{"id":16322,"date":"2019-10-03T03:21:54","date_gmt":"2019-10-03T03:21:54","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/wm-developmentalemporium\/chapter\/read-factor-a-trinomial-with-leading-coefficient-1\/"},"modified":"2024-04-30T23:20:14","modified_gmt":"2024-04-30T23:20:14","slug":"read-factor-a-trinomial-with-leading-coefficient-1","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/wm-developmentalemporium\/chapter\/read-factor-a-trinomial-with-leading-coefficient-1\/","title":{"raw":"Factoring a Trinomial with a Leading Coefficient of 1","rendered":"Factoring a Trinomial with a Leading Coefficient of 1"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Factor a trinomial with a leading coefficient of [latex]1[\/latex]<\/li>\r\n \t<li>Use a shortcut to factor trinomials of the form\u00a0[latex]x^2+bx+c[\/latex]<\/li>\r\n<\/ul>\r\n<\/div>\r\nNow that we have shown you how to factor by grouping, We are going to show you how to factor a trinomial whose leading coefficient is\u00a0[latex]1[\/latex].\u00a0 That is, trinomials of the form\u00a0[latex]x^2+bx+c[\/latex].\u00a0 Polynomials whose leading coefficients are 1 can be factored using the grouping method that we showed you in the previous section.\u00a0 We will begin by showing a few examples using the grouping method and then we will show you a shortcut that will make factoring trinomials with a leading coefficient of 1 even easier!\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\nSpecial Note:\u00a0 Does the summary above look familiar?\u00a0 It should!\u00a0 How we factor polynomials of the form\u00a0[latex]x^{2}+bx+c[\/latex] using grouping is exactly the same as for polynomials of the form\u00a0[latex]ax^{2}+bx+c[\/latex].\u00a0 The only difference is that, previously, we were looking for two integers, <i>r<\/i> and <i>s<\/i>, whose sum is <i>b<\/i> and whose product is <i>ac.\u00a0<\/i> Notice that if our polynomial is of the form\u00a0 [latex]x^{2}+bx+c[\/latex], then [latex]a=1[\/latex], making [latex]ac=1\\cdot{c}=c[\/latex].\u00a0 As a result, we can skip the step of multiplying <em>a<\/em> by <em>c<\/em>!\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 \u00a0[latex]5[\/latex] and the <i>c<\/i> term is \u00a0[latex]6[\/latex]. A chart will help us organize possibilities. On the left, list all possible factors of the <i>c<\/i> term, [latex]6[\/latex]; 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 \u00a0[latex]6[\/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\\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, [latex]1[\/latex] and [latex]6[\/latex], and [latex]2[\/latex] and [latex]3[\/latex]. 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 [latex]3[\/latex] 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\n&nbsp;\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 [latex]1[\/latex].\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 [latex]1[\/latex], 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 [latex]x[\/latex] 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 [latex]\u22123[\/latex] \u00a0out 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 three examples.\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 [latex]1[\/latex]), 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&nbsp;\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 [latex]48[\/latex], but we have found the ones that sum to \u00a0[latex]-16[\/latex], 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 with a leading coefficient of 1.\r\n\r\nhttps:\/\/youtu.be\/-SVBVVYVNTM\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question]7900[\/ohm_question]\r\n\r\n<\/div>\r\nTo summarize our process, consider the following steps:\r\n<div class=\"textbox\">\r\n<h3>How To: Given a trinomial in the form [latex]{x}^{2}+bx+c[\/latex], factor it<\/h3>\r\n<ol>\r\n \t<li>List factors of [latex]c[\/latex].<\/li>\r\n \t<li>Find [latex]p[\/latex] and [latex]q[\/latex], a pair of factors of [latex]c[\/latex] with a sum of [latex]b[\/latex].<\/li>\r\n \t<li>Write the factored expression [latex]\\left(x+p\\right)\\left(x+q\\right)[\/latex].<\/li>\r\n<\/ol>\r\n<\/div>\r\nWe will now show an example where the trinomial has a negative c term. Pay attention to the signs of the numbers that are considered for p and q.\u00a0 We will show that when c is negative, either p or q will be negative.\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nFactor\u00a0[latex]x^{2}+x\u201312[\/latex].\r\n\r\n[reveal-answer q=\"205738\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"205738\"]\r\n\r\nConsider all the combinations of numbers whose product is\u00a0[latex]-12[\/latex] and list their sum.\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\nChoose the values whose sum is\u00a0[latex]+1[\/latex]:\u00a0\u00a0[latex]p=4[\/latex] and [latex]q=\u22123[\/latex], and place them into a product of binomials.\r\n\r\n&nbsp;\r\n<p style=\"text-align: center;\">[latex]\\left(x+4\\right)\\left(x-3\\right)[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Think About It<\/h3>\r\nWhich property of multiplication can be used to describe why\u00a0[latex]\\left(x+4\\right)\\left(x-3\\right) =\\left(x-3\\right)\\left(x+4\\right)[\/latex]. Use the textbox below to write down your ideas before you look at the answer.\r\n\r\n[practice-area rows=\"2\"][\/practice-area]\r\n[reveal-answer q=\"177955\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"177955\"]\r\n\r\nThe <strong>commutative property of multiplication<\/strong> states that factors may be multiplied in any order without affecting the product.\r\n<div class=\"bcc-box bcc-success\">\r\n<div style=\"text-align: center;\">[latex]a\\cdot b=b\\cdot a[\/latex]<\/div>\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nIn our last example, we will show how to factor a trinomial whose b term is negative.\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nFactor [latex]{x}^{2}-7x+6[\/latex].\r\n[reveal-answer q=\"662468\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"662468\"]\r\n\r\nList the factors of\u00a0[latex]6[\/latex]. Note that the b term is negative, so we will need to consider negative numbers in our list.\r\n<table style=\"width: 20%;\" summary=\"A table with five rows and two columns. The first row has columns labeled: Factors of -15 and Sum of Factors. The entries in the first column are: 1, -15; -1, 15; 3, -5; and -3,5. The entries in the second column are: -14, 14, -2, and 2.\">\r\n<thead>\r\n<tr>\r\n<th>Factors of [latex]6[\/latex]<\/th>\r\n<th>Sum of Factors<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr>\r\n<td>[latex]1,6[\/latex]<\/td>\r\n<td>[latex]7[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]2, 3[\/latex]<\/td>\r\n<td>[latex]5[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]-1, -6[\/latex]<\/td>\r\n<td>[latex]-7[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]-2, -3[\/latex]<\/td>\r\n<td>[latex]-5[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nChoose the pair that sum to [latex]-7[\/latex], which is\u00a0[latex]-1, -6[\/latex]\r\n\r\nWrite the pair as constant terms in a product of binomials.\r\n\r\n[latex]\\left(x-1\\right)\\left(x-6\\right)[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nIn the last example, the b\u00a0term was negative and the c term was positive. This will always mean that if it can be factored, p and q\u00a0will both be negative.\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Think About It<\/h3>\r\nCan every trinomial be factored as a product of binomials?\r\n\r\nMathematicians often use a counterexample to prove\u00a0or disprove a question. A counterexample means you provide an example where a proposed rule or definition is not true. Can you create a trinomial with leading coefficient\u00a0[latex]1[\/latex] that\u00a0<em>cannot\u00a0<\/em>be factored as a product of binomials?\r\n\r\nUse the textbox below to write your ideas.\r\n\r\n[practice-area rows=\"2\"][\/practice-area]\r\n[reveal-answer q=\"776075\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"776075\"]\r\n\r\nCan every trinomial be factored as a product of binomials? No. Some polynomials cannot be factored. These polynomials are said to be prime.\r\n\r\nA counterexample would be: [latex]x^2+3x+7[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question]7907[\/ohm_question]\r\n\r\n<\/div>","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Factor a trinomial with a leading coefficient of [latex]1[\/latex]<\/li>\n<li>Use a shortcut to factor trinomials of the form\u00a0[latex]x^2+bx+c[\/latex]<\/li>\n<\/ul>\n<\/div>\n<p>Now that we have shown you how to factor by grouping, We are going to show you how to factor a trinomial whose leading coefficient is\u00a0[latex]1[\/latex].\u00a0 That is, trinomials of the form\u00a0[latex]x^2+bx+c[\/latex].\u00a0 Polynomials whose leading coefficients are 1 can be factored using the grouping method that we showed you in the previous section.\u00a0 We will begin by showing a few examples using the grouping method and then we will show you a shortcut that will make factoring trinomials with a leading coefficient of 1 even easier!<\/p>\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>Special Note:\u00a0 Does the summary above look familiar?\u00a0 It should!\u00a0 How we factor polynomials of the form\u00a0[latex]x^{2}+bx+c[\/latex] using grouping is exactly the same as for polynomials of the form\u00a0[latex]ax^{2}+bx+c[\/latex].\u00a0 The only difference is that, previously, we were looking for two integers, <i>r<\/i> and <i>s<\/i>, whose sum is <i>b<\/i> and whose product is <i>ac.\u00a0<\/i> Notice that if our polynomial is of the form\u00a0 [latex]x^{2}+bx+c[\/latex], then [latex]a=1[\/latex], making [latex]ac=1\\cdot{c}=c[\/latex].\u00a0 As a result, we can skip the step of multiplying <em>a<\/em> by <em>c<\/em>!<\/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 \u00a0[latex]5[\/latex] and the <i>c<\/i> term is \u00a0[latex]6[\/latex]. A chart will help us organize possibilities. On the left, list all possible factors of the <i>c<\/i> term, [latex]6[\/latex]; 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 \u00a0[latex]6[\/latex]<\/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, [latex]1[\/latex] and [latex]6[\/latex], and [latex]2[\/latex] and [latex]3[\/latex]. 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 [latex]3[\/latex] 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-1\" 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>&nbsp;<\/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 [latex]1[\/latex].<\/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 [latex]1[\/latex], 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 [latex]x[\/latex] 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 [latex]\u22123[\/latex] \u00a0out 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 three examples.<\/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 [latex]1[\/latex]), 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<p>&nbsp;<\/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 [latex]48[\/latex], but we have found the ones that sum to \u00a0[latex]-16[\/latex], 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 with a leading coefficient of 1.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-2\" 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<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm7900\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=7900&theme=oea&iframe_resize_id=ohm7900&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<p>To summarize our process, consider the following steps:<\/p>\n<div class=\"textbox\">\n<h3>How To: Given a trinomial in the form [latex]{x}^{2}+bx+c[\/latex], factor it<\/h3>\n<ol>\n<li>List factors of [latex]c[\/latex].<\/li>\n<li>Find [latex]p[\/latex] and [latex]q[\/latex], a pair of factors of [latex]c[\/latex] with a sum of [latex]b[\/latex].<\/li>\n<li>Write the factored expression [latex]\\left(x+p\\right)\\left(x+q\\right)[\/latex].<\/li>\n<\/ol>\n<\/div>\n<p>We will now show an example where the trinomial has a negative c term. Pay attention to the signs of the numbers that are considered for p and q.\u00a0 We will show that when c is negative, either p or q will be negative.<\/p>\n<div class=\"textbox exercises\">\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=\"q205738\">Show Solution<\/span><\/p>\n<div id=\"q205738\" class=\"hidden-answer\" style=\"display: none\">\n<p>Consider all the combinations of numbers whose product is\u00a0[latex]-12[\/latex] and list their sum.<\/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>Choose the values whose sum is\u00a0[latex]+1[\/latex]:\u00a0\u00a0[latex]p=4[\/latex] and [latex]q=\u22123[\/latex], and place them into a product of binomials.<\/p>\n<p>&nbsp;<\/p>\n<p style=\"text-align: center;\">[latex]\\left(x+4\\right)\\left(x-3\\right)[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Think About It<\/h3>\n<p>Which property of multiplication can be used to describe why\u00a0[latex]\\left(x+4\\right)\\left(x-3\\right) =\\left(x-3\\right)\\left(x+4\\right)[\/latex]. Use the textbox below to write down your ideas before you look at the answer.<\/p>\n<p><textarea aria-label=\"Your Answer\" rows=\"2\"><\/textarea><\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q177955\">Show Solution<\/span><\/p>\n<div id=\"q177955\" class=\"hidden-answer\" style=\"display: none\">\n<p>The <strong>commutative property of multiplication<\/strong> states that factors may be multiplied in any order without affecting the product.<\/p>\n<div class=\"bcc-box bcc-success\">\n<div style=\"text-align: center;\">[latex]a\\cdot b=b\\cdot a[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<p>In our last example, we will show how to factor a trinomial whose b term is negative.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Factor [latex]{x}^{2}-7x+6[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q662468\">Show Solution<\/span><\/p>\n<div id=\"q662468\" class=\"hidden-answer\" style=\"display: none\">\n<p>List the factors of\u00a0[latex]6[\/latex]. Note that the b term is negative, so we will need to consider negative numbers in our list.<\/p>\n<table style=\"width: 20%;\" summary=\"A table with five rows and two columns. The first row has columns labeled: Factors of -15 and Sum of Factors. The entries in the first column are: 1, -15; -1, 15; 3, -5; and -3,5. The entries in the second column are: -14, 14, -2, and 2.\">\n<thead>\n<tr>\n<th>Factors of [latex]6[\/latex]<\/th>\n<th>Sum of Factors<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<td>[latex]1,6[\/latex]<\/td>\n<td>[latex]7[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]2, 3[\/latex]<\/td>\n<td>[latex]5[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]-1, -6[\/latex]<\/td>\n<td>[latex]-7[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]-2, -3[\/latex]<\/td>\n<td>[latex]-5[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>Choose the pair that sum to [latex]-7[\/latex], which is\u00a0[latex]-1, -6[\/latex]<\/p>\n<p>Write the pair as constant terms in a product of binomials.<\/p>\n<p>[latex]\\left(x-1\\right)\\left(x-6\\right)[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>In the last example, the b\u00a0term was negative and the c term was positive. This will always mean that if it can be factored, p and q\u00a0will both be negative.<\/p>\n<div class=\"textbox key-takeaways\">\n<h3>Think About It<\/h3>\n<p>Can every trinomial be factored as a product of binomials?<\/p>\n<p>Mathematicians often use a counterexample to prove\u00a0or disprove a question. A counterexample means you provide an example where a proposed rule or definition is not true. Can you create a trinomial with leading coefficient\u00a0[latex]1[\/latex] that\u00a0<em>cannot\u00a0<\/em>be factored as a product of binomials?<\/p>\n<p>Use the textbox below to write your ideas.<\/p>\n<p><textarea aria-label=\"Your Answer\" rows=\"2\"><\/textarea><\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q776075\">Show Solution<\/span><\/p>\n<div id=\"q776075\" class=\"hidden-answer\" style=\"display: none\">\n<p>Can every trinomial be factored as a product of binomials? No. Some polynomials cannot be factored. These polynomials are said to be prime.<\/p>\n<p>A counterexample would be: [latex]x^2+3x+7[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm7907\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=7907&theme=oea&iframe_resize_id=ohm7907&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\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-16322\">\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>Factor a Trinomial Using the Shortcut Method - Form x^2+bx+c. <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com) for Lumen Learning. <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>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><\/ul><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Shared previously<\/div><ul class=\"citation-list\"><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\/dm-opentext\">http:\/\/nrocnetwork.org\/dm-opentext<\/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":169554,"menu_order":12,"template":"","meta":{"_candela_citation":"[{\"type\":\"original\",\"description\":\"Factor a Trinomial Using the Shortcut Method - Form x^2+bx+c\",\"author\":\"James Sousa (Mathispower4u.com) for Lumen Learning\",\"organization\":\"\",\"url\":\"https:\/\/youtu.be\/-SVBVVYVNTM\",\"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\/dm-opentext\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"original\",\"description\":\"Revision and Adaptation\",\"author\":\"\",\"organization\":\"Lumen Learning\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"}]","CANDELA_OUTCOMES_GUID":"8076008d880d480494979e137e1b2ada, b75194d24022471d844d3085742431c7","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-16322","chapter","type-chapter","status-publish","hentry"],"part":16188,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/wm-developmentalemporium\/wp-json\/pressbooks\/v2\/chapters\/16322","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/wm-developmentalemporium\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/wm-developmentalemporium\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/wm-developmentalemporium\/wp-json\/wp\/v2\/users\/169554"}],"version-history":[{"count":13,"href":"https:\/\/courses.lumenlearning.com\/wm-developmentalemporium\/wp-json\/pressbooks\/v2\/chapters\/16322\/revisions"}],"predecessor-version":[{"id":19866,"href":"https:\/\/courses.lumenlearning.com\/wm-developmentalemporium\/wp-json\/pressbooks\/v2\/chapters\/16322\/revisions\/19866"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/wm-developmentalemporium\/wp-json\/pressbooks\/v2\/parts\/16188"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/wm-developmentalemporium\/wp-json\/pressbooks\/v2\/chapters\/16322\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/wm-developmentalemporium\/wp-json\/wp\/v2\/media?parent=16322"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/wm-developmentalemporium\/wp-json\/pressbooks\/v2\/chapter-type?post=16322"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/wm-developmentalemporium\/wp-json\/wp\/v2\/contributor?post=16322"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/wm-developmentalemporium\/wp-json\/wp\/v2\/license?post=16322"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}