{"id":366,"date":"2021-06-04T00:05:52","date_gmt":"2021-06-04T00:05:52","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/uvu-introductoryalgebra\/chapter\/read-factor-a-trinomial-with-leading-coefficient-1\/"},"modified":"2022-01-19T00:22:29","modified_gmt":"2022-01-19T00:22:29","slug":"read-factor-a-trinomial-with-leading-coefficient-1","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/uvu-introductoryalgebra\/chapter\/read-factor-a-trinomial-with-leading-coefficient-1\/","title":{"raw":"8.6.3: Factoring Trinomials with Leading Coefficients of 1","rendered":"8.6.3: Factoring Trinomials with Leading Coefficients 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\n<div class=\"textbox key-takeaways\">\r\n<h3>Key words<\/h3>\r\n<ul>\r\n \t<li><strong>Leading term<\/strong>: the term with the highest degree<\/li>\r\n \t<li><strong>Leading coefficient<\/strong>: the number in the leading term<\/li>\r\n<\/ul>\r\n<\/div>\r\n<h2>Factoring Trinomials<\/h2>\r\nNow that we know how to factor by grouping, we are going to factor a trinomial whose leading coefficient is\u00a0[latex]1[\/latex].\u00a0 In particular, we will look at trinomials of the form\u00a0[latex]x^2+bx+c[\/latex], where [latex]b, c[\/latex] are real numbers.\u00a0 Polynomials whose leading coefficients are 1 can be factored using the grouping method that we used in the previous section.\u00a0 For example, [latex]x^2+5x+4[\/latex] can be written as [latex]x^2+x+4x+4[\/latex], which can then be factored using grouping: [latex]x^2+x+4x+4=x(x+1)+4(x+1)=(x+1)(x+4)[\/latex].\u00a0 The question is, how do we know how to split up the [latex]x[\/latex]-term?\u00a0 If we multiply the generic binomials [latex](x+r)(x+s)[\/latex], we get [latex]x(x+s)+r(x+s)=x^2+sx+rx+rs=x^2+(r+s)x+rs[\/latex].\r\n\r\nIf we now compare this to the generic trinomial [latex]x^2+bx+c[\/latex], we can equate the [latex]x[\/latex]-terms so that [latex]r+s=b[\/latex] and we can equate the constants so that [latex]c=rs[\/latex].\u00a0 In other words, we split up the [latex]x[\/latex]-term [latex]bx[\/latex] into [latex]rx+sx[\/latex] where [latex]r+s=b[\/latex] and [latex]rs=c[\/latex].\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\nLet\u2019s factor the trinomial [latex]x^{2}+5x+6[\/latex]. In this polynomial, [latex]b=5[\/latex] and [latex]c=6[\/latex]. We look for two numbers that add to 5 and multiply to 6. A chart will help us organize possibilities. On the left, list all possible factors of the [latex]c[\/latex]-term, [latex]6[\/latex]; on the right we'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<tr>\r\n<td>[latex]-1\\cdot -6=6[\/latex]<\/td>\r\n<td>[latex]-1+(-6)=-7[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]-2\\cdot -3=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 with positive factors, [latex]1[\/latex] and [latex]6[\/latex], or [latex]2[\/latex] and [latex]3[\/latex], but there are also two possible factor combinations with negative factors,\u00a0[latex]-1[\/latex] and [latex]-6[\/latex], or [latex]-2[\/latex] and [latex]-3[\/latex] . We can see that\u00a0[latex]2+3=5[\/latex]. So [latex]2x+3x=5x[\/latex], giving us the correct [latex]x[\/latex]-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<h4>Solution<\/h4>\r\nFrom above, we know we need to split [latex]5x[\/latex] into [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<\/div>\r\nNote that if we 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\nThe following video presents another example of how to use grouping to factor a quadratic polynomial.\r\n\r\nhttps:\/\/youtu.be\/_Rtp7nSxf6c\r\n\r\nLet\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 we look at all of the combinations of factors whose product is [latex]\u221212[\/latex]. Then see which of these combinations will give us the correct [latex]x[\/latex]-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 -12[\/latex].\r\n<h4>Solution<\/h4>\r\nRewrite the trinomial using the correct 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<\/div>\r\nIn this example, we 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<div class=\"textbox tryit\">\r\n<h3>Try It<\/h3>\r\n1. Factor [latex]x^2+7x-8[\/latex]\r\n\r\n[reveal-answer q=\"hjm471\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"hjm471\"][latex](x+8)(x-1)[\/latex][\/hidden-answer]\r\n\r\n&nbsp;\r\n\r\n2. Factor [latex]x^2-7x+10[\/latex]\r\n\r\n[reveal-answer q=\"hjm362\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"hjm362\"][latex](x-5)(x-2)[\/latex][\/hidden-answer]\r\n\r\n&nbsp;\r\n\r\n3. Factor [latex]x^2+6x+9[\/latex]\r\n\r\n[reveal-answer q=\"hjm199\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"hjm199\"][latex](x+3)^2[\/latex][\/hidden-answer]\r\n\r\n&nbsp;\r\n\r\n4. Factor [latex]x^2-5x-24[\/latex]\r\n\r\n[reveal-answer q=\"hjm878\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"hjm878\"][latex](x-8)(x+3)[\/latex][\/hidden-answer]\r\n\r\n<\/div>\r\n&nbsp;\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. In this case the trinomial is said to be\u00a0<em><strong>prime<\/strong><\/em>.\r\n\r\nHere 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. In other words,\u00a0<i>r<\/i> or <i>s<\/i> will have opposite signs.<\/li>\r\n<\/ul>\r\nLook at the <i style=\"font-size: 1rem; orphans: 1; text-align: initial;\">b<\/i><span style=\"font-size: 1rem; orphans: 1; text-align: initial;\"> term second.<\/span>\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\nThis is only true when the leading coefficient of the trinomial is 1.\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 we can identify the correct <i>r<\/i> and <i>s<\/i> values, we can effectively skip the grouping steps and go straight to the factored form.\u00a0The idea is that we can build factors for a trinomial in this form: [latex]x^2+bx+c[\/latex] by finding [latex]r[\/latex] and [latex]s[\/latex], 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<h4>Solution<\/h4>\r\nFind[latex]r[\/latex] and [latex]s[\/latex]:\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[latex]r[\/latex] and [latex]s[\/latex] 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.<\/p>\r\n<p style=\"text-align: left;\">The nice thing about factoring is we can check our work by multiplying the binomials together to get back to the original trinomial.<\/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<\/div>\r\nIn the next example, the leading coefficient is negative.\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nFactor: [latex]-m^2+16m-48[\/latex]\r\n<h4>Solution<\/h4>\r\nThere is a negative in front of the squared term, so we will start by factoring out a [latex]-1[\/latex] from the whole trinomial. 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 [latex]r[\/latex] and [latex]s[\/latex]. Note that [latex]b[\/latex] is negative, and [latex]c[\/latex] is 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\r\n<\/div>\r\nThe following video presents 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 shaded\">\r\n<h3>FACTORING a trinomial in the form [latex]{x}^{2}+bx+c[\/latex]<\/h3>\r\n<ol>\r\n \t<li>List factors of [latex]c[\/latex].<\/li>\r\n \t<li>Find [latex]r[\/latex] and [latex]s[\/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\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nFactor\u00a0[latex]x^{2}+x\u201312[\/latex].\r\n<h4>Solution<\/h4>\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]r=4[\/latex] and [latex]s=\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\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\n<div class=\"textbox tryit\">\r\n<h3>Try It<\/h3>\r\nFactor [latex]x^2+x-20[\/latex]\r\n\r\n&nbsp;\r\n\r\n[reveal-answer q=\"hjm328\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"hjm328\"][latex](x+5)(x-4)[\/latex][\/hidden-answer]\r\n\r\n<\/div>\r\n&nbsp;\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nFactor [latex]{x}^{2}-7x+6[\/latex].\r\n<h4>Solution<\/h4>\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<\/div>\r\nIn the last example, [latex]b[\/latex] was negative and [latex]c[\/latex] was positive. This will always mean that if it can be factored, [latex]r[\/latex] and [latex]s[\/latex] will 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>\r\nThe more examples you work through, the more likely you are to spot patterns. This will lead to faster factoring, which is useful in later courses.","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<div class=\"textbox key-takeaways\">\n<h3>Key words<\/h3>\n<ul>\n<li><strong>Leading term<\/strong>: the term with the highest degree<\/li>\n<li><strong>Leading coefficient<\/strong>: the number in the leading term<\/li>\n<\/ul>\n<\/div>\n<h2>Factoring Trinomials<\/h2>\n<p>Now that we know how to factor by grouping, we are going to factor a trinomial whose leading coefficient is\u00a0[latex]1[\/latex].\u00a0 In particular, we will look at trinomials of the form\u00a0[latex]x^2+bx+c[\/latex], where [latex]b, c[\/latex] are real numbers.\u00a0 Polynomials whose leading coefficients are 1 can be factored using the grouping method that we used in the previous section.\u00a0 For example, [latex]x^2+5x+4[\/latex] can be written as [latex]x^2+x+4x+4[\/latex], which can then be factored using grouping: [latex]x^2+x+4x+4=x(x+1)+4(x+1)=(x+1)(x+4)[\/latex].\u00a0 The question is, how do we know how to split up the [latex]x[\/latex]-term?\u00a0 If we multiply the generic binomials [latex](x+r)(x+s)[\/latex], we get [latex]x(x+s)+r(x+s)=x^2+sx+rx+rs=x^2+(r+s)x+rs[\/latex].<\/p>\n<p>If we now compare this to the generic trinomial [latex]x^2+bx+c[\/latex], we can equate the [latex]x[\/latex]-terms so that [latex]r+s=b[\/latex] and we can equate the constants so that [latex]c=rs[\/latex].\u00a0 In other words, we split up the [latex]x[\/latex]-term [latex]bx[\/latex] into [latex]rx+sx[\/latex] where [latex]r+s=b[\/latex] and [latex]rs=c[\/latex].<\/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>Let\u2019s factor the trinomial [latex]x^{2}+5x+6[\/latex]. In this polynomial, [latex]b=5[\/latex] and [latex]c=6[\/latex]. We look for two numbers that add to 5 and multiply to 6. A chart will help us organize possibilities. On the left, list all possible factors of the [latex]c[\/latex]-term, [latex]6[\/latex]; on the right we&#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<tr>\n<td>[latex]-1\\cdot -6=6[\/latex]<\/td>\n<td>[latex]-1+(-6)=-7[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]-2\\cdot -3=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 with positive factors, [latex]1[\/latex] and [latex]6[\/latex], or [latex]2[\/latex] and [latex]3[\/latex], but there are also two possible factor combinations with negative factors,\u00a0[latex]-1[\/latex] and [latex]-6[\/latex], or [latex]-2[\/latex] and [latex]-3[\/latex] . We can see that\u00a0[latex]2+3=5[\/latex]. So [latex]2x+3x=5x[\/latex], giving us the correct [latex]x[\/latex]-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<h4>Solution<\/h4>\n<p>From above, we know we need to split [latex]5x[\/latex] into [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<p>Note that if we 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>The following video presents 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>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 we look at all of the combinations of factors whose product is [latex]\u221212[\/latex]. Then see which of these combinations will give us the correct [latex]x[\/latex]-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 -12[\/latex].<\/p>\n<h4>Solution<\/h4>\n<p>Rewrite the trinomial using the correct 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<p>In this example, we 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<div class=\"textbox tryit\">\n<h3>Try It<\/h3>\n<p>1. Factor [latex]x^2+7x-8[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qhjm471\">Show Answer<\/span><\/p>\n<div id=\"qhjm471\" class=\"hidden-answer\" style=\"display: none\">[latex](x+8)(x-1)[\/latex]<\/div>\n<\/div>\n<p>&nbsp;<\/p>\n<p>2. Factor [latex]x^2-7x+10[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qhjm362\">Show Answer<\/span><\/p>\n<div id=\"qhjm362\" class=\"hidden-answer\" style=\"display: none\">[latex](x-5)(x-2)[\/latex]<\/div>\n<\/div>\n<p>&nbsp;<\/p>\n<p>3. Factor [latex]x^2+6x+9[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qhjm199\">Show Answer<\/span><\/p>\n<div id=\"qhjm199\" class=\"hidden-answer\" style=\"display: none\">[latex](x+3)^2[\/latex]<\/div>\n<\/div>\n<p>&nbsp;<\/p>\n<p>4. Factor [latex]x^2-5x-24[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qhjm878\">Show Answer<\/span><\/p>\n<div id=\"qhjm878\" class=\"hidden-answer\" style=\"display: none\">[latex](x-8)(x+3)[\/latex]<\/div>\n<\/div>\n<\/div>\n<p>&nbsp;<\/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. In this case the trinomial is said to be\u00a0<em><strong>prime<\/strong><\/em>.<\/p>\n<p>Here 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. In other words,\u00a0<i>r<\/i> or <i>s<\/i> will have opposite signs.<\/li>\n<\/ul>\n<p>Look at the <i style=\"font-size: 1rem; orphans: 1; text-align: initial;\">b<\/i><span style=\"font-size: 1rem; orphans: 1; text-align: initial;\"> term second.<\/span><\/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<p>This is only true when the leading coefficient of the trinomial is 1.<\/p>\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 we can identify the correct <i>r<\/i> and <i>s<\/i> values, we can effectively skip the grouping steps and go straight to the factored form.\u00a0The idea is that we can build factors for a trinomial in this form: [latex]x^2+bx+c[\/latex] by finding [latex]r[\/latex] and [latex]s[\/latex], 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<h4>Solution<\/h4>\n<p>Find[latex]r[\/latex] and [latex]s[\/latex]:<\/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[latex]r[\/latex] and [latex]s[\/latex] 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.<\/p>\n<p style=\"text-align: left;\">The nice thing about factoring is we can check our work by multiplying the binomials together to get back to the original trinomial.<\/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<p>In the next example, the leading coefficient is negative.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Factor: [latex]-m^2+16m-48[\/latex]<\/p>\n<h4>Solution<\/h4>\n<p>There is a negative in front of the squared term, so we will start by factoring out a [latex]-1[\/latex] from the whole trinomial. 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 [latex]r[\/latex] and [latex]s[\/latex]. Note that [latex]b[\/latex] is negative, and [latex]c[\/latex] is 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<\/div>\n<p>The following video presents 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 shaded\">\n<h3>FACTORING a trinomial in the form [latex]{x}^{2}+bx+c[\/latex]<\/h3>\n<ol>\n<li>List factors of [latex]c[\/latex].<\/li>\n<li>Find [latex]r[\/latex] and [latex]s[\/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<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Factor\u00a0[latex]x^{2}+x\u201312[\/latex].<\/p>\n<h4>Solution<\/h4>\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]r=4[\/latex] and [latex]s=\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 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<div class=\"textbox tryit\">\n<h3>Try It<\/h3>\n<p>Factor [latex]x^2+x-20[\/latex]<\/p>\n<p>&nbsp;<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qhjm328\">Show Answer<\/span><\/p>\n<div id=\"qhjm328\" class=\"hidden-answer\" style=\"display: none\">[latex](x+5)(x-4)[\/latex]<\/div>\n<\/div>\n<\/div>\n<p>&nbsp;<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Factor [latex]{x}^{2}-7x+6[\/latex].<\/p>\n<h4>Solution<\/h4>\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<p>In the last example, [latex]b[\/latex] was negative and [latex]c[\/latex] was positive. This will always mean that if it can be factored, [latex]r[\/latex] and [latex]s[\/latex] will 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<p>The more examples you work through, the more likely you are to spot patterns. This will lead to faster factoring, which is useful in later courses.<\/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-366\">\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><li>Ex 1: Factor a Quadratic Expression by Grouping. <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>Try It hjm471; hjm199; hjm362; hjm878; hjm328. <strong>Authored by<\/strong>: Hazel McKenna. <strong>Provided by<\/strong>: Utah Valley University. <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":23485,"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\":\"\"},{\"type\":\"original\",\"description\":\"Ex 1: Factor a Quadratic Expression by Grouping\",\"author\":\"James Sousa (Mathispower4u.com) for Lumen Learning\",\"organization\":\"\",\"url\":\"https:\/\/youtu.be\/_Rtp7nSxf6c\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"original\",\"description\":\"Try It hjm471; 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