/p>\n<p>To factor the trinomial, you need to figure out how to rewrite [latex]\u221211h[\/latex].\u00a0The product of [latex]rs=4\\cdot\u22123=\u221212[\/latex], and the sum of [latex]rs=\u221211[\/latex].<\/p>\n<table style=\"width: 20%;\">\n<tbody>\n<tr>\n<td>[latex]r\\cdot{s}=\u221212[\/latex]<\/td>\n<td>[latex]r+s=\u221211[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]\u221212\\cdot1=\u221212[\/latex]<\/td>\n<td>[latex]\u221212+1=\u221211[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]\u22126\\cdot2=\u221212[\/latex]<\/td>\n<td>[latex]\u22126+2=\u22124[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]\u22124\\cdot3=\u221212[\/latex]<\/td>\n<td>[latex]\u22124+3=\u22121[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>Rewrite the middle term\u00a0[latex]\u221211h[\/latex] as [latex]\u221212h+1h[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]\u22121\\left(4h^{2}\u201312h+1h\u20133\\right)[\/latex]<\/p>\n<p>Group terms.<\/p>\n<p style=\"text-align: center;\">[latex]\u22121\\left[\\left(4h^{2}\u201312h\\right)+\\left(1h\u20133\\right)\\right][\/latex]<\/p>\n<p>Factor out 4<i>h<\/i> from the first pair. The second group cannot be factored further, but you can write it as [latex]+1\\left(h\u20133\\right)[\/latex]\u00a0since [latex]+1\\left(h\u20133\\right)=\\left(h\u20133\\right)[\/latex]. This helps with factoring in the next step.<\/p>\n<p style=\"text-align: center;\">[latex]\u22121\\left[4h\\left(h\u20133\\right)+1\\left(h\u20133\\right)\\right][\/latex]<\/p>\n<p>Factor out a common factor of [latex]\\left(h\u20133\\right)[\/latex]. Notice you are left with [latex]\\left(h\u20133\\right)\\left(4h+1\\right)[\/latex]; the [latex]+1[\/latex] comes from the term [latex]+1\\left(h\u20133\\right)[\/latex]\u00a0in the previous step.<\/p>\n<p style=\"text-align: center;\">[latex]\u22121\\left[\\left(h\u20133\\right)\\left(4h+1\\right)\\right][\/latex]<\/p>\n<h4>Answer<\/h4>\n<p>[latex]\u22121\\left(h\u20133\\right)\\left(4h+1\\right)[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>Note that the answer above can also be written as [latex]\\left(\u2212h+3\\right)\\left(4h+1\\right)[\/latex] or [latex]\\left(h\u20133\\right)\\left(\u22124h\u20131\\right)[\/latex] if you multiply [latex]\u22121[\/latex]\u00a0times one of the other factors.<\/p>\n<p>In the following video we present another example of factoring a trinomial in the form [latex]-ax^2+bx+c[\/latex] using the grouping technique.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-7\" title=\"Factor a Trinomial in the Form -ax^2+bx+c Using the Grouping Technique\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/zDAMjdBfkDs?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<h2>Difference of Squares<\/h2>\n<p>We would be remiss if we failed to introduce one more type of polynomial that can be factored. This polynomial can be factored into two binomials but has only two terms. \u00a0Let&#8217;s start from the product of two binomials to see the pattern.<\/p>\n<p>Given the product of two binomials: [latex]\\left(x-2\\right)\\left(x+2\\right)[\/latex], if we multiply them together, we lose the middle term that we are used to seeing as a result.<\/p>\n<p style=\"text-align: left;\">Multiply:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}\\left(x-2\\right)\\left(x+2\\right)\\\\\\text{}\\\\=x^2-2x+2x-2^2\\\\\\text{}\\\\=x^2-2^2\\\\\\text{}\\\\=x^2-4\\end{array}[\/latex]<\/p>\n<p>\u00a0The polynomial [latex]x^2-4[\/latex] is called a difference of squares because teach term can be written as something squared. \u00a0A difference of squares will always factor in the following way:<\/p>\n<div class=\"textbox shaded\">\n<h3>Factor a Difference of Squares<\/h3>\n<p>Given [latex]a^2-b^2[\/latex], it&#8217;s factored form will be [latex]\\left(a+b\\right)\\left(a-b\\right)[\/latex]<\/p>\n<\/div>\n<p>Let\u2019s factor [latex]x^{2}\u20134[\/latex]\u00a0by writing it as a trinomial, [latex]x^{2}+0x\u20134[\/latex]. \u00a0This\u00a0is similar in format to the\u00a0trinomials we have been factoring so far, so let\u2019s\u00a0use the same method.<\/p>\n<p style=\"text-align: center;\">Find the factors of [latex]a\\cdot{c}[\/latex]\u00a0whose sum is <i>b, <\/i>in this case, 0<i>:<\/i><\/p>\n<table style=\"width: 20%;\">\n<thead>\n<tr>\n<th>Factors of [latex]\u22124[\/latex]<\/th>\n<th>Sum of the factors<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<td>[latex]1\\cdot-4=\u22124[\/latex]<\/td>\n<td>[latex]1-4=\u22123[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]2\\cdot\u22122=\u22124[\/latex]<\/td>\n<td>[latex]2-2=0[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]-1\\cdot4=\u22124[\/latex]<\/td>\n<td>[latex]-1+4=3[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>2, and -2 have a sum of 0. You can use these to factor [latex]x^{2}\u20134[\/latex].<\/p>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>Factor [latex]x^{2}\u20134[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q23133\">Show Solution<\/span><\/p>\n<div id=\"q23133\" class=\"hidden-answer\" style=\"display: none\">Rewrite\u00a0[latex]0x[\/latex] as [latex]\u22122x+2x[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}x^{2}+0x-4\\\\x^{2}-2x+2x-4\\end{array}[\/latex]<\/p>\n<p>Group pairs.<\/p>\n<p style=\"text-align: center;\">[latex]\\left(x^{2}\u20132x\\right)+\\left(2x\u20134\\right)[\/latex]<\/p>\n<p>Factor <i>x<\/i> out of the first group. Factor 2 out of the second group.<\/p>\n<p style=\"text-align: center;\">[latex]x\\left(x\u20132\\right)+2\\left(x\u20132\\right)[\/latex]<\/p>\n<p>Factor out [latex]\\left(x\u20132\\right)[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]\\left(x\u20132\\right)\\left(x+2\\right)[\/latex]<\/p>\n<h4>Answer<\/h4>\n<p>[latex]\\left(3x\u20132\\right)\\left(3x+2\\right)[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>Since order doesn&#8217;t matter with multiplication, the answer can also be written as [latex]\\left(x+2\\right)\\left(x\u20132\\right)[\/latex].<\/p>\n<p>You can check the answer by multiplying [latex]\\left(x\u20132\\right)\\left(x+2\\right)=x^{2}+2x\u20132x\u20134=x^{2}\u20134[\/latex].<\/p>\n<p>The following video show two more examples of factoring a difference of squares.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-8\" title=\"Ex:  Factor a Difference of Squares\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/Li9IBp5HrFA?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<h2>Summary<\/h2>\n<p>When a trinomial is in the form of [latex]ax^{2}+bx+c[\/latex], where <i>a<\/i> is a coefficient other than 1, look first for common factors for all three terms. Factor out the common factor first, then factor the remaining simpler trinomial. If the remaining trinomial is still of the form [latex]ax^{2}+bx+c[\/latex], find two integers, <i>r<\/i> and <i>s<\/i>, whose sum is <i>b<\/i> and whose product is <i>ac.<\/i> Then rewrite the trinomial as [latex]ax^{2}+rx+sx+c[\/latex]\u00a0and use grouping and the distributive property to factor the polynomial.<\/p>\n<p>When [latex]ax^{2}[\/latex] is negative, you can factor [latex]\u22121[\/latex] out of the whole trinomial before continuing.<\/p>\n<p>A difference of squares [latex]a^2-b^2[\/latex] will factor in this way: [latex]\\left(a+b\\right)\\left(a-b\\right)[\/latex].<\/p>\n<h2>Summary<\/h2>\n<p>Trinomials in the form [latex]x^{2}+bx+c[\/latex] can be factored by finding two integers, <i>r<\/i> and <i>s<\/i>, whose sum is <i>b<\/i> and whose product is <i>c.<\/i> Rewrite the trinomial as [latex]x^{2}+rx+sx+c[\/latex]\u00a0and then use grouping and the distributive property to factor the polynomial.<\/p>\n<p>&nbsp;<\/p>\n\n\t\t\t <section class=\"citations-section\" role=\"contentinfo\">\n\t\t\t <h3>Candela Citations<\/h3>\n\t\t\t\t\t <div>\n\t\t\t\t\t\t <div id=\"citation-list-965\">\n\t\t\t\t\t\t\t <div class=\"licensing\"><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Original<\/div><ul class=\"citation-list\"><li>Screenshot: Why Should I Care?. <strong>Provided by<\/strong>: Lumen Learning. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Revision and Adaptation. <strong>Provided by<\/strong>: Lumen Learning. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Screenshot: Method to the Madness. <strong>Provided by<\/strong>: Lumen Learning. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Image: Shortcut This Way. <strong>Provided by<\/strong>: Lumen Learning. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><\/ul><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Shared previously<\/div><ul class=\"citation-list\"><li>Ex 2: Intro to Factor By Grouping Technique. <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com) for Lumen Learning. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/RR5nj7RFSiU\">https:\/\/youtu.be\/RR5nj7RFSiU<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Ex 1: Intro to Factor By Grouping Technique Mathispower4u . <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com) . <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/0dvGmDGVC5U\">https:\/\/youtu.be\/0dvGmDGVC5U<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Unit 12: Factoring, from Developmental Math: An Open Program. <strong>Provided by<\/strong>: Monterey Institute of Technology and Education. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"http:\/\/nrocnetwork.org\/resources\/downloads\/nroc-math-open-textbook-units-1-12-pdf-and-word-formats\/\">http:\/\/nrocnetwork.org\/resources\/downloads\/nroc-math-open-textbook-units-1-12-pdf-and-word-formats\/<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Ex 1: Factor a Quadratic Expression Using Grouping When a = 1 . <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com) for Lumen Learning. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/_Rtp7nSxf6c\">https:\/\/youtu.be\/_Rtp7nSxf6c<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Factor a Trinomial Using the Shortcut Method - Form x^2+bx+c. <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com) . <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/-SVBVVYVNTM\">https:\/\/youtu.be\/-SVBVVYVNTM<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Ex: Factor a Difference of Squares. <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com) . <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/Li9IBp5HrFA\">https:\/\/youtu.be\/Li9IBp5HrFA<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><\/ul><\/div>\n\t\t\t\t\t\t <\/div>\n\t\t\t\t\t <\/div>\n\t\t\t <\/section>","protected":false},"author":115,"menu_order":4,"template":"","meta":{"_candela_citation":"[{\"type\":\"original\",\"description\":\"Screenshot: Why Should I Care?\",\"author\":\"\",\"organization\":\"Lumen Learning\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"Ex 2: Intro to Factor By Grouping Technique\",\"author\":\"James Sousa (Mathispower4u.com) for Lumen Learning\",\"organization\":\"\",\"url\":\"https:\/\/youtu.be\/RR5nj7RFSiU\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"Ex 1: Intro to Factor By Grouping Technique Mathispower4u \",\"author\":\"James Sousa (Mathispower4u.com) \",\"organization\":\"\",\"url\":\"https:\/\/youtu.be\/0dvGmDGVC5U\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"Unit 12: Factoring, from Developmental Math: An Open Program\",\"author\":\"\",\"organization\":\"Monterey Institute of Technology and Education\",\"url\":\"http:\/\/nrocnetwork.org\/resources\/downloads\/nroc-math-open-textbook-units-1-12-pdf-and-word-formats\/\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"original\",\"description\":\"Revision and Adaptation\",\"author\":\"\",\"organization\":\"Lumen Learning\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"Ex 1: Factor a Quadratic Expression Using Grouping When a = 1 \",\"author\":\"James Sousa (Mathispower4u.com) for Lumen Learning\",\"organization\":\"\",\"url\":\"https:\/\/youtu.be\/_Rtp7nSxf6c\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"Factor a Trinomial Using the Shortcut Method - 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