7.3 Factor Trinomials of the Form ax²+bx+c

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. 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.

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 1

Identify the best method to use to factor each polynomial.

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

ⓐ

	$6 y squared minus 72$
Is there a greatest common factor?	Yes, 6.
Factor out the 6.	$6 open parenthesis y squared minus 12 close parenthesis$
Is it a binomial, trinomial, or are there more than 3 terms?	Binomial, we have no method to factor binomials yet.

ⓑ

	$r^{2} - 10 r - 24$
Is there a greatest common factor?	No, there is no common factor.
Is it a binomial, trinomial, or are there more than three terms?	Trinomial, with leading coefficient 1, so “undo” FOIL.

ⓒ

	$p^{2} + 5 p + p q + 5 q$
Is there a greatest common factor?	No, there is no common factor.
Is it a binomial, trinomial, or are there more than three terms?	More than three terms, so factor using grouping.

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.

Example 2

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

Use the preliminary strategy.

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.	$2 open parenthesis n close parenthesis open parenthesis n close parenthesis$
Use 3 and −7 as the last terms of the binomials.	$2 (n + 3) (n - 7)$

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

Check.

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

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

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

$2 n^{2} - 8 n - 42 ✓$

Example 3

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

Use the preliminary strategy.

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.	$4 open parenthesis y close parenthesis open parenthesis y close parenthesis$
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.	$4 (y - 2) (y - 7)$

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

Check.

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

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

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

$4 y^{2} - 36 y + 56 ✓$

In the next example the GCF will include a variable.

Example 4

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

Use the preliminary strategy.

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

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

Check.

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

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

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

$4 u^{3} + 16 u^{2} - 20 u ✓$

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.

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

We know the first terms of the binomial factors will multiply to give us $3 x squared$ . The only factors of $3 x squared$ 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.

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

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

Our result of the factoring is:

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

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

Example 6

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

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

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

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

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

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 $14 x squared$ with each pair of factors of $−7.$

Factors of $14 x squared$	Pair with	Factors of $−7$
$x$ , $14 x$		$1$ , $−7$ $−7$ , $1$ (reverse order)
$x$ , $14 x$		$−1$ , $7$ $7$ , $−1$ (reverse order)
$2 x, 7 x$		$1$ , $−7$ $−7$ , $1$ (reverse order)
$2 x, 7 x$		$−1$ , $7$ $7$ , $−1$ (reverse order)

These pairings lead to the following eight combinations.

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

Example 8

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

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 facotrs.

Consider all the combinations of factors.

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

Don’t forget to look for a GCF first.

Example 9

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

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

Consider all the combinations.

The correct factors are those whose product is the original trinomial. Remember to include the factor 5y².	$5 y^{2} (y + 4) (2 y + 3)$
Check by multiplying. $\begin{matrix} 5 y^{2} (y + 4) (2 y + 3) \\ 5 y^{2} (2 y^{2} + 8 y + 3 y + 12) \\ 10 y^{4} + 55 y^{3} + 60 y^{2} ✓ \end{matrix}$

Factor Trinomials using the “ac” Method

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

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

Example 11

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

Is there a greatest common factor? No.
Find $a \cdot c$ .	$a \cdot c$
	$8 open parenthesis −21 close parenthesis$
	$−168$

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

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

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

Example 12

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

Is there a greatest common factor? No.
Find $a \cdot c$ .	$a \cdot c$
	$2 open parenthesis 5 close parenthesis$
	$10$

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.

Don’t forget to look for a common factor!

Example 13

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

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

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

Key Concepts

Factor Trinomials of the Form $a x^{2} + b x + c$ using Trial and Error: See Example 5.
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 10.
1. Factor any GCF.
2. Find the product ac.
3. Find two numbers m and n that:
  $\begin{matrix} Multiply to a c & m \cdot n = a \cdot c \\ Add to b & m + n = b \end{matrix}$
4. Split the middle term using m and n:
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 $x^{2} + b x + c$
     Undo FOIL $(x) (x)$ .
  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.

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.

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

ⓐ factor the GCF, binomial ⓑ Undo FOIL ⓒ factor by grouping

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

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

ⓐ undo FOIL ⓑ factor by grouping ⓒ factor the GCF, binomial

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

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

In the following exercises, factor completely.

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

$5 (x + 1) (x + 6)$

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

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

$2 (z - 4) (z + 3)$

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

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

$7 (v - 1) (v - 8)$

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

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

$p (p - 10) (p + 2)$

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

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

$3 m (m - 5) (m - 2)$

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

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

$5 x^{2} (x - 3) (x + 5)$

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

Factor Trinomials Using Trial and Error

In the following exercises, factor.

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

$(2 t + 5) (t + 1)$

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

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

$(11 x + 1) (x + 3)$

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

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

$(4 w - 1) (w - 1)$

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

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

$(3 p - 2) (2 p - 5)$

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

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

$(4 q + 1) (q - 2)$

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

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

$(4 p - 3) (p + 5)$

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

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

$16 (x - 1) (x - 1)$

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

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

$10 q (3 q + 2) (q + 4)$

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

Factor Trinomials using the ‘ac’ Method

In the following exercises, factor.

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

$(5 n + 1) (n + 4)$

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

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

$(3 z + 1) (3 z + 4)$

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

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

$(2 k - 3) (2 k - 5)$

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

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

$(5 s - 4) (s - 1)$

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

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

$(3 y + 5) (2 y - 3)$

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

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

$(2 n + 3) (n - 15)$

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

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

prime

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

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

$10 (6 y - 1) (y + 5)$

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

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

$3 z (8 z + 3) (2 z - 5)$

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

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

$8 (2 s + 3) (s + 1)$

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

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

$12 (4 y - 3) (y + 1)$

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

Mixed Practice

In the following exercises, factor.

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

$(4 y - 7) (3 y - 2)$

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

$a^{2} - a - 20$

$(a - 5) (a + 4)$

$m^{2} - m - 12$

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

$(2 n - 1) (3 n + 4)$

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

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

prime

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

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

$13 (z^{2} + 3 z - 2)$

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

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

$(x + 7) (x - 4)$

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

$3 p squared plus 21 p$

$3 p open parenthesis p plus 7 close parenthesis$

$7 x squared minus 21 x$

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

$6 (r + 2) (r + 3)$

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

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

$4 (2 n + 1) (3 n + 1)$

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

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

$(x + 6) (x - 4)$

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

Everyday Math

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 $−16 t^{2} + 80 t + 96$ . Completely factor the trinomial.

$−16 (t - 6) (t + 1)$

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 $−16 t^{2} + 12 t + 4$ . Completely factor the trinomial.

Writing Exercises

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

Answers may vary.

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

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?

Answers may vary.

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?

7.3 Factor Trinomials of the Form ax2+bx+c

Recognize a Preliminary Strategy for Factoring

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

Factor Trinomials using Trial and Error

Factor Trinomials using the “ac” Method

Key Concepts

Practice Makes Perfect

Everyday Math

Writing Exercises

Self Check

7.3 Factor Trinomials of the Form ax²+bx+c

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