ORCCA Factoring Trinomials with Leading Coefficient One

Section 10.3 Factoring Trinomials with Leading Coefficient One

Objectives: PCC Course Content and Outcome Guide

[cross-reference to target(s) "ccog-factor-trinomials-when-a-is-one" missing or not unique]

In Chapter 5, we learned how to multiply binomials like

(x + 2) (x + 3)

and obtain the trinomial

x^{2} + 5 x + 6 .

In this section, we will learn how to undo that. So we’ll be starting with a trinomial like

x^{2} + 5 x + 6

and obtaining its factored form

(x + 2) (x + 3) .

The trinomials that we’ll factor in this section all have leading coefficient

1,

but Section 4 will cover some more general trinomials.

🔗

Figure 10.3.1. Alternative Video Lesson

🔗

Subsection 10.3.1 Factoring Trinomials by Listing Factor Pairs

🔗

Consider the example

x^{2} + 5 x + 6 = (x + 2) (x + 3) .

There are at least three things that are important to notice:

The leading coefficient of $x^{2} + 5 x + 6$ is $1 .$
The two factors on the right use the numbers $2$ and $3,$ and when you multiply these you get the $6 .$
The two factors on the right use the numbers $2$ and $3,$ and when you add these you get the $5 .$

🔗

So the idea is that if you need to factor

x^{2} + 5 x + 6

and you somehow discover that

2

and

3

are special numbers (because

2 \cdot 3 = 6

and

2 + 3 = 5

), then you can conclude that

(x + 2) (x + 3)

is the factored form of the given polynomial.

🔗

Example 10.3.2.

🔗

Factor

x^{2} + 13 x + 40 .

Since the leading coefficient is

1,

we are looking to write this polynomial as

(x + ?) (x + ?)

where the question marks are two possibly different, possibly negative, numbers. We need these two numbers to multiply to

40

and add to

13 .

How can you track these two numbers down? Since the numbers need to multiply to

40,

one method is to list all factor pairs of

40

in a table just to see what your options are. We’ll write every pair of factors that multiply to

40 .

1 \cdot 40

2 \cdot 20

4 \cdot 10

5 \cdot 8

- 1 \cdot (- 40)

- 2 \cdot (- 20)

- 4 \cdot (- 10)

- 5 \cdot (- 8)

🔗

We wanted to find all factor pairs. To avoid missing any, we started using

1

as a factor, and then slowly increased that first factor. The table skips over using

3

as a factor, because

3

is not a factor of

40 .

Similarly the table skips using

6

and

7

as a factor. And there would be no need to continue with

8

and beyond, because we already found “large” factors like

8

as the partners of “small” factors like

5 .

🔗

There is an entire second column where the signs are reversed, since these are also ways to multiply two numbers to get

40 .

In the end, there are eight factor pairs.

🔗

We need a pair of numbers that also adds to

13 .

So we check what each of our factor pairs add up to:

Factor Pair	Sum of the Pair
$1 \cdot 40$	$41$
$2 \cdot 20$	$22$
$4 \cdot 10$	$14$
$5 \cdot 8$	$13$ (what we wanted)

Factor Pair	Sum of the Pair
$- 1 \cdot (- 40)$	(no need to go this far)
$- 2 \cdot (- 20)$	(no need to go this far)
$- 4 \cdot (- 10)$	(no need to go this far)
$- 5 \cdot (- 8)$	(no need to go this far)

🔗

The winning pair of numbers is

5

and

8 .

Again, what matters is that

5 \cdot 8 = 40,

and

5 + 8 = 13 .

So we can conclude that

x^{2} + 13 x + 40 = (x + 5) (x + 8) .

🔗

To ensure that we made no mistakes, here are some possible checks.

🔗

Multiply it Out.

🔗

Multiplying out our answer

(x + 5) (x + 8)

should give us

x^{2} + 13 x + 40 .

🔗

\begin{aligned} (x + 5) (x + 8) & = (x + 5) \cdot x + (x + 5) \cdot 8 \\ = x^{2} + 5 x + 8 x + 40 \\ \overset{✓}{=} x^{2} + 13 x + 40 \end{aligned}

🔗

We could also use a rectangular area diagram to verify the factorization is correct:

	$x$	$5$
$x$	$x^{2}$	$5 x$
$8$	$8 x$	$40$

🔗

Evaluating.

🔗

If the answer really is

(x + 5) (x + 8),

then notice how evaluating at

- 5

would result in

0 .

So the original expression should also result in

0

if we evaluate at

- 5 .

And similarly, if we evaluate it at

- 8,

x^{2} + 13 x + 40

should be

0 .

\begin{aligned} (- 5)^{2} + 13 (- 5) + 40 & \overset{?}{=} 0 & (- 8)^{2} + 13 (- 8) + 40 & \overset{?}{=} 0 \\ 25 - 65 + 40 & \overset{?}{=} 0 & 64 - 104 + 40 & \overset{?}{=} 0 \\ 0 & \overset{✓}{=} 0 & 0 & \overset{✓}{=} 0 . \end{aligned}

🔗

This also gives us evidence that the factoring was correct.

🔗

Example 10.3.3.

🔗

Factor

y^{2} - 11 y + 24 .

The negative coefficient is a small complication from Example 2, but the process is actually still the same.

Explanation.

We need a pair of numbers that multiply to

24

and add to

- 11 .

Note that we do care to keep track that they sum to a negative total.

Factor Pair	Sum of the Pair
$1 \cdot 24$	$25$
$2 \cdot 12$	$14$
$3 \cdot 8$	$11$ (close; wrong sign)
$4 \cdot 6$	$10$

Factor Pair	Sum of the Pair
$- 1 \cdot (- 24)$	$- 25$
$- 2 \cdot (- 12)$	$- 14$
$- 3 \cdot (- 8)$	$- 11$ (what we wanted)
$- 4 \cdot (- 6)$	(no need to go this far)

y^{2} - 11 y + 24 = (y - 3) (y - 8) .

To confirm that this is correct, we should check. Either by multiplying out the factored form:

\begin{aligned} (y - 3) (y - 8) & = (y - 3) \cdot y - (y - 3) \cdot 8 \\ = y^{2} - 3 y - 8 y + 24 \\ \overset{✓}{=} y^{2} - 11 y + 24 \end{aligned}

	$y$	$- 3$
$y$	$y^{2}$	$- 3 y$
$- 8$	$- 8 y$	$24$

Or by evaluating the original expression at

3

and

8 :

\begin{aligned} 3^{2} - 11 (3) + 24 & \overset{?}{=} 0 & 8^{2} - 11 (8) + 24 & \overset{?}{=} 0 \\ 9 - 33 + 24 & \overset{?}{=} 0 & 64 - 88 + 24 & \overset{?}{=} 0 \\ 0 & \overset{✓}{=} 0 & 0 & \overset{✓}{=} 0 . \end{aligned}

Our factorization passes the tests.

🔗

Example 10.3.4.

🔗

Factor

z^{2} + 5 z - 6 .

The negative coefficient is again a small complication from Example 2, but the process is actually still the same.

Explanation.

We need a pair of numbers that multiply to

- 6

and add to

5 .

Note that we do care to keep track that they multiply to a negative product.

Factor Pair	Sum of the Pair
$1 \cdot (- 6)$	$- 5$ (close; wrong sign)
$2 \cdot (- 3)$	$14$

Factor Pair	Sum of the Pair
$- 1 \cdot 6$	$5$ (what we wanted)
$- 2 \cdot 3$	(no need to go this far)

z^{2} + 5 z - 6 = (z - 1) (z + 6) .

To confirm that this is correct, we should check. Either by multiplying out the factored form:

\begin{aligned} (z - 1) (z + 6) & = (z - 1) \cdot z + (z - 1) \cdot 6 \\ = z^{2} - z + 6 z - 6 \\ \overset{✓}{=} z^{2} + 5 z - 6 \end{aligned}

	$z$	$- 1$
$z$	$z^{2}$	$- z$
$6$	$6 z$	$- 6$

Or by evaluating the original expression at

1

and

- 6 :

\begin{aligned} 1^{2} + 5 (1) - 6 & \overset{?}{=} 0 & (- 6)^{2} + 5 (- 6) - 6 & \overset{?}{=} 0 \\ 1 + 5 - 6 & \overset{?}{=} 0 & 36 - 30 - 6 & \overset{?}{=} 0 \\ 0 & \overset{✓}{=} 0 & 0 & \overset{✓}{=} 0 . \end{aligned}

Our factorization passes the tests.

🔗

Checkpoint 10.3.5.

🔗

Factor

m^{2} - 6 m - 40 .

Explanation.

We need a pair of numbers that multiply to

- 40

and add to

- 6 .

Note that we do care to keep track that they multiply to a negative product and sum to a negative total.

Factor Pair	Sum of the Pair
$1 \cdot (- 40)$	$- 39$
$2 \cdot (- 20)$	$- 18$
$4 \cdot (- 10)$	$- 6$ (what we wanted)
(no need to continue)	...

m^{2} - 6 m - 40 = (m + 4) (m - 10) .

🔗

Subsection 10.3.2 Connection to Grouping

🔗

The factoring method we just learned is actually taking a shortcut compared to what we will learn in Section 4. To prepare yourself for that more complicated factoring technique, you may want to try taking the “scenic route” instead of that shortcut.

🔗

Example 10.3.6.

🔗

Let’s factor

x^{2} + 13 x + 40

again (the polynomial from Example 2). As before, it is important to discover that

5

and

8

are important numbers, because they multiply to

40

and add to

13 .

As before, listing out all of the factor pairs is one way to discover the

5

and the

8 .

🔗

Instead of jumping to the factored answer, we can show how

x^{2} + 13 x + 40

factors in a more step-by-step fashion using

5

and

8 .

Since they add up to

13,

we can write:

\begin{aligned} x^{2} + \overset{↓}{13} x + 40 & = x^{2} + \overset{↓}{\overset{⏞}{5 x + 8 x}} + 40 \end{aligned}

🔗

We have intentionally split up the trinomial into an unsimplified polynomial with four terms. In Section 2, we handled such four-term polynomials by grouping:

\begin{aligned} = (x^{2} + 5 x) + (8 x + 40) \end{aligned}

🔗

Now we can factor out each group’s greatest common factor:

\begin{aligned} = x (x + 5) + 8 (x + 5) \\ = x \overset{↓}{\overset{⏞}{(x + 5)}} + 8 \overset{↓}{\overset{⏞}{(x + 5)}} \\ = (x + 5) (x + 8) \end{aligned}

🔗

And we have found that

x^{2} + 13 x + 40

factors as

(x + 5) (x + 8)

without taking the shortcut.

🔗

This approach takes more time, and ultimately you may not use it much. However, if you try a few examples this way, it may make you more comfortable with the more complicated technique in Section 4.

🔗

Subsection 10.3.3 Trinomials with Higher Powers

🔗

So far we have only factored examples of quadratic trinomials: trinomials whose highest power of the variable is

2 .

However, this technique can also be used to factor trinomials where there is a larger highest power of the variable. It only requires that the highest power is even, that the next highest power is half of the highest power, and that the third term is a constant term.

🔗

In the four examples below, check:

if the highest power is even
if the next highest power is half of the highest power
if the last term is constant

🔗

Factor pairs will help with…

$y^{6} - 23 y^{3} - 50$
$h^{16} + 22 h^{8} + 105$

🔗

Factor pairs won’t help with…

$y^{5} - 23 y^{3} - 50$
$h^{16} + 22 h^{8} + 105 h^{2}$

🔗

Example 10.3.7.

🔗

Factor

h^{16} + 22 h^{8} + 105 .

This polynomial is one of the examples above where using factor pairs will help. We find that

7 \cdot 15 = 105,

and

7 + 15 = 22,

so the numbers

7

and

15

can be used:

\begin{aligned} h^{16} + 22 h^{8} + 105 & = h^{16} + \overset{⏞}{7 h^{8} + 15 h^{8}} + 105 \\ = (h^{16} + 7 h^{8}) + (15 h^{8} + 105) \\ = h^{8} (h^{8} + 7) + 15 (h^{8} + 7) \\ = (h^{8} + 7) (h^{8} + 15) \end{aligned}

🔗

Actually, once we settled on using

7

and

15,

we could have concluded that

h^{16} + 22 h^{8} + 105

factors as

(h^{8} + 7) (h^{8} + 15),

if we know which power of

h

to use. We’ll always use half the highest power in these factorizations.

🔗

In any case, to confirm that this is correct, we should check by multiplying out the factored form:

🔗

\begin{aligned} (h^{8} + 7) (h^{8} + 15) & = (h^{8} + 7) \cdot h^{8} + (h^{8} + 7) \cdot 15 \\ = h^{16} + 7 h^{8} + 15 h^{8} + 105 \\ \overset{✓}{=} h^{16} + 22 h^{8} + 15 \end{aligned}

	$h^{8}$	$7$
$h^{8}$	$h^{16}$	$7 h^{8}$
$15$	$15 h^{8}$	$105$

🔗

Our factorization passes the tests.

🔗

Checkpoint 10.3.8.

🔗

Factor

y^{6} - 23 y^{3} - 50 .

Explanation.

We need a pair of numbers that multiply to

- 50

and add to

- 23 .

Note that we do care to keep track that they multiply to a negative product and sum to a negative total.

Factor Pair	Sum of the Pair
$1 \cdot (- 50)$	$- 49$
$2 \cdot (- 25)$	$- 23$ (what we wanted)
(no need to continue)	...

y^{6} - 23 y^{3} - 50 = (y^{3} - 25) (y^{3} + 2) .

🔗

Subsection 10.3.4 Factoring in Stages

🔗

Sometimes factoring a polynomial will take two or more “stages.” Always begin factoring a polynomial by factoring out its greatest common factor, and then apply a second stage where you use a technique from this section. The process of factoring a polynomial is not complete until each of the factors cannot be factored further.

🔗

Example 10.3.9.

🔗

Factor

2 z^{2} - 6 z - 80 .

Explanation.

We will first factor out the common factor,

2 :

2 z^{2} - 6 z - 80 = 2 (z^{2} - 3 z - 40)

Now we are left with a factored expression that might factor more. Looking inside the parentheses, we ask ourselves, “what two numbers multiply to be

- 40

and add to be

- 3 ?

” Since

5

and

- 8

do the job the full factorization is:

\begin{aligned} 2 z^{2} - 6 z - 80 & = 2 (z^{2} - 3 z - 40) \\ = 2 (z + 5) (z - 8) \end{aligned}

🔗

Example 10.3.10.

🔗

Factor

- r^{2} + 2 r + 24 .

Explanation.

The three terms don’t exactly have a common factor, but as discussed in Section 1, when the leading term has a negative sign, it is often helpful to factor out that negative sign:

- r^{2} + 2 r + 24 = - (r^{2} - 2 r - 24) .

Looking inside the parentheses, we ask ourselves, “what two numbers multiply to be

- 24

and add to be

- 2 ?

” Since

- 6

and

4

work here and the full factorization is shown:

\begin{aligned} - r^{2} + 2 r + 24 & = - (r^{2} - 2 r - 24) \\ = - (r - 6) (r + 4) \end{aligned}

🔗

Example 10.3.11.

🔗

Factor

p^{2} q^{3} + 4 p^{2} q^{2} - 60 p^{2} q .

Explanation.

First, always look for the greatest common factor: in this trinomial it is

p^{2} q .

After factoring this out, we have

p^{2} q^{3} + 4 p^{2} q^{2} - 60 p^{2} q = p^{2} q (q^{2} + 4 q - 60) .

Looking inside the parentheses, we ask ourselves, “what two numbers multiply to be

- 60

and add to be

4 ?

” Since

10

and

- 6

fit the bill, the full factorization can be shown below:

\begin{aligned} p^{2} q^{3} + 4 p^{2} q^{2} - 60 p^{2} q & = p^{2} q (q^{2} + 4 q - 60) \\ = p^{2} q (q + 10) (q - 6) \end{aligned}

🔗

Subsection 10.3.5 More Trinomials with Two Variables

🔗

You might encounter a trinomial with two variables that can be factored using the methods we’ve discussed in this section. It can be tricky though:

x^{2} + 5 x y + 6 y^{2}

has two variables and it can factor using the methods from this section, but

x^{2} + 5 x + 6 y^{2}

also has two variables and it cannot be factored. So in examples of this nature, it is even more important to check that factorizations you find actually work.

🔗