{"id":6780,"date":"2020-10-09T18:42:55","date_gmt":"2020-10-09T18:42:55","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/slcc-beginalgebra\/?post_type=chapter&p=6780"},"modified":"2023-03-17T16:48:35","modified_gmt":"2023-03-17T16:48:35","slug":"6-3-factoring-trinomials-with-leading-coefficient-of-1","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/slcc-elementaryalgebra\/chapter\/6-3-factoring-trinomials-with-leading-coefficient-of-1\/","title":{"raw":"6.3: Factoring Trinomials with Leading Coefficient of 1","rendered":"6.3: Factoring Trinomials with Leading Coefficient of 1"},"content":{"raw":"
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section 6.3 Learning Objectives<\/h3>\r\n6.3: Factoring Trinomials with Leading Coefficient of 1<\/strong>\r\n
    \r\n \t
  • Factor trinomials of the type [latex]x^2 + bx + c[\/latex] by grouping<\/li>\r\n \t
  • Factor trinomials of the type [latex]x^2+bx+c[\/latex] using the Product and Sum method<\/li>\r\n \t
  • Rewrite and factor trinomials of the above type by first factoring out the GCF<\/li>\r\n \t
  • Factor two-variable trinomials of the type [latex]x^2 + bxy + cy^2[\/latex]<\/li>\r\n<\/ul>\r\n<\/div>\r\n \r\n

    Factor Trinomials Part I - Using Grouping<\/h2>\r\nIn the last section we introduced the technique of factoring by grouping as a means to be able to factor a four-term polynomial. However, many of the polynomials we encountered were actually trinomials in disguise. Now we will get to the work of starting with a three-term polynomial, and rewriting it as a four term polynomial so it can be factored using our grouping process.\r\n\r\nWe will start with factoring trinomials of the form [latex]x^{2}+bx+c[\/latex], and hence with a leading coefficient of 1.\r\n\r\nRemember that when\u00a0[latex]\\left(x+2\\right)[\/latex] and [latex]\\left(x+5\\right)[\/latex], are multiplied, the result is a four term polynomial and then it is\u00a0simplified into a trinomial:\r\n

    \u00a0[latex]\\left(x+2\\right)\\left(x+5\\right)=x^2+5x+2x+10=x^2+7x+10[\/latex]<\/p>\r\nFactoring is the reverse of multiplying, so let\u2019s go in reverse and factor the trinomial [latex]x^{2}+7x+10[\/latex]. The individual terms [latex]x^{2}[\/latex], [latex]7x[\/latex], and 10 share no common factors. If we\u00a0rewrite the middle term as the sum of the two terms [latex]7x=5x+2x[\/latex] then we can use the grouping technique:\r\n

    [latex](x^{2}+5x)+(2x+10)[\/latex]<\/p>\r\n

    Factor each pair:\u00a0[latex]\\begin{array}{l}x\\underbrace{\\left(x+5\\right)}+2\\underbrace{\\left(x+5\\right)}\\\\\\text{common binomial factor}\\end{array}[\/latex]<\/p>\r\n

    Then pull\u00a0out the common binomial factor: [latex]\\left(x+5\\right)\\left(x+2\\right)[\/latex]<\/p>\r\nBut how would we know to rewrite [latex]7x[\/latex] as [latex]5x+2x[\/latex]? What would have happened if we had\u00a0rewritten [latex]7x[\/latex] as [latex]6x+x[\/latex]?\r\n

    [latex](x^{2}+6x)+(x+10)[\/latex]<\/p>\r\n

    Factor each pair:\u00a0[latex]x\\left(x+6\\right)+1\\left(x+10\\right)[\/latex]<\/p>\r\nThen we don't have a\u00a0common binomial! There is a method to the madness of choosing how to rewrite the middle terms so that you will end up with a common binomial factor.\r\n\r\n[caption id=\"attachment_4833\" align=\"aligncenter\" width=\"300\"]\"Shakespeare Method to the Madness[\/caption]\r\n\r\nThe following is a summary of the method, then we will show some examples of how to use it.\r\n

    \r\n

    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, r<\/i> and s<\/i>, whose product is c <\/i>and whose sum is b<\/i>.\r\n

    [latex]\\begin{array}{l}r\\cdot{s}=c\\\\\\text{ and }\\\\r+s=b\\end{array}[\/latex]<\/p>\r\nRewrite the trinomial as [latex]x^{2}+rx+sx+c[\/latex]\u00a0and then use grouping and the distributive property to factor the polynomial. The resulting factors will be [latex]\\left(x+r\\right)[\/latex] and [latex]\\left(x+s\\right)[\/latex].\r\n\r\n<\/div>\r\nFor example, to factor [latex]x^{2}+7x+10[\/latex], you are looking for two numbers whose sum is 7 (the coefficient of the middle term) and whose product is 10 (the last term).\r\n\r\nLook at factor pairs of 10: 1 and 10, 2, and 5. Do either of these pairs have a sum of 7? Yes, 2 and 5. So you can rewrite\u00a0[latex]7x[\/latex] as [latex]2x+5x[\/latex], and continue factoring as in the example above. Note that you can also rewrite [latex]7x[\/latex] as [latex]5x+2x[\/latex]. Both will work.\r\n\r\nWith all terms positive in the last example, we did not really need to consider any negative factors. Let\u2019s take a look at the trinomial [latex]x^{2}+x\u201312[\/latex]. In this trinomial, the 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 b<\/i> is 1.\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n
    Factors whose product is [latex]\u221212[\/latex]<\/th>\r\nSum of the factors<\/th>\r\n<\/tr>\r\n<\/thead>\r\n
    [latex]1\\cdot\u221212=\u221212[\/latex]<\/td>\r\n[latex]1+(\u221212)=\u221211[\/latex]<\/td>\r\n<\/tr>\r\n
    [latex]2\\cdot\u22126=\u221212[\/latex]<\/td>\r\n[latex]2+(\u22126)=\u22124[\/latex]<\/td>\r\n<\/tr>\r\n
    [latex]3\\cdot\u22124=\u221212[\/latex]<\/td>\r\n[latex]3+(\u22124)=\u22121[\/latex]<\/td>\r\n<\/tr>\r\n
    [latex]4\\cdot\u22123=\u221212[\/latex]<\/td>\r\n[latex]4+(\u22123)=1[\/latex]<\/td>\r\n<\/tr>\r\n
    [latex]6\\cdot\u22122=\u221212[\/latex]<\/td>\r\n[latex]6+(\u22122)=4[\/latex]<\/td>\r\n<\/tr>\r\n
    [latex]12\\cdot\u22121=\u221212[\/latex]<\/td>\r\n[latex]12+(\u22121)=11[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nThere is only one combination where the product is [latex]\u221212[\/latex] and the sum is 1, and that is when [latex]r=4[\/latex], and [latex]s=\u22123[\/latex]. Let\u2019s use these to factor our original trinomial.\r\n
    \r\n

    Example 1<\/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

    [latex]x^{2}+4x\u22123x\u201312[\/latex]<\/p>\r\nGroup pairs of terms.\r\n

    [latex]\\left(x^{2}+4x\\right)+\\left(\u22123x\u201312\\right)[\/latex]<\/p>\r\nFactor x<\/em> out of the first group.\r\n

    [latex]x\\left(x+4\\right)+\\left(-3x-12\\right)[\/latex]<\/p>\r\nFactor \u22123 out of the second group.\r\n

    [latex]x\\left(x+4\\right)\u20133\\left(x+4\\right)[\/latex]<\/p>\r\nFactor out [latex]\\left(x+4\\right)[\/latex].\r\n

    [latex]\\left(x+4\\right)\\left(x-3\\right)[\/latex]<\/p>\r\n\r\n

    Answer<\/h4>\r\n[latex]x^2+x-12=\\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\r\nIn the following video, we present another example of how to use grouping to factor a trinomial of this form.\r\n\r\nhttps:\/\/youtu.be\/_Rtp7nSxf6c\r\n

    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
    \r\n

    Tips for Finding Values that Work when factoring a trinomial<\/span><\/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 c<\/i> term first.\r\n
      \r\n \t
    • If the c<\/i> term is a positive number, then the factors of c<\/i> will both be positive or both be negative. In other words, r<\/i> and s<\/i> will have the same sign.<\/li>\r\n \t
    • If the c<\/i> term is a negative number, then one factor of c<\/i> will be positive, and one factor of c<\/i> will be negative. Either r<\/i> or s<\/i> will be negative, but not both.<\/li>\r\n<\/ul>\r\nLook at the b<\/i> term second.\r\n
        \r\n \t
      • If the c<\/i> term is positive and the b<\/i> term is positive, then both r<\/i> and s<\/i> are positive.<\/li>\r\n \t
      • If the c<\/i> term is positive and the b<\/i> term is negative, then both r <\/i>and s<\/i> are negative.<\/li>\r\n \t
      • If the c<\/i> term is negative and the 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 r<\/i> is positive and s <\/i>is negative.<\/li>\r\n \t
      • If the c<\/i> term is negative and the 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>then r<\/i> is negative and 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 r<\/i> and s<\/i> end up being included in the factored form of the trinomial. Have a look at the following chart, which includes some of the problems you have seen so far.\r\n\r\n\r\n\r\n\r\n\r\n
        Trinomial<\/th>\r\n[latex]x^{2}+7x+10[\/latex]<\/th>\r\n[latex]x^{2}+5x+6[\/latex]<\/th>\r\n[latex]x^{2}+x-12[\/latex]<\/th>\r\n<\/tr>\r\n
        r<\/i> and s<\/i> values<\/th>\r\n[latex]r=+5,s=+2[\/latex]<\/td>\r\n[latex]r=+2,s=+3[\/latex]<\/td>\r\n[latex]r=+4,s=\u20133[\/latex]<\/td>\r\n<\/tr>\r\n
        Factored form<\/th>\r\n[latex]\\left(x+5\\right)\\left(x+2\\right)[\/latex]<\/td>\r\n[latex]\\left(x+2\\right)\\left(x+3\\right)[\/latex]<\/td>\r\n[latex](x+4)(x\u20133)[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n

        Factoring Trinomials II - The Shortcut (Product and Sum Method)<\/h2>\r\nNotice that in all of the examples, the r<\/i> and s\u00a0<\/i>values are repeated in the factored form of the trinomial.\u00a0So what does this mean? It means that in trinomials of the form [latex]x^{2}+bx+c[\/latex] (where the coefficient in front of [latex]x^{2}[\/latex]\u00a0is 1), if you can identify the correct r<\/i> and s<\/i> values, you can effectively skip the grouping steps and go right to the factored form. For those of you that like shortcuts, let's look at some examples where we use this idea.\r\n\r\n[caption id=\"attachment_4884\" align=\"aligncenter\" width=\"530\"]\"Picture 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

        [latex]\\left(x+r\\right)\\left(x+s\\right)\\text{ OR }\\left(x+s\\right)\\left(x+r\\right)[\/latex]<\/p>\r\nThis technique is sometime referred to as the \"Product and Sum Method.\"\r\n\r\nLet us revisit the problem [latex]x^2+7x+10[\/latex]. We determined that the factors [latex]2[\/latex] and [latex]5[\/latex] multiply to [latex]10[\/latex] while simultaneously adding to [latex]7[\/latex]. We proceeded by rewriting the problem and transforming it into a grouping problem. Alternatively, we could have jumped to the final answer immediately using Product and Sum, factoring the polynomial as\r\n

        [latex]x^2+7x+10=(x+2)(x+5)[\/latex].<\/p>\r\nNote that the order does not matter, and we could have also written [latex]x^2+7x+10=(x+5)(x+2)[\/latex].\r\n\r\nWe give several more examples demonstrating the Product and Sum Method.\r\n

        \r\n

        Example 2<\/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\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n
        Factors whose product is -27<\/th>\r\nSum of the factors<\/th>\r\n<\/tr>\r\n<\/thead>\r\n
        [latex]1\\cdot{-27}=-27[\/latex]<\/td>\r\n[latex]1-27=-26[\/latex]<\/td>\r\n<\/tr>\r\n
        [latex]-1\\cdot{27}=-27[\/latex]<\/td>\r\n[latex]-1+27=26[\/latex]<\/td>\r\n<\/tr>\r\n
        [latex]3\\cdot{-9}=-27[\/latex]<\/td>\r\n[latex]3-9=-6[\/latex]<\/td>\r\n<\/tr>\r\n
        [latex]-3\\cdot{9}=-27[\/latex]<\/td>\r\n\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

        [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

        [latex]\\left(\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\right)\\left(\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\right)[\/latex]<\/p>\r\n

        The squared term is y, so we will place a y in each set of parentheses:<\/p>\r\n

        [latex]\\left(y\\,\\,\\,\\,\\,\\,\\,\\,\\right)\\left(y\\,\\,\\,\\,\\,\\,\\,\\,\\right)[\/latex]<\/p>\r\n

        Now we can fill in the rest of each binomial with the values we found for r and s.<\/p>\r\n

        [latex]\\left(y-3\\right)\\left(y+9\\right)[\/latex]<\/p>\r\n

        Note how we kept the sign on each of the values.\u00a0 We also could have reversed the order and written [latex](y+9)(y-3)[\/latex] instead due to the commutative property of multiplication.<\/p>\r\n

        One nice thing about factoring is you can check your work. Multiply the binomials together to see if you did it correctly.<\/p>\r\n

        [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

        Answer<\/span><\/h4>\r\n[latex]y^2+6y-27=\\left(y-3\\right)\\left(y+9\\right)[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nLet us try another.\r\n
        \r\n

        Example 3<\/h3>\r\nFactor: [latex]x^2-x-12[\/latex]\r\n\r\n[reveal-answer q=\"78176\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"78176\"]\r\n\r\nWe have [latex]b=-1[\/latex] and [latex]c=-12[\/latex].\u00a0 Instead of listing from possible pair of factors for [latex]-12[\/latex], let us utilize the tips shared earlier. Since both\u00a0b<\/em> and\u00a0c<\/em> are negative, we know\u00a0r<\/em> and\u00a0s<\/em> must have opposite signs (one positive and one negative), and that the absolute of the negative value must be the larger of the two.\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n
        Factors whose product is -12<\/th>\r\nSum of the factors<\/th>\r\n<\/tr>\r\n<\/thead>\r\n
        [latex]1\\cdot{-12}=-12[\/latex]<\/td>\r\n[latex]1-12=-11[\/latex]<\/td>\r\n<\/tr>\r\n
        [latex]2\\cdot(-6)=-12[\/latex]<\/td>\r\n[latex]-2-6=-4[\/latex]<\/td>\r\n<\/tr>\r\n
        [latex]3\\cdot(-4)=-12[\/latex]<\/td>\r\n[latex]3-4=-1[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nHence, we should use [latex]r=3[\/latex] and [latex]s=-4[\/latex].\r\n\r\nWe know the factored form will look like\r\n

        [latex](x\\hspace{.25in})(x\\hspace{.25in})[\/latex]<\/p>\r\nInserting the values we found for\u00a0r<\/em> and\u00a0s<\/em> gives\r\n

        [latex](x+3)(x-4)[\/latex]<\/p>\r\n\r\n

        Answer<\/span><\/h4>\r\n[latex]x^2-x-12=(x+3)(x-4)[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nIn the following video, we present two more examples of factoring a trinomial using the shortcut presented here.\r\n\r\nhttps:\/\/youtu.be\/-SVBVVYVNTM\r\n\r\nAnother situation we sometime encounter is a trinomial with two variables of the form [latex]x^2+bxy+cy^2[\/latex]. Applying the same logic behind beginning each parenthetical expression with [latex]x[\/latex] and [latex]x[\/latex] to produce the leading term of [latex]x^2[\/latex], we will similarly end each expression with [latex]y[\/latex].\u00a0 We will then be able to deal with the coefficients using the same approach as previous examples.\r\n
        \r\n

        Example 4<\/h3>\r\nFactor: [latex]x^2-12xy+20y^2[\/latex]\r\n\r\n[reveal-answer q=\"52477\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"52477\"]\r\n\r\nTo ensure this can FOIL to produce the original expression, we will factor it as\r\n

        [latex](x\\hspace{.3in}y)(x\\hspace{.3in}y)[\/latex].<\/p>\r\nWe simply need to fill in the gaps with the corrects numbers and operations, which we can use the normal Product and Sum approach for. So, we are looking for two numbers that multiply to [latex]20[\/latex] and add to [latex]-12[\/latex]. Note that in order to multiply to a positive but add to a negative, this will require two negative factors.\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n
        Factors whose product is 20<\/th>\r\nSum of the factors<\/th>\r\n<\/tr>\r\n<\/thead>\r\n
        [latex]-1\\cdot{-20}=20[\/latex]<\/td>\r\n[latex]-1-20=-21[\/latex]<\/td>\r\n<\/tr>\r\n
        [latex]-2\\cdot(-10)=20[\/latex]<\/td>\r\n[latex]-2-10=-12[\/latex]<\/td>\r\n<\/tr>\r\n
        [latex]-4\\cdot(-5)=20[\/latex]<\/td>\r\n[latex]-4-5=-9[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nWe conclude that the correct pair is [latex]-2[\/latex] and [latex]-12[\/latex]. Filling in the blanks in our above factored form gives us\r\n

        [latex](x-2y)(x-10y)[\/latex].<\/p>\r\nIt may not be obvious that this would match the original polynomial. Let us check the answer by performing the FOIL.\r\n\r\n[latex](x-2y)(x-10y)=x^2-10xy-2xy+20y^2=x^2-12xy+20y^2[\/latex]\r\n\r\nThis matches the original problem and therefore the factoring was correct.\r\n

        Answer<\/span><\/h4>\r\n[latex]x^2-12xy+20y^2=(x-2y)(x-10y)[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n \r\n

        Its Okay To Be Prime<\/h2>\r\nMany students become frustrated when searching for the proper pair of factors, only to discover that no such pair existed! Remember, not every polynomial can be factored.\r\n
        \r\n

        ExAMPLE 5<\/h3>\r\nFactor, if possible: [latex]x^2-7x+18[\/latex]\r\n\r\n[reveal-answer q=\"241682\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"241682\"]\r\n\r\nWhen looking at the coefficient, we may be tempted to think that we can somehow use 9 and 2 to get the desired results. But this stresses how crucial it is to consider the signs as well. Let's list all the possible pairs of factors.\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n
        Factors whose product is 18<\/th>\r\nSum of the factors<\/th>\r\n<\/tr>\r\n<\/thead>\r\n
        [latex]-1\\cdot{-18}=18[\/latex]<\/td>\r\n[latex]-1-18=-19[\/latex]<\/td>\r\n<\/tr>\r\n
        [latex]-2\\cdot{-9}=18[\/latex]<\/td>\r\n[latex]-2-9=-11[\/latex]<\/td>\r\n<\/tr>\r\n
        [latex]-3\\cdot -6=18[\/latex]<\/td>\r\n[latex]-3-6=-9[\/latex]<\/td>\r\n<\/tr>\r\n
        [latex]1\\cdot{18}=18[\/latex]<\/td>\r\n[latex]1+18=19[\/latex]<\/td>\r\n<\/tr>\r\n
        [latex]2\\cdot{9}=18[\/latex]<\/td>\r\n[latex]2+9=11[\/latex]<\/td>\r\n<\/tr>\r\n
        [latex]3\\cdot 6=18[\/latex]<\/td>\r\n[latex]3+6=9[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nSo, there is no pair of numbers that multiply to 18 and add to -7, meaning this polynomial cannot be factored.\r\n

        Answer<\/span><\/h4>\r\nCannot be factored (Prime)\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n \r\n

        Factor Trinomials Part III - Factoring Out A GCF First<\/h2>\r\nNot all trinomials look like [latex]x^{2}+5x+6[\/latex], where the coefficient in front of the [latex]x^{2}[\/latex]\u00a0term is 1. In these cases, your first step should be to look for common factors for the three terms.\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n
        Trinomial<\/th>\r\nFactor out Common Factor<\/th>\r\nFactored<\/th>\r\n<\/tr>\r\n<\/thead>\r\n
        [latex]2x^{2}+10x+12[\/latex]<\/td>\r\n[latex]2(x^{2}+5x+6)[\/latex]<\/td>\r\n[latex]2\\left(x+2\\right)\\left(x+3\\right)[\/latex]<\/td>\r\n<\/tr>\r\n
        [latex]\u22125a^{2}\u221215a\u221210[\/latex]<\/td>\r\n[latex]\u22125(a^{2}+3a+2)[\/latex]<\/td>\r\n[latex]\u22125\\left(a+2\\right)\\left(a+1\\right)[\/latex]<\/td>\r\n<\/tr>\r\n
        [latex]c^{3}\u20138c^{2}+15c[\/latex]<\/td>\r\n[latex]c\\left(c^{2}\u20138c+15\\right)[\/latex]<\/td>\r\n[latex]c\\left(c\u20135\\right)\\left(c\u20133\\right)[\/latex]<\/td>\r\n<\/tr>\r\n
        [latex]y^{4}\u20139y^{3}\u201310y^{2}[\/latex]<\/td>\r\n[latex]y^{2}\\left(y^{2}\u20139y\u201310\\right)[\/latex]<\/td>\r\n[latex]y^{2}\\left(y\u201310\\right)\\left(y+1\\right)[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nNotice that once you have identified and pulled out the common factor, if the resulting trinomial has a leading coefficient of 1, you can factor the remaining trinomial as usual.\r\n\r\nLet's begin with an example where the leading coefficient is [latex]-1[\/latex]. Recall that in Section 6.1, we discussed that it could be useful to factor out the negative in this case. Here, it is imperative in order to directly apply the shortcut we have introduced.\r\n
        \r\n

        Example 6<\/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

        [latex]-m^2+16m-48=-1\\left(m^2-16m+48\\right)[\/latex]<\/p>\r\n

        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 looking for two negative numbers:<\/p>\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n
        Factors whose product is 48<\/th>\r\nSum of the factors<\/th>\r\n<\/tr>\r\n<\/thead>\r\n
        [latex]-1\\cdot{-48}=48[\/latex]<\/td>\r\n[latex]-1-48=-49[\/latex]<\/td>\r\n<\/tr>\r\n
        [latex]-2\\cdot{-12}=48[\/latex]<\/td>\r\n[latex]-2-12=-14[\/latex]<\/td>\r\n<\/tr>\r\n
        [latex]-3\\cdot{-16}=48[\/latex]<\/td>\r\n[latex]-3-16=-19[\/latex]<\/td>\r\n<\/tr>\r\n
        [latex]-4\\cdot{-12}=48[\/latex]<\/td>\r\n[latex]-4-12=-16[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n

        There are more factors whose product is 48, but we have found the ones that sum to -16, so we can stop.<\/p>\r\n

        [latex]\\begin{array}{l}r=-4\\\\s=-12\\end{array}[\/latex]<\/p>\r\n

        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

        [latex]\\left(m-4\\right)\\left(m-12\\right)[\/latex]<\/p>\r\n

        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

        \u00a0[latex]-1\\left(m-4\\right)\\left(m-12\\right)[\/latex]<\/p>\r\n\r\n

        Answer<\/span><\/h4>\r\n[latex]-m^2+16m-48=-\\left(m-4\\right)\\left(m-12\\right)[\/latex]\r\n

        [\/hidden-answer]<\/p>\r\n\r\n<\/div>\r\nIn the next example, we must first factor out a more substantial GCF.\r\n

        \r\n

        Example 7<\/h3>\r\nFactor\u00a0[latex]3x^{3}\u20133x^{2}\u201390x[\/latex].\r\n\r\n[reveal-answer q=\"298928\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"298928\"]Since [latex]3x[\/latex] is a common factor for the three terms, factor it out first.\r\n

        [latex]3x\\left(x^{2}\u2013x\u201330\\right)[\/latex]<\/p>\r\nNow you can factor the trinomial\u00a0[latex]x^{2}\u2013x\u201330[\/latex]. To find r<\/i> and s<\/i>, identify two numbers whose product is [latex]\u221230[\/latex] and whose sum is [latex]\u22121[\/latex].\r\n\r\nThe pair of factors is [latex]\u22126[\/latex] and [latex]5[\/latex]. Using our shortcut, [latex]x^2-x-30[\/latex] factors as\r\n

        [latex](x-6)(x+5)[\/latex]<\/p>\r\nHowever, do not forget the GCF of [latex]3x[\/latex] that we factored out. This needs to remain in front, implying that the original expression factors as\r\n

        [latex]3x\\left(x\u20136\\right)\\left(x+5\\right)[\/latex]<\/p>\r\n\r\n

        Answer<\/h4>\r\n[latex]3x^3-3x^2-90x=3x\\left(x\u20136\\right)\\left(x+5\\right)[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nThe following video contains two more examples of factoring a quadratic trinomial where the first step is to factor out a GCF.\r\n\r\nhttps:\/\/youtu.be\/pgH77rAtsbs\r\n\r\nThe general form of trinomials with a leading coefficient of a<\/i> is [latex]ax^{2}+bx+c[\/latex]. Sometimes the factor of a<\/i> can be factored out as you saw above; this happens when a <\/i>is in common to\u00a0all three terms. The remaining trinomial that still needs factoring (if possible) will then be simpler, with the leading term only being an [latex]x^{2}[\/latex]\u00a0term, instead of an [latex]ax^{2}[\/latex]\u00a0term.\r\n\r\nBut what do we do if\u00a0a<\/em> is not 1 and yet is not a common factor? We explore this in the next section.","rendered":"
        \n

        section 6.3 Learning Objectives<\/h3>\n

        6.3: Factoring Trinomials with Leading Coefficient of 1<\/strong><\/p>\n

          \n
        • Factor trinomials of the type [latex]x^2 + bx + c[\/latex] by grouping<\/li>\n
        • Factor trinomials of the type [latex]x^2+bx+c[\/latex] using the Product and Sum method<\/li>\n
        • Rewrite and factor trinomials of the above type by first factoring out the GCF<\/li>\n
        • Factor two-variable trinomials of the type [latex]x^2 + bxy + cy^2[\/latex]<\/li>\n<\/ul>\n<\/div>\n

           <\/p>\n

          Factor Trinomials Part I – Using Grouping<\/h2>\n

          In the last section we introduced the technique of factoring by grouping as a means to be able to factor a four-term polynomial. However, many of the polynomials we encountered were actually trinomials in disguise. Now we will get to the work of starting with a three-term polynomial, and rewriting it as a four term polynomial so it can be factored using our grouping process.<\/p>\n

          We will start with factoring trinomials of the form [latex]x^{2}+bx+c[\/latex], and hence with a leading coefficient of 1.<\/p>\n

          Remember that when\u00a0[latex]\\left(x+2\\right)[\/latex] and [latex]\\left(x+5\\right)[\/latex], are multiplied, the result is a four term polynomial and then it is\u00a0simplified into a trinomial:<\/p>\n

          \u00a0[latex]\\left(x+2\\right)\\left(x+5\\right)=x^2+5x+2x+10=x^2+7x+10[\/latex]<\/p>\n

          Factoring is the reverse of multiplying, so let\u2019s go in reverse and factor the trinomial [latex]x^{2}+7x+10[\/latex]. The individual terms [latex]x^{2}[\/latex], [latex]7x[\/latex], and 10 share no common factors. If we\u00a0rewrite the middle term as the sum of the two terms [latex]7x=5x+2x[\/latex] then we can use the grouping technique:<\/p>\n

          [latex](x^{2}+5x)+(2x+10)[\/latex]<\/p>\n

          Factor each pair:\u00a0[latex]\\begin{array}{l}x\\underbrace{\\left(x+5\\right)}+2\\underbrace{\\left(x+5\\right)}\\\\\\text{common binomial factor}\\end{array}[\/latex]<\/p>\n

          Then pull\u00a0out the common binomial factor: [latex]\\left(x+5\\right)\\left(x+2\\right)[\/latex]<\/p>\n

          But how would we know to rewrite [latex]7x[\/latex] as [latex]5x+2x[\/latex]? What would have happened if we had\u00a0rewritten [latex]7x[\/latex] as [latex]6x+x[\/latex]?<\/p>\n

          [latex](x^{2}+6x)+(x+10)[\/latex]<\/p>\n

          Factor each pair:\u00a0[latex]x\\left(x+6\\right)+1\\left(x+10\\right)[\/latex]<\/p>\n

          Then we don’t have a\u00a0common binomial! There is a method to the madness of choosing how to rewrite the middle terms so that you will end up with a common binomial factor.<\/p>\n

          \"Shakespeare<\/p>\n

          Method to the Madness<\/p>\n<\/div>\n

          The following is a summary of the method, then we will show some examples of how to use it.<\/p>\n

          \n

          Factoring Trinomials in the form\u00a0[latex]x^{2}+bx+c[\/latex]<\/h3>\n

          To factor a trinomial in the form [latex]x^{2}+bx+c[\/latex], find two integers, r<\/i> and s<\/i>, whose product is c <\/i>and whose sum is b<\/i>.<\/p>\n

          [latex]\\begin{array}{l}r\\cdot{s}=c\\\\\\text{ and }\\\\r+s=b\\end{array}[\/latex]<\/p>\n

          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

          For example, to factor [latex]x^{2}+7x+10[\/latex], you are looking for two numbers whose sum is 7 (the coefficient of the middle term) and whose product is 10 (the last term).<\/p>\n

          Look at factor pairs of 10: 1 and 10, 2, and 5. Do either of these pairs have a sum of 7? Yes, 2 and 5. So you can rewrite\u00a0[latex]7x[\/latex] as [latex]2x+5x[\/latex], and continue factoring as in the example above. Note that you can also rewrite [latex]7x[\/latex] as [latex]5x+2x[\/latex]. Both will work.<\/p>\n

          With all terms positive in the last example, we did not really need to consider any negative factors. Let\u2019s take a look at the trinomial [latex]x^{2}+x\u201312[\/latex]. In this trinomial, the 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 b<\/i> is 1.<\/p>\n\n\n\n\n\n\n\n\n\n\n
          Factors whose product is [latex]\u221212[\/latex]<\/th>\nSum of the factors<\/th>\n<\/tr>\n<\/thead>\n
          [latex]1\\cdot\u221212=\u221212[\/latex]<\/td>\n[latex]1+(\u221212)=\u221211[\/latex]<\/td>\n<\/tr>\n
          [latex]2\\cdot\u22126=\u221212[\/latex]<\/td>\n[latex]2+(\u22126)=\u22124[\/latex]<\/td>\n<\/tr>\n
          [latex]3\\cdot\u22124=\u221212[\/latex]<\/td>\n[latex]3+(\u22124)=\u22121[\/latex]<\/td>\n<\/tr>\n
          [latex]4\\cdot\u22123=\u221212[\/latex]<\/td>\n[latex]4+(\u22123)=1[\/latex]<\/td>\n<\/tr>\n
          [latex]6\\cdot\u22122=\u221212[\/latex]<\/td>\n[latex]6+(\u22122)=4[\/latex]<\/td>\n<\/tr>\n
          [latex]12\\cdot\u22121=\u221212[\/latex]<\/td>\n[latex]12+(\u22121)=11[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

          There is only one combination where the product is [latex]\u221212[\/latex] and the sum is 1, and that is when [latex]r=4[\/latex], and [latex]s=\u22123[\/latex]. Let\u2019s use these to factor our original trinomial.<\/p>\n

          \n

          Example 1<\/h3>\n

          Factor\u00a0[latex]x^{2}+x\u201312[\/latex].<\/p>\n

          Show Solution<\/span><\/p>\n
          Rewrite the trinomial using the values from the chart above. Use values [latex]r=4[\/latex] and [latex]s=\u22123[\/latex].<\/p>\n

          [latex]x^{2}+4x\u22123x\u201312[\/latex]<\/p>\n

          Group pairs of terms.<\/p>\n

          [latex]\\left(x^{2}+4x\\right)+\\left(\u22123x\u201312\\right)[\/latex]<\/p>\n

          Factor x<\/em> out of the first group.<\/p>\n

          [latex]x\\left(x+4\\right)+\\left(-3x-12\\right)[\/latex]<\/p>\n

          Factor \u22123 out of the second group.<\/p>\n

          [latex]x\\left(x+4\\right)\u20133\\left(x+4\\right)[\/latex]<\/p>\n

          Factor out [latex]\\left(x+4\\right)[\/latex].<\/p>\n

          [latex]\\left(x+4\\right)\\left(x-3\\right)[\/latex]<\/p>\n

          Answer<\/h4>\n

          [latex]x^2+x-12=\\left(x+4\\right)\\left(x-3\\right)[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n

          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

          In the following video, we present another example of how to use grouping to factor a trinomial of this form.<\/p>\n