Contents

Factor Quadratic Trinomials with Leading Coefficient Other than 1

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 $a x^{2} + b x + c a x^{2} + b x + c$ with a GCF
Factor trinomials using trial and error
Factor trinomials using the ‘ac’ method

Be Prepared 7.3

Before you get started, take this readiness quiz.

Find the GCF of $45 p^{2} and 30 p^{6} 45 p^{2}$ .
If you missed this problem, review Example 7.2.
Multiply $(3 y + 4) (2 y + 5) (3 y + 4) (2 y + 5)$ .
If you missed this problem, review Example 6.40.
Combine like terms $12 x^{2} + 3 x + 5 x + 9 12 x^{2} + 3 x + 5 x + 9$ .
If you missed this problem, review Example 1.24.

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, $x^{2} + b x + c x^{2} + b x + c$ , 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. These are detailed in Choose a strategy to factor polynomials completely.

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.

How To

Choose a strategy to factor polynomials completely.

Step 1.
Is there a greatest common factor?
- Factor it out.
Step 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 $x^{2} + b x + c x^{2} + b x + c$ : Undo FOIL $(x) (x) (x) (x)$
- If it has more than three terms: Use the grouping method.
Step 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.

Example 7.29

Identify the best method to use to factor each polynomial.

ⓐ $6 y^{2} - 72 6 y^{2} - 72$
ⓑ $r^{2} - 10 r - 24 r^{2} - 10 r - 24$
ⓒ $p^{2} + 5 p + p q + 5 q p^{2} + 5 p + p q + 5 q$

Show solution

Solution

ⓐ
$\begin{matrix} 6 y^{2} - 72 \\ Is there a greatest common factor? & Yes, 6. \\ Factor out the 6. & 6 (y^{2} - 12) \\ Is it a binomial, trinomial, or are there & Binomial, we have no method to factor \\ more than 3 terms? & binomials yet. \end{matrix} \begin{matrix} 6 y^{2} - 72 \\ Is there a greatest common factor? & Yes, 6. \\ Factor out the 6. & 6 (y^{2} - 12) \\ Is it a binomial, trinomial, or are there & Binomial, we have no method to factor \\ more than 3 terms? & binomials yet. \end{matrix}$
ⓑ
$\begin{matrix} r^{2} - 10 r - 24 \\ Is there a greatest common factor? & No, there is no common factor. \\ Is it a binomial, trinomial, or are there & Trinomial, with leading coefficient 1, so \\ more than three terms? & “undo” FOIL. \end{matrix} \begin{matrix} r^{2} - 10 r - 24 \\ Is there a greatest common factor? & No, there is no common factor. \\ Is it a binomial, trinomial, or are there & Trinomial, with leading coefficient 1, so \\ more than three terms? & “undo” FOIL. \end{matrix}$
ⓒ
$\begin{matrix} p^{2} + 5 p + p q + 5 q \\ Is there a greatest common factor? & No, there is no common factor. \\ Is it a binomial, trinomial, or are there & More than three terms, so factor using \\ more than three terms? & grouping. \end{matrix} \begin{matrix} p^{2} + 5 p + p q + 5 q \\ Is there a greatest common factor? & No, there is no common factor. \\ Is it a binomial, trinomial, or are there & More than three terms, so factor using \\ more than three terms? & grouping. \end{matrix}$

Try It 7.57

Identify the best method to use to factor each polynomial:

ⓐ $4 y^{2} + 32 4 y^{2} + 32$
ⓑ $y^{2} + 10 y + 21 y^{2} + 10 y + 21$
ⓒ $y z + 2 y + 3 z + 6 y z + 2 y + 3 z + 6$

Try It 7.58

Identify the best method to use to factor each polynomial:

ⓐ $a b + a + 4 b + 4 a b + a + 4 b + 4$
ⓑ $3 k^{2} + 15 3 k^{2} + 15$
ⓒ $p^{2} + 9 p + 8 p^{2} + 9 p + 8$

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 $a x^{2} + b x + c a x^{2} + b x + c$ .

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.

Example 7.30

Factor completely: $2 n^{2} - 8 n - 42 2 n^{2} - 8 n - 42$ .

Show solution

Solution

Use the preliminary strategy.

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

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

Check.

$2 (n + 3) (n - 7)$

$2 (n^{2} - 7 n + 3 n - 21)$

$2 (n^{2} - 4 n - 21)$

$2 n^{2} - 8 n - 42 ✓$

Try It 7.59

Factor completely: $4 m^{2} - 4 m - 8 4 m^{2} - 4 m - 8$ .

Try It 7.60

Factor completely: $5 k^{2} - 15 k - 50 5 k^{2} - 15 k - 50$ .

Example 7.31

Factor completely: $4 y^{2} - 36 y + 56 4 y^{2} - 36 y + 56$ .

Show solution

Solution

Use the preliminary strategy.

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

Factors of $1414$	Sum of factors
$−1, −14 −1, −14$	$−1 + (−14) = −15 −1 + (−14) = −15$
$−2, −7 −2, −7$	$−2 + (−7) = −9 * −2 + (−7) = −9 *$

Check.

$4 (y - 2) (y - 7)$

$4 (y^{2} - 7 y - 2 y + 14)$

$4 (y^{2} - 9 y + 14)$

$4 y^{2} - 36 y + 42 ✓$

Try It 7.61

Factor completely: $3 r^{2} - 9 r + 6 3 r^{2} - 9 r + 6$ .

Try It 7.62

Factor completely: $2 t^{2} - 10 t + 12 2 t^{2} - 10 t + 12$ .

In the next example the GCF will include a variable.

Example 7.32

Factor completely: $4 u^{3} + 16 u^{2} - 20 u 4 u^{3} + 16 u^{2} - 20 u$ .

Show solution

Solution

Use the preliminary strategy.

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

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

Check.

$4 u (u - 1) (u + 5)$

$4 u (u^{2} + 5 u - u - 5)$

$4 u (u^{2} + 4 u - 5)$

$4 u^{3} + 16 u^{2} - 20 u ✓$

Try It 7.63

Factor completely: $5 x^{3} + 15 x^{2} - 20 x 5 x^{3} + 15 x^{2} - 20 x$ .

Try It 7.64

Factor completely: $6 y^{3} + 18 y^{2} - 60 y 6 y^{3} + 18 y^{2} - 60 y$ .

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 $3 x^{2} + 5 x + 2 3 x^{2} + 5 x + 2$ .

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

\begin{matrix} 3 x^{2} + 5 x + 2 \\ () () \end{matrix} \begin{matrix} 3 x^{2} + 5 x + 2 \\ () () \end{matrix}

We know the first terms of the binomial factors will multiply to give us $3 x^{2} 3 x^{2}$ . The only factors of $3 x^{2} 3 x^{2}$ are $1 x, 3 x 1 x, 3 x$ . 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 $1 x \cdot 3 x = 3 x^{2} 1 x \cdot 3 x = 3 x^{2}$ ?

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.

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

\begin{matrix} (x + 1) (3 x + 2) \\ 3 x^{2} + 2 x + 3 x + 2 \\ 3 x^{2} + 5 x + 2 ✓ \end{matrix} \begin{matrix} (x + 1) (3 x + 2) \\ 3 x^{2} + 2 x + 3 x + 2 \\ 3 x^{2} + 5 x + 2 \end{matrix}

Our result of the factoring is:

\begin{matrix} 3 x^{2} + 5 x + 2 \\ (x + 1) (3 x + 2) \end{matrix} \begin{matrix} 3 x^{2} + 5 x + 2 & (x + 1) (3 x + 2) \end{matrix}

Example 7.33

How to Factor Trinomials of the Form $a x^{2} + b x + c a x^{2} + b x + c$ Using Trial and Error

Factor completely: $3 y^{2} + 22 y + 7 3 y^{2} + 22 y + 7$ .

Show solution

Solution

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.

Try It 7.65

Factor completely: $2 a^{2} + 5 a + 3 2 a^{2} + 5 a + 3$ .

Try It 7.66

Factor completely: $4 b^{2} + 5 b + 1 4 b^{2} + 5 b + 1$ .

How To

Factor trinomials of the form $a x^{2} + b x + c a x^{2} + b x + c$ using trial and error.

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

Example 7.34

Factor completely: $6 b^{2} - 13 b + 5 6 b^{2} - 13 b + 5$ .

Show solution

Solution

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.

$6 b^{2} - 13 b + 5 6 b^{2} - 13 b + 5$
Possible factors	Product
$(b - 1) (6 b - 5) (b - 1) (6 b - 5)$	$6 b^{2} - 11 b + 5 6 b^{2} - 11 b + 5$
$(b - 5) (6 b - 1) (b - 5) (6 b - 1)$	$6 b^{2} - 31 b + 5 6 b^{2} - 31 b + 5$
$(2 b - 1) (3 b - 5) (2 b - 1) (3 b - 5)$	$6 b^{2} - 13 b + 5 6 b^{2} - 13 b + 5$ *
$(2 b - 5) (3 b - 1) (2 b - 5) (3 b - 1)$	$6 b^{2} - 17 b + 5 6 b^{2} - 17 b + 5$

$\begin{matrix} The correct factors are those whose product \\ is the original trinomial. & (2 b - 1) (3 b - 5) \\ Check by multiplying. \\ (2 b - 1) (3 b - 5) \\ 6 b^{2} - 10 b - 3 b + 5 \\ 6 b^{2} - 13 b + 5 ✓ \end{matrix} \begin{matrix} The correct factors are those whose product \\ is the original trinomial. \\ Check by multiplying. \end{matrix}$

Try It 7.67

Factor completely: $8 x^{2} - 13 x + 3 8 x^{2} - 13 x + 3$ .

Try It 7.68

Factor completely: $10 y^{2} - 37 y + 7 10 y^{2} - 37 y + 7$ .

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.

Example 7.35

Factor completely: $14 x^{2} - 47 x - 7 14 x^{2} - 47 x - 7$ .

Show solution

Solution

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 $14 x^{2} 14 x^{2}$ with each pair of factors of $−7 . −7 .$

Factors of $14 x^{2} 14 x^{2}$	Pair with	Factors of $−7 −7$
$x x$ , $14 x 14 x$		$11$ , $−7 −7$ $−7 −7$ , $11$ (reverse order)
$x x$ , $14 x 14 x$		$−1 −1$ , $77$ $77$ , $−1 −1$ (reverse order)
$2 x, 7 x 2 x, 7 x$		$11$ , $−7 −7$ $−7 −7$ , $11$ (reverse order)
$2 x, 7 x 2 x, 7 x$		$−1 −1$ , $77$ $77$ , $−1 −1$ (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.”

$\begin{matrix} The correct factors are those whose product is the & (2 x - 7) (7 x + 1) \\ original trinomial. \\ Check by multiplying. \\ (2 x - 7) (7 x + 1) \\ 14 x^{2} + 2 x - 49 x - 7 \\ 14 x^{2} - 47 x - 7 ✓ \end{matrix} \begin{matrix} The correct factors are those whose product is the \\ original trinomial. \\ Check by multiplying. \end{matrix}$

Try It 7.69

Factor completely: $8 a^{2} - 3 a - 5 8 a^{2} - 3 a - 5$ .

Try It 7.70

Factor completely: $6 b^{2} - b - 15 6 b^{2} - b - 15$ .

Example 7.36

Factor completely: $18 n^{2} - 37 n + 15 18 n^{2} - 37 n + 15$ .

Show solution

Solution

The trinomial is already in descending order.	$18 n^{2} - 37 n + 15 18 n^{2} - 37 n + 15$
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 facotrs.

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.

$\begin{matrix} The correct factors are those whose product is & (2 n - 3) (9 n - 5) \\ the original trinomial. \\ Check by multiplying. \\ (2 n - 3) (9 n - 5) \\ 18 n^{2} - 10 n - 27 n + 15 \\ 18 n^{2} - 37 n + 15 ✓ \end{matrix} \begin{matrix} The correct factors are those whose product is \\ the original trinomial. \\ Check by multiplying. \end{matrix}$

Try It 7.71

Factor completely: $18 x^{2} - 3 x - 10 18 x^{2} - 3 x - 10$ .

Try It 7.72

Factor completely: $30 y^{2} - 53 y - 21 30 y^{2} - 53 y - 21$ .

Don’t forget to look for a GCF first.

Example 7.37

Factor completely: $10 y^{4} + 55 y^{3} + 60 y^{2} 10 y^{4} + 55 y^{3} + 60 y^{2}$ .

Show solution

Solution

	$10 y^{4} + 55 y^{3} + 60 y^{2} 10 y^{4} + 55 y^{3} + 60 y^{2}$
Notice the greatest common factor, and factor it first.	$15 y^{2} (2 y^{2} + 11 y + 12) 15 y^{2} (2 y^{2} + 11 y + 12)$
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.

$\begin{matrix} The correct factors are those whose product & 5 y^{2} (y + 4) (2 y + 3) \\ is the original trinomial. Remember to include \\ the factor 5 y^{2} . \\ Check by multiplying. \\ 5 y^{2} (y + 4) (2 y + 3) \\ 5 y^{2} (2 y^{2} + 8 y + 3 y + 12) \\ 10 y^{4} + 55 y^{3} + 60 y^{2} ✓ \end{matrix} \begin{matrix} The correct factors are those whose product \\ is the original trinomial. Remember to include \\ the factor \\ Check by multiplying. \end{matrix}$

Try It 7.73

Factor completely: $15 n^{3} - 85 n^{2} + 100 n 15 n^{3} - 85 n^{2} + 100 n$ .

Try It 7.74

Factor completely: $56 q^{3} + 320 q^{2} - 96 q 56 q^{3} + 320 q^{2} - 96 q$ .

Factor Trinomials using the “ac” Method

Another way to factor trinomials of the form $a x^{2} + b x + c a x^{2} + b x + c$ 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!

Example 7.38

How to Factor Trinomials Using the “ac” Method

Factor: $6 x^{2} + 7 x + 2 6 x^{2} + 7 x + 2$ .

Show solution

Solution

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.

Try It 7.75

Factor: $6 x^{2} + 13 x + 2 6 x^{2} + 13 x + 2$ .

Try It 7.76

Factor: $4 y^{2} + 8 y + 3 4 y^{2} + 8 y + 3$ .

How To

Factor trinomials of the form using the “ac” method.

Step 1. Factor any GCF.
Step 2. Find the product ac.
Step 3. Find two numbers m and n that:
$\begin{matrix} Multiply to a c & m \cdot n = a \cdot c \\ Add to b & m + n = b \end{matrix} \begin{matrix} Multiply to & m \cdot n = a \cdot c \\ Add to & m + n = b \end{matrix}$
Step 4. Split the middle term using m and n:
Step 5. Factor by grouping.
Step 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.

Example 7.39

Factor: $8 u^{2} - 17 u - 21 8 u^{2} - 17 u - 21$ .

Show solution

Solution

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

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

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

$\begin{matrix} Split the middle term using 7 u and −24 u . & 8 u^{2} - \underset{↙}{17} \underset{↘}{u} - 21 \\ \underset{⎵}{8 u^{2} + 7 u} \underset{⎵}{−24 u - 21} \\ Factor by grouping. & u (8 u + 7) - 3 (8 u + 7) \\ (8 u + 7) (u - 3) \\ Check by multiplying. \\ (8 u + 7) (u - 3) \\ 8 u^{2} - 24 u + 7 u - 21 \\ 8 u^{2} - 17 u - 21 ✓ \end{matrix} \begin{matrix} Split the middle term using \\ Factor by grouping. & Check by multiplying. \end{matrix}$

Try It 7.77

Factor: $20 h^{2} + 13 h - 15 20 h^{2} + 13 h - 15$ .

Try It 7.78

Factor: $6 g^{2} + 19 g - 20 6 g^{2} + 19 g - 20$ .

Example 7.40

Factor: $2 x^{2} + 6 x + 5 2 x^{2} + 6 x + 5$ .

Show solution

Solution

Is there a greatest common factor? No.
Find $a \cdot c a \cdot c$ .	$a \cdot c a \cdot c$
	$2 (5) 2 (5)$
	$1010$

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

Factors of $1010$	Sum of factors
$1, 10 1, 10$	$1 + 10 = 11 1 + 10 = 11$
2, 5	$2 + 5 = 7 2 + 5 = 7$

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

Try It 7.79

Factor: $10 t^{2} + 19 t - 15 10 t^{2} + 19 t - 15$ .

Try It 7.80

Factor: $3 u^{2} + 8 u + 5 3 u^{2} + 8 u + 5$ .

Don’t forget to look for a common factor!

Example 7.41

Factor: $10 y^{2} - 55 y + 70 10 y^{2} - 55 y + 70$ .

Show solution

Solution

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 mulitplying all three factors.
$5 (2 y^{2} - 2 y - 4 y + 14) 5 (2 y^{2} - 2 y - 4 y + 14)$
$5 (2 y^{2} - 11 y + 14) 5 (2 y^{2} - 11 y + 14)$
$10 y^{2} - 55 y + 70 ✓ 10 y^{2} - 55 y + 70 ✓$

Try It 7.81

Factor: $16 x^{2} - 32 x + 12 16 x^{2} - 32 x + 12$ .

Try It 7.82

Factor: $18 w^{2} - 39 w + 18 18 w^{2} - 39 w + 18$ .

We can now update the Preliminary Factoring Strategy, as shown in Figure 7.2 and detailed in Choose a strategy to factor polynomials completely (updated), to include trinomials of the form $a x^{2} + b x + c a x^{2} + b x + c$ . 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 7.2

How To

Choose a strategy to factor polynomials completely (updated).

Step 1.
Is there a greatest common factor?
- Factor it.
Step 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 $x^{2} + b x + c x^{2} + b x + c$
  Undo FOIL $(x) (x) (x) (x)$ .
- If it is a trinomial of the form $a x^{2} + b x + c a x^{2} + b x + c$
  Use Trial and Error or the “ac” method.
- If it has more than three terms
  Use the grouping method.
Step 3. Check by multiplying the factors.

Media

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

Factoring Trinomials, a is not 1

Practice Makes Perfect

Recognize a Preliminary Strategy to Factor Polynomials Completely

In the following exercises, identify the best method to use to factor each polynomial.

135.

ⓐ $10 q^{2} + 50 10 q^{2} + 50$
ⓑ $a^{2} - 5 a - 14 a^{2} - 5 a - 14$
ⓒ $u v + 2 u + 3 v + 6 u v + 2 u + 3 v + 6$

136.

ⓐ $n^{2} + 10 n + 24 n^{2} + 10 n + 24$
ⓑ $8 u^{2} + 16 8 u^{2} + 16$
ⓒ $p q + 5 p + 2 q + 10 p q + 5 p + 2 q + 10$

137.

ⓐ $x^{2} + 4 x - 21 x^{2} + 4 x - 21$
ⓑ $a b + 10 b + 4 a + 40 a b + 10 b + 4 a + 40$
ⓒ $6 c^{2} + 24 6 c^{2} + 24$

138.

ⓐ $20 x^{2} + 100 20 x^{2} + 100$
ⓑ $u v + 6 u + 4 v + 24 u v + 6 u + 4 v + 24$
ⓒ $y^{2} - 8 y + 15 y^{2} - 8 y + 15$

Factor Trinomials of the form $a x^{2} + b x + c a x^{2} + b x + c$ with a GCF

In the following exercises, factor completely.

139.

$5 x^{2} + 35 x + 30 5 x^{2} + 35 x + 30$

140.

$12 s^{2} + 24 s + 12 12 s^{2} + 24 s + 12$

141.

$2 z^{2} - 2 z - 24 2 z^{2} - 2 z - 24$

142.

$3 u^{2} - 12 u - 36 3 u^{2} - 12 u - 36$

143.

$7 v^{2} - 63 v + 56 7 v^{2} - 63 v + 56$

144.

$5 w^{2} - 30 w + 45 5 w^{2} - 30 w + 45$

145.

$p^{3} - 8 p^{2} - 20 p p^{3} - 8 p^{2} - 20 p$

146.

$q^{3} - 5 q^{2} - 24 q q^{3} - 5 q^{2} - 24 q$

147.

$3 m^{3} - 21 m^{2} + 30 m 3 m^{3} - 21 m^{2} + 30 m$

148.

$11 n^{3} - 55 n^{2} + 44 n 11 n^{3} - 55 n^{2} + 44 n$

149.

$5 x^{4} + 10 x^{3} - 75 x^{2} 5 x^{4} + 10 x^{3} - 75 x^{2}$

150.

$6 y^{4} + 12 y^{3} - 48 y^{2} 6 y^{4} + 12 y^{3} - 48 y^{2}$

Factor Trinomials Using Trial and Error

In the following exercises, factor.

151.

$2 t^{2} + 7 t + 5 2 t^{2} + 7 t + 5$

152.

$5 y^{2} + 16 y + 11 5 y^{2} + 16 y + 11$

153.

$11 x^{2} + 34 x + 3 11 x^{2} + 34 x + 3$

154.

$7 b^{2} + 50 b + 7 7 b^{2} + 50 b + 7$

155.

$4 w^{2} - 5 w + 1 4 w^{2} - 5 w + 1$

156.

$5 x^{2} - 17 x + 6 5 x^{2} - 17 x + 6$

157.

$6 p^{2} - 19 p + 10 6 p^{2} - 19 p + 10$

158.

$21 m^{2} - 29 m + 10 21 m^{2} - 29 m + 10$

159.

$4 q^{2} - 7 q - 2 4 q^{2} - 7 q - 2$

160.

$10 y^{2} - 53 y - 11 10 y^{2} - 53 y - 11$

161.

$4 p^{2} + 17 p - 15 4 p^{2} + 17 p - 15$

162.

$6 u^{2} + 5 u - 14 6 u^{2} + 5 u - 14$

163.

$16 x^{2} - 32 x + 16 16 x^{2} - 32 x + 16$

164.

$81 a^{2} + 153 a - 18 81 a^{2} + 153 a - 18$

165.

$30 q^{3} + 140 q^{2} + 80 q 30 q^{3} + 140 q^{2} + 80 q$

166.

$5 y^{3} + 30 y^{2} - 35 y 5 y^{3} + 30 y^{2} - 35 y$

Factor Trinomials using the ‘ac’ Method

In the following exercises, factor.

167.

$5 n^{2} + 21 n + 4 5 n^{2} + 21 n + 4$

168.

$8 w^{2} + 25 w + 3 8 w^{2} + 25 w + 3$

169.

$9 z^{2} + 15 z + 4 9 z^{2} + 15 z + 4$

170.

$3 m^{2} + 26 m + 48 3 m^{2} + 26 m + 48$

171.

$4 k^{2} - 16 k + 15 4 k^{2} - 16 k + 15$

172.

$4 q^{2} - 9 q + 5 4 q^{2} - 9 q + 5$

173.

$5 s^{2} - 9 s + 4 5 s^{2} - 9 s + 4$

174.

$4 r^{2} - 20 r + 25 4 r^{2} - 20 r + 25$

175.

$6 y^{2} + y - 15 6 y^{2} + y - 15$

176.

$6 p^{2} + p - 22 6 p^{2} + p - 22$

177.

$2 n^{2} - 27 n - 45 2 n^{2} - 27 n - 45$

178.

$12 z^{2} - 41 z - 11 12 z^{2} - 41 z - 11$

179.

$3 x^{2} + 5 x + 4 3 x^{2} + 5 x + 4$

180.

$4 y^{2} + 15 y + 6 4 y^{2} + 15 y + 6$

181.

$60 y^{2} + 290 y - 50 60 y^{2} + 290 y - 50$

182.

$6 u^{2} - 46 u - 16 6 u^{2} - 46 u - 16$

183.

$48 z^{3} - 102 z^{2} - 45 z 48 z^{3} - 102 z^{2} - 45 z$

184.

$90 n^{3} + 42 n^{2} - 216 n 90 n^{3} + 42 n^{2} - 216 n$

185.

$16 s^{2} + 40 s + 24 16 s^{2} + 40 s + 24$

186.

$24 p^{2} + 160 p + 96 24 p^{2} + 160 p + 96$

187.

$48 y^{2} + 12 y - 36 48 y^{2} + 12 y - 36$

188.

$30 x^{2} + 105 x - 60 30 x^{2} + 105 x - 60$

Mixed Practice

In the following exercises, factor.

189.

$12 y^{2} - 29 y + 14 12 y^{2} - 29 y + 14$

190.

$12 x^{2} + 36 y - 24 z 12 x^{2} + 36 y - 24 z$

191.

$a^{2} - a - 20 a^{2} - a - 20$

192.

$m^{2} - m - 12 m^{2} - m - 12$

193.

$6 n^{2} + 5 n - 4 6 n^{2} + 5 n - 4$

194.

$12 y^{2} - 37 y + 21 12 y^{2} - 37 y + 21$

195.

$2 p^{2} + 4 p + 3 2 p^{2} + 4 p + 3$

196.

$3 q^{2} + 6 q + 2 3 q^{2} + 6 q + 2$

197.

$13 z^{2} + 39 z - 26 13 z^{2} + 39 z - 26$

198.

$5 r^{2} + 25 r + 30 5 r^{2} + 25 r + 30$

199.

$x^{2} + 3 x - 28 x^{2} + 3 x - 28$

200.

$6 u^{2} + 7 u - 5 6 u^{2} + 7 u - 5$

201.

$3 p^{2} + 21 p 3 p^{2} + 21 p$

202.

$7 x^{2} - 21 x 7 x^{2} - 21 x$

203.

$6 r^{2} + 30 r + 36 6 r^{2} + 30 r + 36$

204.

$18 m^{2} + 15 m + 3 18 m^{2} + 15 m + 3$

205.

$24 n^{2} + 20 n + 4 24 n^{2} + 20 n + 4$

206.

$4 a^{2} + 5 a + 2 4 a^{2} + 5 a + 2$

207.

$x^{2} + 2 x - 24 x^{2} + 2 x - 24$

208.

$2 b^{2} - 7 b + 4 2 b^{2} - 7 b + 4$

Everyday Math

209.

Height of a toy rocket The height of a toy rocket launched with an initial speed of 80 feet per second from the balcony of an apartment building is related to the number of seconds, t, since it is launched by the trinomial $−16 t^{2} + 80 t + 96 −16 t^{2} + 80 t + 96$ . Completely factor the trinomial.

210.

Height of a beach ball The height of a beach ball tossed up with an initial speed of 12 feet per second from a height of 4 feet is related to the number of seconds, t, since it is tossed by the trinomial $−16 t^{2} + 12 t + 4 −16 t^{2} + 12 t + 4$ . Completely factor the trinomial.

Writing Exercises

211.

List, in order, all the steps you take when using the “ac” method to factor a trinomial of the form $a x^{2} + b x + c . a x^{2} + b x + c .$

212.

How is the “ac” method similar to the “undo FOIL” method? How is it different?

213.

What are the questions, in order, that you ask yourself as you start to factor a polynomial? What do you need to do as a result of the answer to each question?

214.

On your paper draw the chart that summarizes the factoring strategy. Try to do it without looking at the book. When you are done, look back at the book to finish it or verify it.

Self Check

ⓐ After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.

This table has the following statements all to be preceded by “I can…”. The first row is “recognize a preliminary strategy to factor polynomials completely”. The second row is “factor trinomials of the form a x ^ 2 + b x + c with a GCF”. The third row is “factor trinomials using trial and error”. And the fourth row is “factor trinomials using the “ac” method”. In the columns beside these statements are the headers, “confidently”, “with some help”, and “no-I don’t get it!”.

ⓑ What does this checklist tell you about your mastery of this section? What steps will you take to improve?

Learning Objectives

Be Prepared 7.3

Recognize a Preliminary Strategy for Factoring

How To

Choose a strategy to factor polynomials completely.

Example 7.29

Solution

Try It 7.57

Try It 7.58

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

Example 7.30

Solution

Try It 7.59

Try It 7.60

Example 7.31

Solution

Try It 7.61

Try It 7.62

Example 7.32

Solution

Try It 7.63

Try It 7.64

Factor Trinomials using Trial and Error

Example 7.33

How to Factor Trinomials of the Form ax2+bx+cax2+bx+c Using Trial and Error

Solution

Try It 7.65

Try It 7.66

How To

Factor trinomials of the form ax2+bx+cax2+bx+c using trial and error.

Example 7.34

Solution

Try It 7.67

Try It 7.68

Example 7.35

Solution

Try It 7.69

Try It 7.70

Example 7.36

Solution

Try It 7.71

Try It 7.72

Example 7.37

Solution

Try It 7.73

Try It 7.74

Factor Trinomials using the “ac” Method

Example 7.38

How to Factor Trinomials Using the “ac” Method

Solution

Try It 7.75

Try It 7.76

How To

Factor trinomials of the form using the “ac” method.

Example 7.39

Solution

Try It 7.77

Try It 7.78

Example 7.40

Solution

Try It 7.79

Try It 7.80

Example 7.41

Solution

Try It 7.81

Try It 7.82

How To

Choose a strategy to factor polynomials completely (updated).

Media

Practice Makes Perfect

Everyday Math

Writing Exercises

Self Check

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

How to Factor Trinomials of the Form $a x^{2} + b x + c a x^{2} + b x + c$ Using Trial and Error

Factor trinomials of the form $a x^{2} + b x + c a x^{2} + b x + c$ using trial and error.