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Mathematics LibreTexts

7.3: Factor Quadratic Trinomials with Leading Coefficient Other than 1

  • Page ID
    18972
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    Learning Objectives

    By the end of this section, you will be able to:

    • Recognize a preliminary strategy to factor polynomials completely
    • Factor trinomials of the form \(ax^{2}+bx+c\) with a GCF
    • Factor trinomials using trial and error
    • Factor trinomials using the ‘ac’ method

    Note

    Before you get started, take this readiness quiz.

    1. Find the GCF of 45\(p^{2}\) and 30\(p^{6}\)
      If you missed this problem, review Exercise 7.1.4.
    2. Multiply \((3 y+4)(2 y+5)\)
      If you missed this problem, review Exercise 6.3.37.
    3. Combine like terms \(12 x^{2}+3 x+5 x+9\)
      If you missed this problem, review Exercise 1.3.37.

    Recognize a Preliminary Strategy for Factoring

    Let’s summarize where we are so far with factoring polynomials. In the first two sections of this chapter, we used three methods of factoring: factoring the GCF, factoring by grouping, and factoring a trinomial by “undoing” FOIL. More methods will follow as you continue in this chapter, as well as later in your studies of algebra.

    How will you know when to use each factoring method? As you learn more methods of factoring, how will you know when to apply each method and not get them confused? It will help to organize the factoring methods into a strategy that can guide you to use the correct method.

    As you start to factor a polynomial, always ask first, “Is there a greatest common factor?” If there is, factor it first.

    The next thing to consider is the type of polynomial. How many terms does it have? Is it a binomial? A trinomial? Or does it have more than three terms?

    • If it is a trinomial where the leading coefficient is one, \(x^{2}+b x+c\), use the “undo FOIL” method.
    • If it has more than three terms, try the grouping method. This is the only method to use for polynomials of more than three terms.

    Some polynomials cannot be factored. They are called “prime.” Below we summarize the methods we have so far.

    This figure lists strategies for factoring polynomials. At the top of the figure is G C F, where factoring always starts. From there, the figure has three branches. The first is binomial, the second is trinomial with the form x ^ 2 + b x +c, and the third is “more than three terms”, which is labeled with grouping.

    CHOOSE A STRATEGY TO FACTOR POLYNOMIALS COMPLETELY.

    1. Is there a greatest common factor?
      • Factor it out.
    2. Is the polynomial a binomial, trinomial, or are there more than three terms?
      • If it is a binomial, right now we have no method to factor it.
      • If it is a trinomial of the form \(x^{2}+b x+c\): Undo FOIL \((x\qquad)(x\qquad)\)
      • If it has more than three terms: Use the grouping method.
    3. Check by multiplying the factors.

    Use the preliminary strategy to completely factor a polynomial. A polynomial is factored completely if, other than monomials, all of its factors are prime.

    Exercise \(\PageIndex{1}\)

    Identify the best method to use to factor each polynomial.

    1. \(6 y^{2}-72\)
    2. \(r^{2}-10 r-24\)
    3. \(p^{2}+5 p+p q+5 q\)
    Answer a

    \[\begin{array}{ll} &6 y^{2}-72\\ \text { Is there a greatest common factor? } & \text {Yes, 6. } \\ \text { Factor out the } 6 &6\left(y^{2}-12\right) \\ \text { Is it a binomial, trinomial, or are there } & \text {Binomial, we have no method to factor } \\ \text { more than } 3 \text { terms? } & \text {binomials yet. } \end{array} \nonumber\]

    Answer b

    \[\begin{array}{ll} &r^{2}-10 r-24\\ \text { Is there a greatest common factor? }& \text {No, there is no common factor. } \\\text { Is it a binomial, trinomial, or are there } &\text {Trinomial, with leading coefficient } 1, \text { so } \\ \text { more than three terms? }& \text {"undo" FOIL. }\end{array} \nonumber\]

    Answer c

    \[\begin{array}{ll} &p^{2}+5 p+p q+5 q\\ \text { Is there a greatest common factor? }& \text {No, there is no common factor. } \\\text { Is it a binomial, trinomial, or are there } &\text {More than three terms, so factor using }\\ \text { more than three terms? }& \text {grouping. }\end{array} \nonumber\]

    Exercise \(\PageIndex{2}\)

    Identify the best method to use to factor each polynomial:

    1. \(4 y^{2}+32\)
    2. \(y^{2}+10 y+21\)
    3. \(y z+2 y+3 z+6\)
    Answer a

    no method

    Answer b

    undo using FOIL

    Answer c

    factor with grouping

    Exercise \(\PageIndex{3}\)

    Identify the best method to use to factor each polynomial:

    1. \(a b+a+4 b+4\)
    2. \(3 k^{2}+15\)
    3. \(p^{2}+9 p+8\)
    Answer a

    factor using grouping

    Answer b

    no method

    Answer c

    undo using FOIL

    Factor Trinomials of the form ax2 + bx + c with a GCF

    Now that we have organized what we’ve covered so far, we are ready to factor trinomials whose leading coefficient is not 1, trinomials of the form \(a x^{2}+b x+c\). Remember to always check for a GCF first! Sometimes, after you factor the GCF, the leading coefficient of the trinomial becomes 1 and you can factor it by the methods in the last section. Let’s do a few examples to see how this works. Watch out for the signs in the next two examples.

    Exercise \(\PageIndex{4}\)

    Factor completely: \(2 n^{2}-8 n-42\).

    Answer

    Use the preliminary strategy.

    \(\begin{array}{ll} \text { Is there a greatest common factor? }&2 n^{2}-8 n-42\\ \text { Yes, GCF }=2 . \text { Factor it out. }& 2\left(n^{2}-4 n-21\right) \\\text { Inside the parentheses, is it a binomial, trinomial, or are there }&\\ \text { more than three terms? }& \\ \text { It is a trinomial whose coefficient is } 1, \text { so undo FOIL. } & 2(n\qquad )(n\qquad) \\ \text { Use } 3 \text { and }-7 \text { as the last terms of the binomials. } & 2(n+3)(n-7) \end{array}\)

    Factors of −21 Sum of factors
    1,−21 1+(−21)=−20
    3,−7 3+(−7)=−4*

    \(\begin{array}{l}{\text {Check. }} \\ {2(n+3)(n-7)} \\ {2\left(n^{2}-7 n+3 n-21\right)} \\ {2\left(n^{2}-4 n-21\right)} \\ {2 n^{2}-8 n-42 }\checkmark \end{array}\)

    Exercise \(\PageIndex{5}\)

    Factor completely: \(4 m^{2}-4 m-8\)

    Answer

    4\((m+1)(m-2)\)

    Exercise \(\PageIndex{6}\)

    Factor completely: \(5 k^{2}-15 k-50\)

    Answer

    5\((k+2)(k-5)\)

    Exercise \(\PageIndex{7}\)

    Factor completely: \(4 y^{2}-36 y+56\)

    Answer

    Use the preliminary strategy.
    \(\begin{array}{ll} \text { Is there a greatest common factor? }&4 y^{2}-36 y+56\\ \text { Yes, GCF }=4 . \text { Factor it out. }&4\left(y^{2}-9 y+14\right) \\\text { Inside the parentheses, is it a binomial, trinomial, or are there }&\\ \text { more than three terms? }& \\ \text { It is a trinomial whose coefficient is } 1, \text { so undo FOIL. } & 4(y\qquad )(y\qquad) \\\text { Use a table like the one below to find two numbers that multiply to }&\\ 14 \text { and add to }-9\\ \text { Both factors of } 14 \text { must be negative. } & 4(y-2)(y-7) \end{array}\)

    Factors of 14 Sum of factors
    −1,−14 −1+(−14)=−15
    −2,−7 −2+(−7)=−9*

    \(\begin{array}{l}{\text { Check. }} \\ {4(y-2)(y-7)} \\ {4\left(y^{2}-7 y-2 y+14\right)} \\ {4\left(y^{2}-9 y+14\right)} \\ {4 y^{2}-36 y+42 } \checkmark \end{array}\)

    Exercise \(\PageIndex{8}\)

    Factor completely: \(3 r^{2}-9 r+6\)

    Answer

    3\((r-1)(r-2)\)

    Exercise \(\PageIndex{9}\)

    Factor completely: \(2 t^{2}-10 t+12\)

    Answer

    2\((t-2)(t-3)\)

    In the next example the GCF will include a variable.

    Exercise \(\PageIndex{10}\)

    Factor completely: \(4 u^{3}+16 u^{2}-20 u\)

    Answer

    Use the preliminary strategy.
    \(\begin{array}{ll} \text { Is there a greatest common factor? }&4 u^{3}+16 u^{2}-20 u\\ \text { Yes, GCF }=4 u . \text { Factor it. }&4 u\left(u^{2}+4 u-5\right) \\\text { Binomial, trinomial, or more than three terms? }&\\ \text { more than three terms? }& \\ \text { It is a trinomial. So "undo FOIL." } & 4u(u\qquad )(u\qquad) \\\text { Use a table like the table below to find two numbers that }&4 u(u-1)(u+5)\\ \text { multiply to }-5 \text { and add to } 4\end{array}\)

    Factors of −5 Sum of factors
    −1,5 −1+5=4*
    1,−5 1+(−5)=−4

    Check.

    \(\begin{array}{l}{4 u(u-1)(u+5)} \\ {4 u\left(u^{2}+5 u-u-5\right)} \\ {4 u\left(u^{2}+4 u-5\right)} \\ {4 u^{3}+16 u^{2}-20 u }\checkmark \end{array}\)

    Exercise \(\PageIndex{11}\)

    Factor completely: \(5 x^{3}+15 x^{2}-20 x\)

    Answer

    5\(x(x-1)(x+4)\)

    Exercise \(\PageIndex{12}\)

    Factor completely: \(6 y^{3}+18 y^{2}-60 y\)

    Answer

    6\(y(y-2)(y+5)\)

    Factor Trinomials using Trial and Error

    What happens when the leading coefficient is not 1 and there is no GCF? There are several methods that can be used to factor these trinomials. First we will use the Trial and Error method.

    Let’s factor the trinomial \(3 x^{2}+5 x+2\)

    From our earlier work we expect this will factor into two binomials.

    \[\begin{array}{c}{3 x^{2}+5 x+2} \\ {( \qquad)( \qquad)}\end{array}\]

    We know the first terms of the binomial factors will multiply to give us 3\(x^{2}\). The only factors of 3\(x^{2}\) are \(1 x, 3 x\). We can place them in the binomials.

    This figure has the polynomial 3 x^ 2 +5 x +2. Underneath there are two terms, 1 x, and 3 x. Below these are the two factors x and (3 x) being shown multiplied.

    Check. Does \(1 x \cdot 3 x=3 x^{2}\)?

    We know the last terms of the binomials will multiply to 2. Since this trinomial has all positive terms, we only need to consider positive factors. The only factors of 2 are 1 and 2. But we now have two cases to consider as it will make a difference if we write 1, 2, or 2, 1.

    This figure demonstrates the possible factors of the polynomial 3x^2 +5x +2. The polynomial is written twice. Underneath both, there are the terms 1x, 3x under the 3x^2. Also, there are the factors 1,2 under the 2 term. At the bottom of the figure there are two possible factorizations of the polynomial. The first is (x + 1)(3x + 2) and the next is (x + 2)(3x + 1).

    Which factors are correct? To decide that, we multiply the inner and outer terms.

    This figure demonstrates the possible factors of the polynomial 3 x^ 2 + 5 x +2. The polynomial is written twice. Underneath both, there are the terms 1 x, 3 x under the 3 x ^ 2. Also, there are the factors 1, 2 under the 2 term. At the bottom of the figure there are two possible factorizations of the polynomial. The first is (x + 1)(3 x + 2). Underneath this factorization are the products 3 x from multiplying the middle terms 1 and 3 x. Also there is the product of 2 x from multiplying the outer terms x and 2. These products of 3 x and 2 x add to 5 x. Underneath the second factorization are the products 6 x from multiplying the middle terms 2 and 3 x. Also there is the product of 1 x from multiplying the outer terms x and 1. These two products of 6 x and 1 x add to 7 x.

    Since the middle term of the trinomial is 5x, the factors in the first case will work. Let’s FOIL to check.

    \[\begin{array}{l}{(x+1)(3 x+2)} \\ {3 x^{2}+2 x+3 x+2} \\ {3 x^{2}+5 x+2}\checkmark \end{array}\]

    Our result of the factoring is:

    \[\begin{array}{l}{3 x^{2}+5 x+2} \\ {(x+1)(3 x+2)}\end{array}\]

    Exercise \(\PageIndex{13}\): How to Factor Trinomials of the Form \(ax^2+bx+c\) Using Trial and Error

    Factor completely: \(3 y^{2}+22 y+7\)

    Answer

    This table summarizes the steps for factoring 3 y ^ 2 + 22 y + 7. The first row states write the trinomial in descending order. The polynomial is written 3 y ^ 2 +22 y + 7.The second row states find all the factor pairs of the first term. The only pairs listed are 1 y, 3 y. Then, since there is only one pair, they are in the parentheses written (1 y ) and (3 y ).The third row states “find all the factored pairs of the third term”. It also states the only factors of 7 are 1 and 7.The fourth row states test all the possible combinations of the factors until the correct product is found. The possible factors are shown (y + 1)(3 y + 7) and (y + 7)(3y + 1). Under each factor is the products of the outer terms and the inner terms. For the first it is 7y and 3y. For the second it is 21 y and y. The combination (y + 7)(3 y + 1) is the correct factoring.

    The last row states to check by multiplying. The product of (y + 7)(3 y + 1) is shown as 3 y ^ 2 + 22 y + 7.

    Exercise \(\PageIndex{14}\)

    Factor completely: \(2 a^{2}+5 a+3\)

    Answer

    \((a+1)(2 a+3)\)

    Exercise \(\PageIndex{15}\)

    Factor completely: \(4 b^{2}+5 b+1\)

    Answer

    \((b+1)(4 b+1)\)

    FACTOR TRINOMIALS OF THE FORM \(ax^2+bx+c\) USING TRIAL AND ERROR.

    1. Write the trinomial in descending order of degrees.
    2. Find all the factor pairs of the first term.
    3. Find all the factor pairs of the third term.
    4. Test all the possible combinations of the factors until the correct product is found.
    5. Check by multiplying.

    When the middle term is negative and the last term is positive, the signs in the binomials must both be negative.

    Exercise \(\PageIndex{16}\)

    Factor completely: \(6 b^{2}-13 b+5\)

    Answer
    The trinomial is already in descending order. .
    Find the factors of the first term. .
    Find the factors of the last term. Consider the signs. Since the last term, 5 is positive its factors must both be positive or both be negative. The coefficient of the middle term is negative, so we use the negative factors. .
    Consider all the combinations of factors.
    \(6 b^{2}-13 b+5\)
    Possible factors Product
    (b−1)(6b−5) \(6 b^{2}-11 b+5\)
    (b−5)(6b−1) \(6 b^{2}-31 b+5\)
    (2b−1)(3b−5) \(6 b^{2}-13 b+5\) *
    (2b−5)(3b−1) \(6 b^{2}-17 b+5\)
    \(\begin{array}{ll}\text{The correct factors are those whose product} & \\ \text{is the original trinomial.} & (2 b-1)(3 b-5)\\\\\text {Check by multiplying. } \\\\\begin{array}{l}{(2 b-1)(3 b-5)} \\ {6 b^{2}-10 b-3 b+5} \\ {6 b^{2}-13 b+5 v}\checkmark \end{array}\end{array}\)

    Exercise \(\PageIndex{17}\)

    Factor completely: \(8 x^{2}-14 x+3\)

    Answer

    \((2 x-3)(4 x-1)\)

    Exercise \(\PageIndex{18}\)

    Factor completely: \(10 y^{2}-37 y+7\)

    Answer

    \((2 y-7)(5 y-1)\)

    When we factor an expression, we always look for a greatest common factor first. If the expression does not have a greatest common factor, there cannot be one in its factors either. This may help us eliminate some of the possible factor combinations.

    Exercise \(\PageIndex{19}\)

    Factor completely: \(14 x^{2}-47 x-7\)

    Answer
    The trinomial is already in descending order. .
    Find the factors of the first term. .
    Find the factors of the last term. Consider the signs. Since it is negative, one factor must be positive and one negative. .
    Consider all the combinations of factors. We use each pair of the factors of 14\(x^{2}\) with each pair of factors of −7.
    Factors of \(14x^2\) Pair with Factors of −7
    \(x, 14 x\) 11, −7
    −7, 11
    (reverse order)
    \(x, 14 x\) −1, 77
    77, −1
    (reverse order)
    \(2x,7x\) 11, −7
    −7, 11
    (reverse order)
    \(2x,7x\) −1, 77
    77, −1
    (reverse order)

    These pairings lead to the following eight combinations.

    This table has the heading 14 x ^ 2 – 47 x minus 7. This table has two columns. The first column is labeled “possible factors” and the second column is labeled “product”. The first column lists all the combinations of possible factors and the second column has the products. In the first row under “possible factors” it reads (x+1) and (14 x minus 7). Under product, in the next column, it says “not an option”. In the next row down, it shows (x minus 7) and (14 x plus 1). In the next row down, it shows (x minus 1) and (14 x plus 7). Next to this in the product column, it says “not an option.” The next row down under “possible factors”, it has the equation (x plus 7 and 14 x minus 1. Next to this in the product column it has 14 x ^2 plus 97 x minus 7. The next row down under possible factors, it has 2 x plus 1 and 7 x minus 7. Next to this under the product column, is says “not an option”. The next row down reads 2 x minus 7 and 7x plus 1. Next to this under the product column, it has 14 x ^2 minus 47 x minus 7 with the asterisk following the 7. The next row down reads 2 x minus 1 and 7 x plus 7. Next to this in the product column it reads “not an option”. The final row reads 2 x plus 7 and 7 x minus 1. Next to this in the product column it reads 14, x, ^ 2 plus 47 x minus 7. Next to the table is a box with four arrows point to each “not an option” row. The reason given in the textbox is “if the trinomial has no common factors, then neither factor can contain a common factor. That means that each of these combinations is not an option.”
    \(\begin{array}{ll}\text{The correct factors are those whose product} & \\ \text{is the original trinomial.} & (2 x-7)(7 x+1)\\\\\text {Check by multiplying. } \\\\\begin{array}{l}{(2 b-1)(3 b-5)} \\ {6 b^{2}-10 b-3 b+5} \\ {6 b^{2}-13 b+5 }\checkmark \end{array}\end{array}\)

    Exercise \(\PageIndex{20}\)

    Factor completely: \(8 a^{2}-3 a-5\)

    Answer

    \((a-1)(8 a+5)\)

    Exercise \(\PageIndex{21}\)

    Factor completely: \(6 b^{2}-b-15\)

    Answer

    \((2 b+3)(3 b-5)\)

    Exercise \(\PageIndex{22}\)

    Factor completely: \(18 n^{2}-37 n+15\)

    Answer
    The trinomial is already in descending order. \(18 n^{2}-37 n+15\)
    Find the factors of the first term. .
    Find the factors of the last term. Consider the signs. Since 15 is positive and the coefficient of the middle term is negative, we use the negative factors. .

    Consider all the combinations of factors.

    This table has the heading 18 n ^ 2 – 37n + 15. This table has two columns. The first column is labeled possible factors and the second column is labeled product. The first column lists all the combinations of possible factors and the second column has the products. Eight rows list the product is not an option. There is a textbox giving the reason for no option. The reason in the textbox is “if the trinomial has no common factors, then neither factor can contain a common factor”. The row containing the factors (2n – 3)(9n – 5) with the product 18n^2 minus 37 n + 15 has an asterisk.
    \(\begin{array}{ll}\text{The correct factors are those whose product} & \\ \text{is the original trinomial.} & (2 n-3)(9 n-5)\\\\\text {Check by multiplying. } \\\\ \begin{array}{l}{(2 n-3)(9 n-5)} \\ {18 n^{2}-10 n-27 n+15} \\ {18 n^{2}-37 n+15 } \checkmark\end{array} \end{array}\)

    Exercise \(\PageIndex{23}\)

    Factor completely: \(18 x^{2}-3 x-10\)

    Answer

    \((3 x+2)(6 x-5)\)

    Exercise \(\PageIndex{24}\)

    Factor completely: \(30 y^{2}-53 y-21\)

    Answer

    \((3 y+1)(10 y-21)\)

    Don’t forget to look for a GCF first.

    Exercise \(\PageIndex{25}\)

    Factor completely: \(10 y^{4}+55 y^{3}+60 y^{2}\)

    Answer
    \(10 y^{4}+55 y^{3}+60 y^{2}\)
    Notice the greatest common factor, and factor it first. 5\(y^{2}\left(2 y^{2}+11 y+12\right)\)
    Factor the trinomial. .

    Consider all the combinations.

    This table has the heading 2 y squared + 11 y + 12 This table has two columns. The first column is labeled “possible factors” and the second column is labeled “product”. The first column lists all the combinations of possible factors and the second column has the products. Four rows list the product is not an option. There is a textbox giving the reason for no option. The reason in the textbox is “if the trinomial has no common factors, then neither factor can contain a common factor”. The row containing the factors (y + 4)(2y + 3) with the product 2 y squared + 11 y + 12 has an asterisk.
    \(\begin{array}{ll}\text{The correct factors are those whose product} &5 y^{2}(y+4)(2 y+3) \\ \text{is the original trinomial. Remember to include} & \\\text {the factor } 5 y^{2}\\\text {Check by multiplying. } \\\\ \begin{array}{l}{5 y^{2}(y+4)(2 y+3)} \\ {5 y^{2}\left(2 y^{2}+8 y+3 y+12\right)} \\ {10 y^{4}+55 y^{3}+60 y^{2}}\checkmark\end{array} \end{array}\)

    Exercise \(\PageIndex{26}\)

    Factor completely: \(15 n^{3}-85 n^{2}+100 n\)

    Answer

    5\(n(n-4)(3 n-5)\)

    Exercise \(\PageIndex{27}\)

    Factor completely: \(56 q^{3}+320 q^{2}-96 q\)

    Answer

    8\(q(q+6)(7 q-2)\)

    Factor Trinomials using the “ac” Method

    Another way to factor trinomials of the form \(ax^2+bx+c\) is the “ac” method. (The “ac” method is sometimes called the grouping method.) The “ac” method is actually an extension of the methods you used in the last section to factor trinomials with leading coefficient one. This method is very structured (that is step-by-step), and it always works!

    Exercise \(\PageIndex{28}\): How to Factor Trinomials Using the “ac” Method

    Factor: \(6 x^{2}+7 x+2\)

    Answer

    This table lists the steps for factoring 6 x ^ 2 + 7 x + 2. The first step is to factor the GCF. This polynomial has none.The second row states to find the product a c. Then, it lists a c as 6 times 2 = 12.The third step is to find two numbers m and n in which m times n = a c and m + n = b. The middle column reads, “find two numbers that add to 7. Both factors must be positive”. The numbers are 3 and 4. 3 times 4 is 12 and 3 + 4 is 7.The next step is to split the middle term using m and n. That is, to write 7 x as 3 x + 4 x. Therefore, 6 x ^ 2 + 7 x + 2 is rewritten as 6 x ^ 2 +3 x + 4 x + 2.The next step is to factor by grouping. 3 x(2 x + 1) + 2(2 x + 1) then factor again (2 x + 1)(3 x + 2).The last step is to check by multiplying. Multiply the factors (2 x + 1)(3 x + 2) to get 6 x ^ 2 + 7 x + 2.

    Exercise \(\PageIndex{29}\)

    Factor: \(6 x^{2}+13 x+2\)

    Answer

    \((x+2)(6 x+1)\)

    Exercise \(\PageIndex{30}\)

    Factor: \(4 y^{2}+8 y+3\)

    Answer

    \((2 y+1)(2 y+3)\)

    FACTOR TRINOMIALS OF THE FORM USING THE “AC” METHOD.

    1. Factor any GCF.
    2. Find the product ac.
    3. Find two numbers m and n that:
      \(\begin{array}{ll}{\text { Multiply to } a c} & {m \cdot n=a \cdot c} \\ {\text { Add to } b} & {m+n=b}\end{array}\)
    4. Split the middle term using m and n: This figure shows two equations. The top equation reads a times x squared plus b times x plus c. Under this, is the equation a times x squared plus m times x plus n times x plus c. Above the m times x plus n times x is a bracket with b times x above it.
    5. Factor by grouping.
    6. Check by multiplying the factors.

    When the third term of the trinomial is negative, the factors of the third term will have opposite signs.

    Exercise \(\PageIndex{31}\)

    Factor: \(8 u^{2}-17 u-21\)

    Answer
    Is there a greatest common factor? No. .
    Find \(a\cdot c\) \(a\cdot c\)
    8(−21)
    −168

    Find two numbers that multiply to −168 and add to −17. The larger factor must be negative.

    Factors of −168 Sum of factors
    1,−168 1+(−168)=−167
    2,−84 2+(−84)=−82
    3,−56 3+(−56)=−53
    4,−42 4+(−42)=−38
    6,−28 6+(−28)=−22
    7,−24 7+(−24)=−17*
    8,−21 8+(−21)=−13

    \(\begin{array}{lc}\text { Split the middle term using } 7 u \text { and }-24 u &8 u^{2}-17 u-21 \\ & \qquad\space \swarrow\searrow \\ & \underbrace{8 u^{2}+7 u} \underbrace{-24 u-21} \\ \text { Factor by grouping. } & u(8 u+7)-3(8 u+7) \\ & (8 u+7)(u-3) \\ \text { Check by multiplying. } & \begin{array}{l}{(8 u+7)(u-3)} \\ {8 u^{2}-24 u+7 u-21} \\ {8 u^{2}-17 u-21} \checkmark \end{array} \end{array}\)

    Exercise \(\PageIndex{32}\)

    Factor: \(20 h^{2}+13 h-15\)

    Answer

    \((4 h-5)(5 h+3)\)

    Exercise \(\PageIndex{33}\)

    Factor: \(6 g^{2}+19 g-20\)

    Answer

    \((q+4)(6 q-5)\)

    Exercise \(\PageIndex{34}\)

    Factor: \(2 x^{2}+6 x+5\)

    Answer
    Is there a greatest common factor? No. .
    Find a⋅c a⋅c
    2(5)
    10

    Find two numbers that multiply to 10 and add to 6.

    Factors of 10 Sum of factors
    1,10 1+10=11
    2, 5 2+5=7

    There are no factors that multiply to 10 and add to 6. The polynomial is prime.

    Exercise \(\PageIndex{35}\)

    Factor: \(10 t^{2}+19 t-15\)

    Answer

    \((2 t+5)(5 t-3)\)

    Exercise \(\PageIndex{36}\)

    Factor: \(3 u^{2}+8 u+5\)

    Answer

    \((u+1)(3 u+5)\)

    Don’t forget to look for a common factor!

    Exercise \(\PageIndex{37}\)

    Factor: \(10 y^{2}-55 y+70\)

    Answer
    Is there a greatest common factor? Yes. The GCF is 5. .
    Factor it. Be careful to keep the factor of 5 all the way through the solution! .
    The trinomial inside the parentheses has a leading coefficient that is not 1. .
    Factor the trinomial. .
    Check by multiplying all three factors.
    5\(\left(2 y^{2}-2 y-4 y+14\right)\)
    5\(\left(2 y^{2}-11 y+14\right)\)
    \(10 y^{2}-55 y+70\) ✓

    Exercise \(\PageIndex{38}\)

    Factor: \(16 x^{2}-32 x+12\)

    Answer

    4\((2 x-3)(2 x-1)\)

    Exercise \(\PageIndex{39}\)

    Factor: \(18 w^{2}-39 w+18\)

    Answer

    3\((3 w-2)(2 w-3)\)

    We can now update the Preliminary Factoring Strategy, as shown in Figure \(\PageIndex{1}\) and detailed in Choose a strategy to factor polynomials completely (updated), to include trinomials of the form \(a x^{2}+b x+c\). Remember, some polynomials are prime and so they cannot be factored.

    This figure has the strategy for factoring polynomials. At the top of the figure is GCF. Below this, there are three options. The first is binomial. The second is trinomial. Under trinomial there are x squared + b x + c and a x squared + b x +c. The two methods here are trial and error and the “a c” method. The third option is for more than three terms. It is grouping.
    Figure \(\PageIndex{1}\)

    CHOOSE A STRATEGY TO FACTOR POLYNOMIALS COMPLETELY (UPDATED).

    1. Is there a greatest common factor?
      • Factor it.
    2. Is the polynomial a binomial, trinomial, or are there more than three terms?
      • If it is a binomial, right now we have no method to factor it.
      • If it is a trinomial of the form \(x^{2}+b x+c\)
        Undo FOIL \\((x\qquad)(x\qquad)\).
      • If it is a trinomial of the form \(a x^{2}+b x+c\)
        Use Trial and Error or the “ac” method.
      • If it has more than three terms
        Use the grouping method.
    3. Check by multiplying the factors.

    Note

    Access these online resources for additional instruction and practice with factoring trinomials of the form \(a x^{2}+b x+c\)

    Key Concepts

    • Factor Trinomials of the Form \(a x^{2}+b x+c\) using Trial and Error: See Example.
      1. Write the trinomial in descending order of degrees.
      2. Find all the factor pairs of the first term.
      3. Find all the factor pairs of the third term.
      4. Test all the possible combinations of the factors until the correct product is found.
      5. Check by multiplying.
    • Factor Trinomials of the Form \(a x^{2}+b x+c\) Using the “ac” Method: See Example.
      1. Factor any GCF.
      2. Find the product ac.
      3. Find two numbers m and n that: \(\begin{array}{ll}{\text { Multiply to } a c} & {m \cdot n=a \cdot c} \\ {\text { Add to } b} & {m+n=b}\end{array}\)
      4. Split the middle term using m and n:
        This figure shows two equations. The top equation reads a times x squared plus b times x plus c. Under this, is the equation a times x squared plus m times x plus n times x plus c. Above the m times x plus n times x is a bracket with b times x above it.
      5. Factor by grouping.
      6. Check by multiplying the factors.
    • Choose a strategy to factor polynomials completely (updated):
      1. Is there a greatest common factor? Factor it.
      2. Is the polynomial a binomial, trinomial, or are there more than three terms?
        If it is a binomial, right now we have no method to factor it.
        If it is a trinomial of the form \(x^2+bx+c\)
        Undo FOIL \((x\qquad)(x\qquad)\).
        If it is a trinomial of the form \(ax^2+bx+c\)
        Use Trial and Error or the “ac” method.
        If it has more than three terms
        Use the grouping method.
      3. Check by multiplying the factors.

    Glossary

    prime polynomials
    Polynomials that cannot be factored are prime polynomials.