{"id":6785,"date":"2020-10-09T18:58:52","date_gmt":"2020-10-09T18:58:52","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/slcc-beginalgebra\/?post_type=chapter&p=6785"},"modified":"2023-05-22T19:24:09","modified_gmt":"2023-05-22T19:24:09","slug":"6-4-factoring-trinomials-with-leading-coefficient-other-than-1","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/slcc-elementaryalgebra\/chapter\/6-4-factoring-trinomials-with-leading-coefficient-other-than-1\/","title":{"raw":"6.4: Factoring Trinomials with Leading Coefficient other than 1","rendered":"6.4: Factoring Trinomials with Leading Coefficient other than 1"},"content":{"raw":"
\r\n

section 6.4 Learning Objectives<\/h3>\r\n6.4: Factoring Trinomials with Leading Coefficient Other Than 1<\/strong>\r\n
    \r\n \t
  • Factor trinomials of the type ax2<\/sup> + bx + c, where a\u00a0\u2260 1<\/li>\r\n \t
  • Factor trinomials of the above type where the GCF must first be factored out<\/li>\r\n<\/ul>\r\n<\/div>\r\n \r\n\r\nIn Section 6.3, we factored trinomials with a leading coefficient of 1, or were able to factor out a common factor so that the leading coefficient became 1. It was this leading coefficient of 1 that allowed for the nice shortcut provided by the \"Product and Sum Method.\" However, if the leading coefficient is something other than 1 and the coefficients of all three terms of a trinomial don\u2019t have a common factor (other than 1), then you will need to factor a trinomial with a leading coefficient of something other than 1.\r\n
    \r\n

    Factoring Trinomials in the form\u00a0 \u00a0[latex]ax^{2}+bx+c[\/latex]<\/h3>\r\nTo factor a trinomial in the form [latex]ax^{2}+bx+c[\/latex], find two integers, r<\/i> and s<\/i>, whose sum is b<\/i> and whose product is ac.<\/i>\r\n

    [latex]\\begin{array}{l}r\\cdot{s}=a\\cdot{c}\\\\r+s=b\\end{array}[\/latex]<\/p>\r\nRewrite the trinomial as [latex]ax^{2}+rx+sx+c[\/latex] and then use grouping and the distributive property to factor the polynomial.\r\n\r\n<\/div>\r\nThis is very similar to factoring trinomials in the form [latex]x^{2}+bx+c[\/latex],\u00a0except\u00a0n<\/i>ow you are looking for two factors whose product is [latex]a\\cdot{c}[\/latex],\u00a0and whose sum is b<\/i>. Additionally, we will find that there is not a simple shortcut like we had when the leading coefficient was 1, and will stick with the full grouping strategy.\r\n\r\nBecause our first step is to multiply the values for\u00a0a<\/em> and\u00a0c<\/em>, this technique is sometimes referred to as the \"AC-Method.\"\r\n\r\nLet\u2019s see how this strategy works by factoring [latex]6z^{2}+11z+4[\/latex].\r\n\r\nIn this trinomial, [latex]a=6[\/latex], [latex]b=11[\/latex], and [latex]c=4[\/latex]. According to the strategy, you need to find two factors, r<\/i> and s<\/i>, whose sum is [latex]b=11[\/latex]\u00a0and whose product is [latex]a\\cdot{c}=6\\cdot4=24[\/latex]. Like before, you can make a chart to organize the possible factor combinations. (Notice that this chart only has positive numbers. Since ac<\/i> is positive and b<\/i> is positive, you can be certain that the two factors you're looking for are also positive numbers.)\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n
    Factors whose product is 24<\/th>\r\nSum of the factors<\/th>\r\n<\/tr>\r\n<\/thead>\r\n
    [latex]1\\cdot24=24[\/latex]<\/td>\r\n[latex]1+24=25[\/latex]<\/td>\r\n<\/tr>\r\n
    [latex]2\\cdot12=24[\/latex]<\/td>\r\n[latex]2+12=14[\/latex]<\/td>\r\n<\/tr>\r\n
    [latex]3\\cdot8=24[\/latex]<\/td>\r\n[latex]3+8=11[\/latex]<\/td>\r\n<\/tr>\r\n
    [latex]4\\cdot6=24[\/latex]<\/td>\r\n[latex]4+6=10[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nThere is only one combination where the product is 24 and the sum is 11, and that is when [latex]r=3[\/latex], and [latex]s=8[\/latex]. Let\u2019s use these values to factor the original trinomial.\r\n
    \r\n

    Example 1<\/h3>\r\nFactor [latex]6z^{2}+11z+4[\/latex].\r\n[reveal-answer q=\"796129\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"796129\"]Rewrite the middle term, [latex]11z[\/latex], as [latex]3z + 8z[\/latex] (from the chart above.)\r\n

    [latex]6z^{2}+3z+8z+4[\/latex]<\/p>\r\nGroup pairs. Use grouping to consider the terms in pairs.\r\n

    [latex]\\left(6z^{2}+3z\\right)+\\left(8z+4\\right)[\/latex]<\/p>\r\nFactor [latex]3z[\/latex] out of the first group and 4 out of the second group.\r\n

    [latex]3z\\left(2z+1\\right)+4\\left(2z+1\\right)[\/latex]<\/p>\r\nFactor out [latex]\\left(2z+1\\right)[\/latex].\r\n

    [latex]\\left(2z+1\\right)\\left(3z+4\\right)[\/latex]<\/p>\r\n\r\n

    Answer<\/h4>\r\n[latex]6z^2+11z+4=\\left(2z+1\\right)\\left(3z+4\\right)[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n
    \r\n

    Example 2<\/h3>\r\nFactor [latex]10x^2-7x-6[\/latex]\r\n\r\n[reveal-answer q=\"530918\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"530918\"]\r\n\r\nFirst we compute the product [latex]a\\cdot c[\/latex] where [latex]a=10[\/latex] and [latex]c=-6[\/latex]. We get\r\n

    [latex]a\\cdot c=10(-6)=-60[\/latex]<\/p>\r\nNext, we must find two integers that multiply to [latex]-60[\/latex] and add to [latex]-7[\/latex] (you can begin by listing out all pairs that multiply to [latex]-60[\/latex] if you find this helpful). The integers will be [latex]-12[\/latex] and [latex]5[\/latex].\r\n\r\nNow we can rewrite the problem and factor by grouping.\r\n

    [latex]10x^2-7x-6[\/latex]<\/p>\r\n

    [latex]=10x^2-12x+5x-6[\/latex]<\/p>\r\n

    [latex]=\\left(10x^2-12x\\right)+\\left(5x-6\\right)[\/latex]<\/p>\r\n

    [latex]=2x\\left(5x-6\\right)+1\\left(5x-6\\right)[\/latex]<\/p>\r\n

    [latex]=\\left(5x-6\\right)\\left(2x+1\\right)[\/latex]<\/p>\r\n\r\n

    Answer<\/h4>\r\n[latex]10x^2-7x-6=\\left(5x-6\\right)\\left(2x+1\\right)[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nIn the following video, we present another example of factoring a trinomial using grouping. \u00a0In this example, the middle term, b, is negative. Note how having a negative coefficient on the middle term and a positive c term influence the options for r and s when factoring.\r\n\r\nhttps:\/\/youtu.be\/agDaQ_cZnNc\r\n\r\nBefore going any further, it is worth mentioning that not all trinomials can be factored using integer pairs. Take the trinomial [latex]2z^{2}+35z+7[\/latex], for instance. Can you think of two integers whose sum is [latex]b=35[\/latex] and whose product is [latex]a\\cdot{c}=2\\cdot7=14[\/latex]? There are none! This type of trinomial, which cannot be factored using integers, is called a prime trinomial.\r\n\r\nIn some situations, a<\/i> is negative, as in [latex]\u22124h^{2}+11h+3[\/latex]. It often makes sense to factor out [latex]\u22121[\/latex] as the first step in factoring, as doing so will change the sign of [latex]ax^{2}[\/latex]\u00a0from negative to positive, making the remaining trinomial easier to factor.\r\n
    \r\n

    Example 3<\/h3>\r\nFactor\u00a0[latex]\u22124h^{2}+11h+3[\/latex].\r\n[reveal-answer q=\"471034\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"471034\"]Factor [latex]\u22121[\/latex] out of the trinomial. Notice that the signs of all three terms have changed.\r\n

    [latex]\u22121\\left(4h^{2}\u201311h\u20133\\right)[\/latex]<\/p>\r\nTo factor the trinomial, you need to figure out how to rewrite [latex]\u221211h[\/latex].\u00a0The product of [latex]rs=4\\cdot\u22123=\u221212[\/latex], and the sum of [latex]rs=\u221211[\/latex].\r\n\r\n\r\n\r\n\r\n\r\n\r\n
    [latex]r\\cdot{s}=\u221212[\/latex]<\/td>\r\n[latex]r+s=\u221211[\/latex]<\/td>\r\n<\/tr>\r\n
    [latex]\u221212\\cdot1=\u221212[\/latex]<\/td>\r\n[latex]\u221212+1=\u221211[\/latex]<\/td>\r\n<\/tr>\r\n
    [latex]\u22126\\cdot2=\u221212[\/latex]<\/td>\r\n[latex]\u22126+2=\u22124[\/latex]<\/td>\r\n<\/tr>\r\n
    [latex]\u22124\\cdot3=\u221212[\/latex]<\/td>\r\n[latex]\u22124+3=\u22121[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nRewrite the middle term\u00a0[latex]\u221211h[\/latex] as [latex]\u221212h+1h[\/latex].\r\n

    [latex]\u22121\\left(4h^{2}\u201312h+1h\u20133\\right)[\/latex]<\/p>\r\nGroup terms.\r\n

    [latex]\u22121\\left[\\left(4h^{2}\u201312h\\right)+\\left(1h\u20133\\right)\\right][\/latex]<\/p>\r\nFactor out [latex]4h[\/latex] from the first pair. The second group cannot be factored further, but you can write it as [latex]+1\\left(h\u20133\\right)[\/latex]\u00a0since [latex]+1\\left(h\u20133\\right)=\\left(h\u20133\\right)[\/latex]. This helps with factoring in the next step.\r\n

    [latex]\u22121\\left[4h\\left(h\u20133\\right)+1\\left(h\u20133\\right)\\right][\/latex]<\/p>\r\nFactor out a common factor of [latex]\\left(h\u20133\\right)[\/latex]. Notice you are left with [latex]\\left(h\u20133\\right)\\left(4h+1\\right)[\/latex]; the [latex]+1[\/latex] comes from the term [latex]+1\\left(h\u20133\\right)[\/latex]\u00a0in the previous step.\r\n

    [latex]\u22121\\left[\\left(h\u20133\\right)\\left(4h+1\\right)\\right][\/latex]<\/p>\r\nor simply\r\n

    [latex]-(h-3)(4h+1)[\/latex]<\/p>\r\n\r\n

    Answer<\/h4>\r\n[latex]-4h^2+11h+3=\u2212\\left(h\u20133\\right)\\left(4h+1\\right)[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nNote that the answer above can also be written as [latex]\\left(\u2212h+3\\right)\\left(4h+1\\right)[\/latex] or [latex]\\left(h\u20133\\right)\\left(\u22124h\u20131\\right)[\/latex] if you multiply [latex]\u22121[\/latex]\u00a0times one of the other factors.\r\n\r\nIn the following video we present another example of factoring a trinomial in the form [latex]-ax^2+bx+c[\/latex] using the grouping technique.\r\n\r\nhttps:\/\/youtu.be\/zDAMjdBfkDs\r\n\r\nSimilar to factoring out a [latex]-1[\/latex], we could encounter a more substantial GCF, which we see in our last example.\r\n
    \r\n

    Example 4<\/h3>\r\nFactor [latex]16x^4-12x^3-10x^2[\/latex].\r\n\r\n[reveal-answer q=\"22057\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"22057\"]\r\n\r\nNote that all three terms share a common factor of [latex]2x^2[\/latex]. We factor this out first.\r\n

    [latex]2x^2(8x^2-6x-5)[\/latex]<\/p>\r\nSince the resulting trinomial still has a leading coefficient other than 1, in this case [latex]a=8[\/latex], we must apply the strategy described in this section. We need two factors that multiply to [latex]a\\cdot c=8\\cdot (-5)=-40[\/latex] while adding to [latex]b=-6[\/latex]. Those two factors are [latex]-10[\/latex] and [latex]4[\/latex].\r\n\r\nNext, rewrite the middle term, [latex]-6x[\/latex], as [latex]-10x+4x[\/latex] and use grouping.\r\n

    [latex]2x^2\\left( 8x^2-10x+4x-5\\right)[\/latex]<\/p>\r\n

    [latex]2x^2\\left[ 2x(4x-5)+1(4x-5)\\right][\/latex]<\/p>\r\n

    [latex]2x^2 (4x-5)(2x+1)[\/latex]<\/p>\r\n\r\n

    Answer<\/h4>\r\n[latex]16x^4-12x^3-10x^2=2x^2 (4x-5)(2x+1)[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n ","rendered":"
    \n

    section 6.4 Learning Objectives<\/h3>\n

    6.4: Factoring Trinomials with Leading Coefficient Other Than 1<\/strong><\/p>\n

      \n
    • Factor trinomials of the type ax2<\/sup> + bx + c, where a\u00a0\u2260 1<\/li>\n
    • Factor trinomials of the above type where the GCF must first be factored out<\/li>\n<\/ul>\n<\/div>\n

       <\/p>\n

      In Section 6.3, we factored trinomials with a leading coefficient of 1, or were able to factor out a common factor so that the leading coefficient became 1. It was this leading coefficient of 1 that allowed for the nice shortcut provided by the “Product and Sum Method.” However, if the leading coefficient is something other than 1 and the coefficients of all three terms of a trinomial don\u2019t have a common factor (other than 1), then you will need to factor a trinomial with a leading coefficient of something other than 1.<\/p>\n

      \n

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

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

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

      Rewrite the trinomial as [latex]ax^{2}+rx+sx+c[\/latex] and then use grouping and the distributive property to factor the polynomial.<\/p>\n<\/div>\n

      This is very similar to factoring trinomials in the form [latex]x^{2}+bx+c[\/latex],\u00a0except\u00a0n<\/i>ow you are looking for two factors whose product is [latex]a\\cdot{c}[\/latex],\u00a0and whose sum is b<\/i>. Additionally, we will find that there is not a simple shortcut like we had when the leading coefficient was 1, and will stick with the full grouping strategy.<\/p>\n

      Because our first step is to multiply the values for\u00a0a<\/em> and\u00a0c<\/em>, this technique is sometimes referred to as the “AC-Method.”<\/p>\n

      Let\u2019s see how this strategy works by factoring [latex]6z^{2}+11z+4[\/latex].<\/p>\n

      In this trinomial, [latex]a=6[\/latex], [latex]b=11[\/latex], and [latex]c=4[\/latex]. According to the strategy, you need to find two factors, r<\/i> and s<\/i>, whose sum is [latex]b=11[\/latex]\u00a0and whose product is [latex]a\\cdot{c}=6\\cdot4=24[\/latex]. Like before, you can make a chart to organize the possible factor combinations. (Notice that this chart only has positive numbers. Since ac<\/i> is positive and b<\/i> is positive, you can be certain that the two factors you’re looking for are also positive numbers.)<\/p>\n\n\n\n\n\n\n\n\n
      Factors whose product is 24<\/th>\nSum of the factors<\/th>\n<\/tr>\n<\/thead>\n
      [latex]1\\cdot24=24[\/latex]<\/td>\n[latex]1+24=25[\/latex]<\/td>\n<\/tr>\n
      [latex]2\\cdot12=24[\/latex]<\/td>\n[latex]2+12=14[\/latex]<\/td>\n<\/tr>\n
      [latex]3\\cdot8=24[\/latex]<\/td>\n[latex]3+8=11[\/latex]<\/td>\n<\/tr>\n
      [latex]4\\cdot6=24[\/latex]<\/td>\n[latex]4+6=10[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

      There is only one combination where the product is 24 and the sum is 11, and that is when [latex]r=3[\/latex], and [latex]s=8[\/latex]. Let\u2019s use these values to factor the original trinomial.<\/p>\n

      \n

      Example 1<\/h3>\n

      Factor [latex]6z^{2}+11z+4[\/latex].<\/p>\n

      Show Solution<\/span><\/p>\n
      Rewrite the middle term, [latex]11z[\/latex], as [latex]3z + 8z[\/latex] (from the chart above.)<\/p>\n

      [latex]6z^{2}+3z+8z+4[\/latex]<\/p>\n

      Group pairs. Use grouping to consider the terms in pairs.<\/p>\n

      [latex]\\left(6z^{2}+3z\\right)+\\left(8z+4\\right)[\/latex]<\/p>\n

      Factor [latex]3z[\/latex] out of the first group and 4 out of the second group.<\/p>\n

      [latex]3z\\left(2z+1\\right)+4\\left(2z+1\\right)[\/latex]<\/p>\n

      Factor out [latex]\\left(2z+1\\right)[\/latex].<\/p>\n

      [latex]\\left(2z+1\\right)\\left(3z+4\\right)[\/latex]<\/p>\n

      Answer<\/h4>\n

      [latex]6z^2+11z+4=\\left(2z+1\\right)\\left(3z+4\\right)[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n

      \n

      Example 2<\/h3>\n

      Factor [latex]10x^2-7x-6[\/latex]<\/p>\n

      Show Solution<\/span><\/p>\n
      \n

      First we compute the product [latex]a\\cdot c[\/latex] where [latex]a=10[\/latex] and [latex]c=-6[\/latex]. We get<\/p>\n

      [latex]a\\cdot c=10(-6)=-60[\/latex]<\/p>\n

      Next, we must find two integers that multiply to [latex]-60[\/latex] and add to [latex]-7[\/latex] (you can begin by listing out all pairs that multiply to [latex]-60[\/latex] if you find this helpful). The integers will be [latex]-12[\/latex] and [latex]5[\/latex].<\/p>\n

      Now we can rewrite the problem and factor by grouping.<\/p>\n

      [latex]10x^2-7x-6[\/latex]<\/p>\n

      [latex]=10x^2-12x+5x-6[\/latex]<\/p>\n

      [latex]=\\left(10x^2-12x\\right)+\\left(5x-6\\right)[\/latex]<\/p>\n

      [latex]=2x\\left(5x-6\\right)+1\\left(5x-6\\right)[\/latex]<\/p>\n

      [latex]=\\left(5x-6\\right)\\left(2x+1\\right)[\/latex]<\/p>\n

      Answer<\/h4>\n

      [latex]10x^2-7x-6=\\left(5x-6\\right)\\left(2x+1\\right)[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n

      In the following video, we present another example of factoring a trinomial using grouping. \u00a0In this example, the middle term, b, is negative. Note how having a negative coefficient on the middle term and a positive c term influence the options for r and s when factoring.<\/p>\n