{"id":1377,"date":"2021-12-02T19:37:43","date_gmt":"2021-12-03T00:37:43","guid":{"rendered":"https:\/\/opentextbc.ca\/intermediatealgebraberg\/chapter\/7-4-factoring-trinomials-where-a-%e2%89%a0-1\/"},"modified":"2023-08-30T18:25:49","modified_gmt":"2023-08-30T22:25:49","slug":"factoring-trinomials-where-a-%e2%89%a0-1","status":"publish","type":"chapter","link":"https:\/\/opentextbc.ca\/intermediatealgebraberg\/chapter\/factoring-trinomials-where-a-%e2%89%a0-1\/","title":{"raw":"7.4 Factoring Trinomials where a \u2260 1","rendered":"7.4 Factoring Trinomials where a \u2260 1"},"content":{"raw":"Factoring trinomials where the leading term is not 1 is only slightly more difficult than when the leading coefficient is 1. The method used to factor the trinomial is unchanged.\r\n
\r\n

Example 7.4.1<\/p>\r\n\r\n<\/header>\r\n

\r\n\r\nFactor the trinomial [latex]3x^2 + 11x + 6[\/latex].\r\n\r\nStart by multiplying the coefficients from the first and the last terms. This is [latex]3\\cdot 6[\/latex], which yields 18.\r\n\r\nThe next task is to find all possible integers that multiply to 18 and their sums.\r\n

[latex]\\begin{array}{cc}\r\n\\text{multiply to }18\\hspace{0.25in}&\\text{sum of these integers} \\\\\r\n1\\cdot 18&19 \\\\\r\n2\\cdot 9&11 \\\\\r\n3\\cdot 6&9 \\\\\r\n6\\cdot 3&9 \\\\\r\n9\\cdot 2&11 \\\\\r\n18\\cdot 1&19\r\n\\end{array}[\/latex]<\/p>\r\nLook for the pair of integers that multiplies to 18 and adds to 11, so that it matches the equation that you started with.\r\n\r\nFor this example, the pair is [latex]2\\cdot 9[\/latex], which adds to 11.\r\n\r\nNow take the original trinomial [latex]3x^2 + 11x + 6[\/latex] and break the [latex]11x[\/latex] into [latex]2x[\/latex] and [latex]9x[\/latex].\r\n\r\nRewrite the original trinomial as [latex]3x^2 + 2x + 9x + 6[\/latex].\r\n\r\nNow, split this into two binomials as done in the previous section and factor.\r\n

[latex]3x^2 + 2x[\/latex] yields [latex]x(3x + 2)[\/latex] and [latex]9x + 6[\/latex] yields [latex]3(3x + 2)[\/latex].<\/p>\r\n

[latex]3x^2 + 2x + 9x + 6[\/latex] becomes [latex]x(3x + 2) + 3(3x + 2)[\/latex].<\/p>\r\n

[latex]x(3x + 2) + 3(3x + 2)[\/latex] factors to [latex](3x + 2)(x + 3)[\/latex].<\/p>\r\n

[latex]3x^2 + 11x + 6 = (3x + 2)(x + 3)[\/latex]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\nThe master product method works for any integer breakup of the polynomial. Slightly more complicated are questions that involve two different variables in the original polynomial.\r\n

\r\n

Example 7.4.2<\/span><\/p>\r\n\r\n<\/header>\r\n

\r\n\r\nFactor the trinomial [latex]4x^2 - xy - 5y^2[\/latex].\r\n\r\nStart by multiplying the coefficients from the first and the last terms. This is [latex]4\\cdot -5[\/latex], which yields \u221220.\r\n\r\nThe next task is to find all possible integers that multiply to \u221220 and their sums.\r\n

[latex]\\begin{array}{cc}\r\n\\text{multiply to }-20\\hspace{0.25in}&\\text{sum of these integers} \\\\\r\n-1\\cdot 20&\\phantom{-}19 \\\\\r\n-2\\cdot 10&\\phantom{-}8 \\\\\r\n-4\\cdot 5&\\phantom{-}1 \\\\\r\n-5\\cdot 4&-1 \\\\\r\n-10\\cdot 2&-8 \\\\\r\n-20\\cdot 1&-19\r\n\\end{array}[\/latex]<\/p>\r\nLook for the pair of integers that multiplies to \u221220 and adds to \u221211, so that it matches the equation that you started with.\r\n\r\nFor this example, the pair is [latex]-5\\cdot 4[\/latex], which adds to \u22121.\r\n\r\nNow take the original trinomial [latex]4x^2 - xy - 5y^2[\/latex] and break the [latex]-xy[\/latex] into [latex]-5xy[\/latex] and [latex]4xy[\/latex].\r\n\r\nRewrite the original trinomial as [latex]4x^2 - 5xy + 4xy - 5y^2[\/latex].\r\n\r\nNow, split this into two binomials as done in the previous section and factor.\r\n

[latex]4x^2 - 5xy[\/latex] yields [latex]x(4x - 5y)[\/latex] and [latex]4xy - 5y^2[\/latex] yields [latex]y(4x - 5y)[\/latex].<\/p>\r\n

[latex]4x^2 - xy - 5y^2[\/latex] becomes [latex]x(4x - 5y) + y(4x - 5y)[\/latex].<\/p>\r\n

[latex]x(4x - 5y) + y(4x - 5y)[\/latex] factors to [latex](x + y) (4x - 5y)[\/latex].<\/p>\r\n

[latex]4x^2 - xy - 5y^2 = (x + y) (4x - 5y)[\/latex]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\nThere are a number of variations potentially encountered when factoring trinomials. For instance, the original terms might be mixed up. There could be something like [latex]-10x + 3x^2 + 8[\/latex] that is not in descending powers of [latex]x[\/latex]. This requires reordering in descending powers before beginning to factor.\r\n

[latex]-10x + 3x^2 + 8 \\longrightarrow 3x^2 -10x + 8 \\text{ (factorable form)}[\/latex]<\/p>\r\nIt might also be necessary to factor out a common factor before starting. The polynomial above can be written as [latex]30x^2 -100x + 80[\/latex], in which a common factor of 10 should be factored out prior to factoring.\r\n

[latex]\\text{This turns }30x^2 -100x + 80 \\text{ into }10(3x^2 -10x + 8)[\/latex].<\/p>\r\nThere are also slight variations on the common factored binomial that can be illustrated by factoring the trinomial [latex]3x^2 -10x + 8[\/latex].\r\n

\r\n

Example 7.4.3<\/p>\r\n\r\n<\/header>\r\n

\r\n\r\nFactor the trinomial [latex]3x^2 - 10x + 8[\/latex].\r\n\r\nStart by multiplying the coefficients from the first and the last terms. This is [latex]3\\cdot 8[\/latex], which yields 24.\r\n\r\nThe next task is to find all possible integers that multiply to 24 and their sums (knowing that the middle coefficient must be negative).\r\n

[latex]\\begin{array}{cc}\r\n\\text{multiply to }24 \\hspace{0.25in}&\\text{sum of these integers} \\\\\r\n-1\\cdot -24&-25 \\\\\r\n-2\\cdot -12&-14 \\\\\r\n-3\\cdot -8&-11 \\\\\r\n-4\\cdot -6&-10 \\\\\r\n-6\\cdot -4 & -10 \\\\\r\n-8\\cdot -3&-11 \\\\\r\n-12\\cdot -2 & -14 \\\\\r\n-24\\cdot -1&-25\r\n\\end{array}[\/latex]<\/p>\r\nLook for the pair of integers that multiplies to 24 and adds to \u221210, so that it matches the equation that you started with.\r\n\r\nFor this example, the pair is [latex]-4\\cdot -6[\/latex], which adds to \u221210.\r\n\r\nNow take the original trinomial [latex]3x^2 - 10x + 8[\/latex] and break the [latex]-10x[\/latex] into [latex]-4x[\/latex] and [latex]-6x[\/latex].\r\n\r\nRewrite the original trinomial as [latex]3x^2 - 4x - 6x + 8[\/latex].\r\n\r\nNow, split this into two binomials as done in the previous section and factor.\r\n

[latex]3x^2 - 4x[\/latex] yields [latex]x(3x - 4)[\/latex], but [latex]-6x + 8[\/latex] yields [latex]2(-3x + 4)[\/latex].<\/p>\r\n

[latex]x(3x - 4)[\/latex] and [latex]2(-3x + 4)[\/latex] are a close match, but their signs are different.<\/p>\r\nThe way to deal with this is to factor out a negative, specifically, \u22122 instead of 2.\r\n

[latex]-6x + 8[\/latex] can be factored two ways: [latex]2(-3x + 4)[\/latex] and [latex]-2(3x - 4)[\/latex].<\/p>\r\nChoose the second factoring, so the common factor matches.\r\n

[latex]3x^2 - 10x + 8[\/latex] becomes [latex]x(3x - 4) + -2(3x - 4)[\/latex].<\/p>\r\n

[latex]x(3x - 4) + -2(3x - 4)[\/latex] factors to [latex](3x - 4)(x - 2)[\/latex].<\/p>\r\n

[latex]3x^2 - 10x + 8 = (3x - 4)(x - 2)[\/latex]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n

\r\n

Example 7.4.4<\/p>\r\n\r\n<\/header>\r\n

\r\n\r\nFactor the following trinomials, which are both variations of the trinomial seen before in 7.4.3:\r\n
    \r\n \t
  1. [latex]3x^2 - 14x + 8[\/latex]\r\nThe pair of numbers that can be used to break it up is \u22122 and \u221212.\r\n[latex]\\begin{array}{lll}\r\n3x^2-14x+8\\text{ breaks into}&3x^2-2x-12x+8& \\\\\r\n&x(3x-2)-4(3x-2)&\\text{Common factor is }(3x-2) \\\\\r\n&(3x-2)(x-4)&\\text{Left over when factored}\r\n\\end{array}[\/latex]<\/li>\r\n \t
  2. [latex]3x^2 - 11x + 8[\/latex]\r\nThe pair of numbers that can be used to break it up is \u22123 and \u22128.\r\n[latex]\\begin{array}{lll}\r\n3x^2-11x+8\\text{ breaks into}&3x^2-3x-8x+8& \\\\\r\n&3x(x-1)-8(x-1)&\\text{Common factor is }(x-1) \\\\\r\n&(x-1)(3x-8)&\\text{Left over when factored}\r\n\\end{array}[\/latex]<\/li>\r\n<\/ol>\r\n<\/div>\r\n<\/div>\r\n

    Questions<\/h1>\r\n

    <\/b>Factor each of the following trinomials.<\/p>\r\n\r\n

      \r\n \t
    1. [latex]7x^2-19x-6[\/latex]<\/li>\r\n \t
    2. [latex]3n^2-2n-8[\/latex]<\/li>\r\n \t
    3. [latex]7b^2+15b+2[\/latex]<\/li>\r\n \t
    4. [latex]21v^2-11v-2[\/latex]<\/li>\r\n \t
    5. [latex]5a^2+13a-6[\/latex]<\/li>\r\n \t
    6. [latex]5n^2-18n-8[\/latex]<\/li>\r\n \t
    7. [latex]2x^2-5x+2[\/latex]<\/li>\r\n \t
    8. [latex]3r^2-4r-4[\/latex]<\/li>\r\n \t
    9. [latex]2x^2+19x+35[\/latex]<\/li>\r\n \t
    10. [latex]3x^2+4x-15[\/latex]<\/li>\r\n \t
    11. [latex]2b^2-b-3[\/latex]<\/li>\r\n \t
    12. [latex]2k^2+5k-12[\/latex]<\/li>\r\n \t
    13. [latex]3x^2+17xy+10y^2[\/latex]<\/li>\r\n \t
    14. [latex]7x^2-2xy-5y^2[\/latex]<\/li>\r\n \t
    15. [latex]3x^2+11xy-20y^2[\/latex]<\/li>\r\n \t
    16. [latex]12u^2+16uv-3v^2[\/latex]<\/li>\r\n \t
    17. [latex]4k^2-17k+4[\/latex]<\/li>\r\n \t
    18. [latex]4r^2+3r-7[\/latex]<\/li>\r\n \t
    19. [latex]4m^2-9mn-9n^2[\/latex]<\/li>\r\n \t
    20. [latex]4x^2-6xy+30y^2[\/latex]<\/li>\r\n \t
    21. [latex]4x^2+13xy+3y^2[\/latex]<\/li>\r\n \t
    22. [latex]6u^2+5uv-4v^2[\/latex]<\/li>\r\n \t
    23. [latex]10x^2+19xy-2y^2[\/latex]<\/li>\r\n \t
    24. [latex]6x^2-13xy-5y^2[\/latex]<\/li>\r\n<\/ol>\r\nAnswer Key 7.4<\/a>","rendered":"

      Factoring trinomials where the leading term is not 1 is only slightly more difficult than when the leading coefficient is 1. The method used to factor the trinomial is unchanged.<\/p>\n

      \n
      \n

      Example 7.4.1<\/p>\n<\/header>\n

      \n

      Factor the trinomial [latex]3x^2 + 11x + 6[\/latex].<\/p>\n

      Start by multiplying the coefficients from the first and the last terms. This is [latex]3\\cdot 6[\/latex], which yields 18.<\/p>\n

      The next task is to find all possible integers that multiply to 18 and their sums.<\/p>\n

      [latex]\\begin{array}{cc} \\text{multiply to }18\\hspace{0.25in}&\\text{sum of these integers} \\\\ 1\\cdot 18&19 \\\\ 2\\cdot 9&11 \\\\ 3\\cdot 6&9 \\\\ 6\\cdot 3&9 \\\\ 9\\cdot 2&11 \\\\ 18\\cdot 1&19 \\end{array}[\/latex]<\/p>\n

      Look for the pair of integers that multiplies to 18 and adds to 11, so that it matches the equation that you started with.<\/p>\n

      For this example, the pair is [latex]2\\cdot 9[\/latex], which adds to 11.<\/p>\n

      Now take the original trinomial [latex]3x^2 + 11x + 6[\/latex] and break the [latex]11x[\/latex] into [latex]2x[\/latex] and [latex]9x[\/latex].<\/p>\n

      Rewrite the original trinomial as [latex]3x^2 + 2x + 9x + 6[\/latex].<\/p>\n

      Now, split this into two binomials as done in the previous section and factor.<\/p>\n

      [latex]3x^2 + 2x[\/latex] yields [latex]x(3x + 2)[\/latex] and [latex]9x + 6[\/latex] yields [latex]3(3x + 2)[\/latex].<\/p>\n

      [latex]3x^2 + 2x + 9x + 6[\/latex] becomes [latex]x(3x + 2) + 3(3x + 2)[\/latex].<\/p>\n

      [latex]x(3x + 2) + 3(3x + 2)[\/latex] factors to [latex](3x + 2)(x + 3)[\/latex].<\/p>\n

      [latex]3x^2 + 11x + 6 = (3x + 2)(x + 3)[\/latex]<\/p>\n<\/div>\n<\/div>\n

      The master product method works for any integer breakup of the polynomial. Slightly more complicated are questions that involve two different variables in the original polynomial.<\/p>\n

      \n
      \n

      Example 7.4.2<\/span><\/p>\n<\/header>\n

      \n

      Factor the trinomial [latex]4x^2 - xy - 5y^2[\/latex].<\/p>\n

      Start by multiplying the coefficients from the first and the last terms. This is [latex]4\\cdot -5[\/latex], which yields \u221220.<\/p>\n

      The next task is to find all possible integers that multiply to \u221220 and their sums.<\/p>\n

      [latex]\\begin{array}{cc} \\text{multiply to }-20\\hspace{0.25in}&\\text{sum of these integers} \\\\ -1\\cdot 20&\\phantom{-}19 \\\\ -2\\cdot 10&\\phantom{-}8 \\\\ -4\\cdot 5&\\phantom{-}1 \\\\ -5\\cdot 4&-1 \\\\ -10\\cdot 2&-8 \\\\ -20\\cdot 1&-19 \\end{array}[\/latex]<\/p>\n

      Look for the pair of integers that multiplies to \u221220 and adds to \u221211, so that it matches the equation that you started with.<\/p>\n

      For this example, the pair is [latex]-5\\cdot 4[\/latex], which adds to \u22121.<\/p>\n

      Now take the original trinomial [latex]4x^2 - xy - 5y^2[\/latex] and break the [latex]-xy[\/latex] into [latex]-5xy[\/latex] and [latex]4xy[\/latex].<\/p>\n

      Rewrite the original trinomial as [latex]4x^2 - 5xy + 4xy - 5y^2[\/latex].<\/p>\n

      Now, split this into two binomials as done in the previous section and factor.<\/p>\n

      [latex]4x^2 - 5xy[\/latex] yields [latex]x(4x - 5y)[\/latex] and [latex]4xy - 5y^2[\/latex] yields [latex]y(4x - 5y)[\/latex].<\/p>\n

      [latex]4x^2 - xy - 5y^2[\/latex] becomes [latex]x(4x - 5y) + y(4x - 5y)[\/latex].<\/p>\n

      [latex]x(4x - 5y) + y(4x - 5y)[\/latex] factors to [latex](x + y) (4x - 5y)[\/latex].<\/p>\n

      [latex]4x^2 - xy - 5y^2 = (x + y) (4x - 5y)[\/latex]<\/p>\n<\/div>\n<\/div>\n

      There are a number of variations potentially encountered when factoring trinomials. For instance, the original terms might be mixed up. There could be something like [latex]-10x + 3x^2 + 8[\/latex] that is not in descending powers of [latex]x[\/latex]. This requires reordering in descending powers before beginning to factor.<\/p>\n

      [latex]-10x + 3x^2 + 8 \\longrightarrow 3x^2 -10x + 8 \\text{ (factorable form)}[\/latex]<\/p>\n

      It might also be necessary to factor out a common factor before starting. The polynomial above can be written as [latex]30x^2 -100x + 80[\/latex], in which a common factor of 10 should be factored out prior to factoring.<\/p>\n

      [latex]\\text{This turns }30x^2 -100x + 80 \\text{ into }10(3x^2 -10x + 8)[\/latex].<\/p>\n

      There are also slight variations on the common factored binomial that can be illustrated by factoring the trinomial [latex]3x^2 -10x + 8[\/latex].<\/p>\n

      \n
      \n

      Example 7.4.3<\/p>\n<\/header>\n

      \n

      Factor the trinomial [latex]3x^2 - 10x + 8[\/latex].<\/p>\n

      Start by multiplying the coefficients from the first and the last terms. This is [latex]3\\cdot 8[\/latex], which yields 24.<\/p>\n

      The next task is to find all possible integers that multiply to 24 and their sums (knowing that the middle coefficient must be negative).<\/p>\n

      [latex]\\begin{array}{cc} \\text{multiply to }24 \\hspace{0.25in}&\\text{sum of these integers} \\\\ -1\\cdot -24&-25 \\\\ -2\\cdot -12&-14 \\\\ -3\\cdot -8&-11 \\\\ -4\\cdot -6&-10 \\\\ -6\\cdot -4 & -10 \\\\ -8\\cdot -3&-11 \\\\ -12\\cdot -2 & -14 \\\\ -24\\cdot -1&-25 \\end{array}[\/latex]<\/p>\n

      Look for the pair of integers that multiplies to 24 and adds to \u221210, so that it matches the equation that you started with.<\/p>\n

      For this example, the pair is [latex]-4\\cdot -6[\/latex], which adds to \u221210.<\/p>\n

      Now take the original trinomial [latex]3x^2 - 10x + 8[\/latex] and break the [latex]-10x[\/latex] into [latex]-4x[\/latex] and [latex]-6x[\/latex].<\/p>\n

      Rewrite the original trinomial as [latex]3x^2 - 4x - 6x + 8[\/latex].<\/p>\n

      Now, split this into two binomials as done in the previous section and factor.<\/p>\n

      [latex]3x^2 - 4x[\/latex] yields [latex]x(3x - 4)[\/latex], but [latex]-6x + 8[\/latex] yields [latex]2(-3x + 4)[\/latex].<\/p>\n

      [latex]x(3x - 4)[\/latex] and [latex]2(-3x + 4)[\/latex] are a close match, but their signs are different.<\/p>\n

      The way to deal with this is to factor out a negative, specifically, \u22122 instead of 2.<\/p>\n

      [latex]-6x + 8[\/latex] can be factored two ways: [latex]2(-3x + 4)[\/latex] and [latex]-2(3x - 4)[\/latex].<\/p>\n

      Choose the second factoring, so the common factor matches.<\/p>\n

      [latex]3x^2 - 10x + 8[\/latex] becomes [latex]x(3x - 4) + -2(3x - 4)[\/latex].<\/p>\n

      [latex]x(3x - 4) + -2(3x - 4)[\/latex] factors to [latex](3x - 4)(x - 2)[\/latex].<\/p>\n

      [latex]3x^2 - 10x + 8 = (3x - 4)(x - 2)[\/latex]<\/p>\n<\/div>\n<\/div>\n

      \n
      \n

      Example 7.4.4<\/p>\n<\/header>\n

      \n

      Factor the following trinomials, which are both variations of the trinomial seen before in 7.4.3:<\/p>\n

        \n
      1. [latex]3x^2 - 14x + 8[\/latex]
        \nThe pair of numbers that can be used to break it up is \u22122 and \u221212.
        \n[latex]\\begin{array}{lll} 3x^2-14x+8\\text{ breaks into}&3x^2-2x-12x+8& \\\\ &x(3x-2)-4(3x-2)&\\text{Common factor is }(3x-2) \\\\ &(3x-2)(x-4)&\\text{Left over when factored} \\end{array}[\/latex]<\/li>\n
      2. [latex]3x^2 - 11x + 8[\/latex]
        \nThe pair of numbers that can be used to break it up is \u22123 and \u22128.
        \n[latex]\\begin{array}{lll} 3x^2-11x+8\\text{ breaks into}&3x^2-3x-8x+8& \\\\ &3x(x-1)-8(x-1)&\\text{Common factor is }(x-1) \\\\ &(x-1)(3x-8)&\\text{Left over when factored} \\end{array}[\/latex]<\/li>\n<\/ol>\n<\/div>\n<\/div>\n

        Questions<\/h1>\n

        <\/b>Factor each of the following trinomials.<\/p>\n

          \n
        1. [latex]7x^2-19x-6[\/latex]<\/li>\n
        2. [latex]3n^2-2n-8[\/latex]<\/li>\n
        3. [latex]7b^2+15b+2[\/latex]<\/li>\n
        4. [latex]21v^2-11v-2[\/latex]<\/li>\n
        5. [latex]5a^2+13a-6[\/latex]<\/li>\n
        6. [latex]5n^2-18n-8[\/latex]<\/li>\n
        7. [latex]2x^2-5x+2[\/latex]<\/li>\n
        8. [latex]3r^2-4r-4[\/latex]<\/li>\n
        9. [latex]2x^2+19x+35[\/latex]<\/li>\n
        10. [latex]3x^2+4x-15[\/latex]<\/li>\n
        11. [latex]2b^2-b-3[\/latex]<\/li>\n
        12. [latex]2k^2+5k-12[\/latex]<\/li>\n
        13. [latex]3x^2+17xy+10y^2[\/latex]<\/li>\n
        14. [latex]7x^2-2xy-5y^2[\/latex]<\/li>\n
        15. [latex]3x^2+11xy-20y^2[\/latex]<\/li>\n
        16. [latex]12u^2+16uv-3v^2[\/latex]<\/li>\n
        17. [latex]4k^2-17k+4[\/latex]<\/li>\n
        18. [latex]4r^2+3r-7[\/latex]<\/li>\n
        19. [latex]4m^2-9mn-9n^2[\/latex]<\/li>\n
        20. [latex]4x^2-6xy+30y^2[\/latex]<\/li>\n
        21. [latex]4x^2+13xy+3y^2[\/latex]<\/li>\n
        22. [latex]6u^2+5uv-4v^2[\/latex]<\/li>\n
        23. [latex]10x^2+19xy-2y^2[\/latex]<\/li>\n
        24. [latex]6x^2-13xy-5y^2[\/latex]<\/li>\n<\/ol>\n

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