<meta http-equiv="refresh" content="1; url=/nojavascript/"> Factoring Quadratic Expressions | CK-12 Foundation
You are reading an older version of this FlexBook® textbook: CK-12 Algebra I Go to the latest version.

Difficulty Level: At Grade Created by: CK-12

## Learning Objectives

• Write quadratic equations in standard form.
• Factor quadratic expressions for different coefficient values.
• Factor when $a = -1$.

## Write Quadratic Expressions in Standard Form

Quadratic polynomials are polynomials of $2^{nd}$ degree. The standard form of a quadratic polynomial is written as

$ax^2 + bx + c$

Here $a, b,$ and $c$ stand for constant numbers. Factoring these polynomials depends on the values of these constants. In this section, we will learn how to factor quadratic polynomials for different values of $a, b,$ and $c$. In the last section, we factored common monomials, so you already know how to factor quadratic polynomials where $c = 0$.

For example for the quadratic $ax^2 + bx$, the common factor is $x$ and this expression is factored as $x(ax + b)$. When all the coefficients are not zero these expressions are also called Quadratic Trinomials, since they are polynomials with three terms.

## Factor when a = 1, b is Positive, and c is Positive

Let’s first consider the case where $a = 1, b$ is positive and $c$ is positive. The quadratic trinomials will take the following form.

$x^2 + bx + c$

You know from multiplying binomials that when you multiply two factors $(x + m) (x + n)$ you obtain a quadratic polynomial. Let’s multiply this and see what happens. We use The Distributive Property.

$(x + m) (x + n) = x^2 + nx + mx + mn$

To simplify this polynomial we would combine the like terms in the middle by adding them.

$(x + m) (x + n) = x^2 + (n + m)x + mn$

To factor we need to do this process in reverse.

$& \text{We see that} && x^2 + (n + m) x + mn \\& \text{Is the same form as} && x^2 + bx + c$

This means that we need to find two numbers $m$ and $n$ where

$n + m = b \quad \quad \text{and} \quad \quad mn = c$

To factor $x^2 + bx + c$, the answer is the product of two parentheses.

$(x + m) (x + n)$

so that $n + m = b$ and $mn = c$

Let’s try some specific examples.

Example 1

Factor $x^2 + 5x +6$

Solution We are looking for an answer that is a product of two binomials in parentheses.

$(x + \underline{\;\;\;\;\;\;\;} )(x + \underline{\;\;\;\;\;\;\;})$

To fill in the blanks, we want two numbers $m$ and $n$ that multiply to 6 and add to 5. A good strategy is to list the possible ways we can multiply two numbers to give us 6 and then see which of these pairs of numbers add to 5. The number six can be written as the product of.

$&6 = 1 \cdot 6 && \text{and} && 1 + 6 = 7 \\&6 = 2 \cdot 3 && \text{and} && 2 + 3 = 5 \qquad \leftarrow \qquad \text{This is the correct choice.}$

So the answer is $(x + 2) (x + 3)$.

We can check to see if this is correct by multiplying $(x + 2) (x + 3)$.

$& \quad \quad \quad x \ \ + \ 2\\& \underline{\;\;\;\;\;\;\;\;\;\;\;x \ \ + \ 3}\\& \quad \quad \quad 3x \ + \ 6\\& \underline{x^2 \ + \ 2x\;\;\;\;\;\;\;\;\;}\\& x^2 \ + \ 5x \ + \ 9$

Example 2

Factor $x ^2 + 7x + 12$

Solution

We are looking for an answer that is a product of two parentheses $(x + \underline{\;\;\;\;\;\;\;} )(x + \underline{\;\;\;\;\;\;\;})$.

The number 12 can be written as the product of the following numbers.

$& 12 = 1 \cdot 12 && \text{and} && 1 + 12 = 13 \\& 12 = 2 \cdot 6 && \text{and} && 2 + 6 = 8 \\& 12 = 3 \cdot 4 && \text{and} && 3 + 4 = 7 \qquad \leftarrow \qquad \text{This is the correct choice.}$

The answer is $(x + 3) (x + 4)$.

Example 3

Factor $x^2 + 8x + 12$.

Solution

We are looking for an answer that is a product of the two parentheses $(x + \underline{\;\;\;\;\;\;\;} )(x + \underline{\;\;\;\;\;\;\;})$.

The number 12 can be written as the product of the following numbers.

$& 12 = 1 \cdot 12 && \text{and} && 1 + 12 = 13 \\& 12 = 2 \cdot 6 && \text{and} && 2 + 6 = 8 \qquad \leftarrow \qquad \text{This is the correct choice}.\\& 12 = 3 \cdot 4 && \text{and} && 3 + 4 = 7$

The answer is $(x + 2) (x + 6).$

Example 4

Factor $x^2 + 12x + 36$.

Solution

We are looking for an answer that is a product of the two parentheses $(x + \underline{\;\;\;\;\;\;\;} )(x + \underline{\;\;\;\;\;\;\;})$.

The number 36 can be written as the product of the following numbers.

$& 36 = 1 \cdot 36 && \text{and} && 1 + 36 = 37 \\& 36 = 2 \cdot 18 && \text{and} && 2 + 18 = 20 \\& 36 = 3 \cdot 12 && \text{and} && 3 + 12 = 15 \\& 36 = 4 \cdot 9 && \text{and} && 4 + 9 = 13 \\& 36 = 6 \cdot 6 && \text{and} && 6 + 6 = 12 \qquad \leftarrow \qquad \text{This is the correct choice}$

The answer is $(x + 6) (x + 6)$.

## Factor when a = 1, b is Negative and c is Positive

Now let’s see how this method works if the middle coefficient $(b)$ is negative.

Example 5

Factor $x^2 - 6x + 8.$

Solution

We are looking for an answer that is a product of the two parentheses $(x + \underline{\;\;\;\;\;\;\;} )(x + \underline{\;\;\;\;\;\;\;})$.

The number 8 can be written as the product of the following numbers.

$8 = 1 \cdot 8$ and $1 + 8 = 9$ Notice that these are two different choices.

But also,

$& 8 = (-1) \cdot (-8) && \text{and} && -1 + (-8) = -9 && \text{Notice that these are two different choices.}\\& 8 = 2 \cdot 4 && \text{and} && 2 + 4 = 6$

But also,

$8 = (-2) \cdot (-4) && \text{and} && -2 + (-4) = -6 \qquad \leftarrow \qquad \text{This is the correct choice.}$

The answer is $(x - 2) (x - 4)$

We can check to see if this is correct by multiplying $(x - 2) (x - 4)$.

$& \quad \quad \quad x \ \ - \ 2\\& \underline{\;\;\;\;\;\;\;\;\;\;\;x \ \ - \ 4}\\& \quad \ - \ 4x \ + \ 8 \\& \underline{x^2 \ - \ 2x\;\;\;\;\;\;\;\;\;}\\& x^2 \ - \ 6x \ + \ 8$

Example 6

Factor $x - 17x + 16$

Solution

We are looking for an answer that is a product of two parentheses: $(x \pm \underline{\;\;\;\;\;\;\;} )(x \pm \underline{\;\;\;\;\;\;\;})$.

The number 16 can be written as the product of the following numbers:

$16 & = 1 \cdot 16 && \text{and} && 1 + 16 = 17 \\16 & = (-1) \cdot (-16) && \text{and} && -1 + (-16) = -17 \qquad \leftarrow \qquad \text{This is the correct choice.}\\16 & = 2 \cdot 8 && \text{and} && 2 + 8 = 10 \\16 & = (-2) \cdot (-8) && \text{and} && -2 + (-8) = -10\\16 & = 4 \cdot 4 && \text{and} && 4 + 4 = 8 \\16 & = (-4) \cdot (-4) && \text{and} && -4 + (-4) = -8$

The answer is $(x - 1) (x - 16)$.

## Factor when a = 1 and c is Negative

Now let’s see how this method works if the constant term is negative.

Example 7

Factor $x^2 + 2x -15$

Solution

We are looking for an answer that is a product of two parentheses $(x \pm \underline{\;\;\;\;\;\;\;} )(x \pm \underline{\;\;\;\;\;\;\;})$.

In this case, we must take the negative sign into account. The number -15 can be written as the product of the following numbers.

$-15 = -1 \cdot 15 && \text{and} && -1 + 15 = 14 && \ \text{Notice that these are two different choices.}$

And also,

$-15 = 1 \cdot (-15) && \text{and} && 1 + (-15) = -14 && \text{Notice that these are two different choices.}$

$& -15 = -3 \cdot 5 && \text{and} && -3 + 5 = 2 \qquad \leftarrow \qquad \text{This is the correct choice.} \\& -15 = 3 \cdot (-5) && \text{and} && 3 + (-5) = -2$

The answer is $(x - 3) (x + 5)$.

We can check to see if this is correct by multiplying $(x - 3) (x + 5)$.

$& \quad \quad \quad \ x \ \ - \ 3\\& \underline{\;\;\;\;\;\;\;\;\;\;\;\;x \ \ + \ 5}\\& \quad \quad \quad 5x \ - \ 15\\& \underline{x^2 \ -\ 3x\;\;\;\;\;\;\;\;\;\;}\\& x^2 \ + \ 2x \ - \ 15$

Example 8

Factor $x^2 - 10x -24$

Solution

We are looking for an answer that is a product of two parentheses $(x \pm \underline{\;\;\;\;\;\;\;} )(x \pm \underline{\;\;\;\;\;\;\;})$.

The number -24 can be written as the product of the following numbers.

$& -24 = -1 \cdot 24 && \text{and} && -1 + 24 = 23 \\& -24 = 1 \cdot (-24) && \text{and} && 1 + (-24) = -23 \\& -24 = -2 \cdot 12 && \text{and} && -2 + 12 = 10 \\& -24 = 2 \cdot (-12) && \text{and} && 2 + (-12) = -10 \qquad \leftarrow \qquad \text{This is the correct choice.}\\& -24 = -3 \cdot 8 && \text{and} && -3 + 8 = 5 \\& -24 = 3 \cdot (-8) && \text{and} && 3 + (-8) = -5 \\& -24 = -4 \cdot 6 && \text{and} && -4 + 6 = 2 \\& -24 = 4 \cdot (-6) && \text{and} && 4 + (-6) = -2$

The answer is $(x - 12)(x + 2).$

Example 9

Factor $x^2 + 34x - 35$

Solution

We are looking for an answer that is a product of two parentheses $(x \pm \underline{\;\;\;\;\;\;\;} )(x \pm \underline{\;\;\;\;\;\;\;})$

The number -35 can be written as the product of the following numbers:

$& -35 = -1 \cdot 35 && \text{and} && -1 + 35 = 34 \qquad \leftarrow \qquad \text{This is the correct choice.}\\& -35 = 1 \cdot (-35) && \text{and} && 1 + (-35) = -34 \\& -35 = -5 \cdot 7 && \text{and} && -5 + 7 = 2 \\& -35 = 5 \cdot (-7) && \text{and} && 5 + (-7) = -2$

The answer is $(x - 1) (x + 35)$.

## Factor when a = - 1

When $a = -1$, the best strategy is to factor the common factor of -1 from all the terms in the quadratic polynomial. Then, you can apply the methods you have learned so far in this section to find the missing factors.

Example 10

Factor $x^2 + x + 6.$

Solution

First factor the common factor of -1 from each term in the trinomial. Factoring -1 changes the signs of each term in the expression.

$- x^2 + x + 6 = - (x^2 - x - 6)$

We are looking for an answer that is a product of two parentheses $(x \pm \underline{\;\;\;\;\;\;\;} )(x \pm \underline{\;\;\;\;\;\;\;})$

Now our job is to factor $x^2 - x - 6$.

The number -6 can be written as the product of the following numbers.

$& -6 = -1 \cdot 6 && \text{and}&& -1 + 6 = 5 \\& -6 = 1 \cdot (-6) && \text{and}&& 1 + (-6) = -5 \\& -6 = -2 \cdot 3 && \text{and}&& -2 + 3 = 1 \\& -6 = 2 \cdot (-3) && \text{and}&& 2 + (-3) = -1 \qquad \leftarrow \qquad \text{This is the correct choice.}$

The answer is $- (x - 3) (x + 2)$.

To Summarize,

A quadratic of the form $x + bx + c$ factors as a product of two parenthesis $(x + m) (x + n)$.

• If $b$ and $c$ are positive then both $m$ and $n$ are positive
• Example $x^2 + 8x + 12$ factors as $(x + 6) (x + 2)$.
• If $b$ is negative and $c$ is positive then both $m$ and $n$ are negative.
• Example $x^2 - 6x + 8$ factors as $(x - 2) (x - 4)$.
• If $c$ is negative then either $m$ is positive and $n$ is negative or vice-versa
• Example $x^2 + 2x -15$ factors as $(x + 5) (x - 3)$.
• Example $x^2 + 34x -35$ factors as $(x + 35) (x - 1)$.
• If $a = -1$, factor a common factor of -1 from each term in the trinomial and then factor as usual. The answer will have the form $- (x + m) (x + n)$.
• Example $- x^2 + x + 6$ factors as $-(x - 3) (x + 2)$.

## Review Questions

1. $x^2 + 10x + 9$
2. $x^2 + 15x + 50$
3. $x^2 + 10x + 21$
4. $x^2 + 16x + 48$
5. $x^2 - 11x + 24$
6. $x^2 - 13x + 42$
7. $x^2 - 14x +33$
8. $x^2 - 9x + 20$
9. $x^2 + 5x - 14$
10. $x^2 + 6x - 27$
11. $x^2 + 7x - 78$
12. $x^2 + 4x - 32$
13. $x^2 - 12x - 45$
14. $x^2 - 5x - 50$
15. $x^2 - 3x - 40$
16. $x^2 - x - 56$
17. $-x^2 - 2x - 1$
18. $-x^2 - 5x + 24$
19. $-x^2 + 18x - 72$
20. $-x^2 + 25x - 150$
21. $x^2 + 21x + 108$
22. $-x^2 + 11x - 30$
23. $x^2 + 12x - 64$
24. $x^2 - 17x - 60$

1. $(x + 1) (x + 9)$
2. $(x + 5) (x + 10)$
3. $(x + 7) (x + 3)$
4. $(x + 12) (x + 4)$
5. $(x - 3) (x - 8)$
6. $(x - 7) (x - 6)$
7. $(x - 11) (x - 3)$
8. $(x - 5) (x - 4)$
9. $(x - 2) (x + 7)$
10. $(x - 3) (x + 9)$
11. $(x - 6) (x + 13)$
12. $(x - 4) (x + 8)$
13. $(x - 15) (x + 3)$
14. $(x - 10) (x + 5)$
15. $(x - 8) (x + 5)$
16. $(x - 8) (x + 7)$
17. $- (x + 1) (x + 1)$
18. $- (x - 3) (x + 8)$
19. $- (x - 6) (x - 12)$
20. $- (x - 15) (x - 10)$
21. $(x + 9) (x + 12)$
22. $- (x - 5) (x - 6)$
23. $(x - 4) (x + 16)$
24. $(x - 20) (x + 3)$

Feb 22, 2012

Aug 22, 2014