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Difficulty Level: At Grade Created by: CK-12

## Learning Objectives

• Write quadratic equations in standard form.
• Factor quadratic expressions for different coefficient values.

## Write Quadratic Expressions in Standard Form

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

$ax^2 + bx + c$

where $a, b,$ and $c$ stand for constant numbers. Factoring these polynomials depends on the values of these constants. In this section we’ll learn how to factor quadratic polynomials for different values of $a, b,$ and $c$. (When none of the coefficients are zero, these expressions are also called quadratic trinomials, since they are polynomials with three terms.)

You’ve already learned 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 $x(ax + b)$. Now we’ll see how to factor quadratics where $c$ is nonzero.

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

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

$x^2 + bx +c$

You know from multiplying binomials that when you multiply two factors $(x + m)(x + n)$, you get a quadratic polynomial. Let’s look at this process in more detail. First we use distribution:

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

Then we simplify by combining the like terms in the middle. We get:

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

So to factor a quadratic, we just need to do this process in reverse.

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

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

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

The factors of $x^2 + bx + c$ are always two binomials

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

such that $n + m = b$ and $mn = c$.

Example 1

Factor $x^2 + 5x + 6$.

Solution

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

$(x\;\;\;\;)(x\;\;\;\;)$

We want two numbers $m$ and $n$ that multiply to 6 and add up to 5. A good strategy is to list the possible ways we can multiply two numbers to get 6 and then see which of these numbers add up to 5:

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

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

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

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

Example 2

Factor $x^2 + 7x + 12$.

Solution

We are looking for an answer that is a product of two binomials in parentheses: $(x\;\;\;\;\;)(x\;\;\;\;)$

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

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

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 two binomials in parentheses: $(x\;\;\;\;\;)(x\;\;\;\;)$

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

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

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 two binomials in parentheses: $(x\;\;\;\;)(x\;\;\;\;)$

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

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

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 is negative.

Example 5

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

Solution

We are looking for an answer that is a product of two binomials in parentheses: $(x\;\;\;\;)(x\;\;\;\;)$

When negative coefficients are involved, we have to remember that negative factors may be involved also. The number 8 can be written as the product of the following numbers:

$8 = 1 \cdot 8 \quad \quad \text{and} \quad \quad 1 + 8 = 9$

but also

$8 = (-1) \cdot (-8) \quad \quad \text{and} \quad \quad -1 + (-8) = -9$

and

$8 = 2 \cdot 4 \quad \quad \text{and} \quad \quad 2 + 4 = 6$

but also

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

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

$& \quad \quad \quad x - 2\\& \underline{\;\;\;\;\;\;\;\;\;\;\;x - 4}\\& \quad \ - \ 4x + 8\\& \underline{x^2 - \ 2x\;\;\;\;\;\;\;}\\& x^2 - \ 6x + 8$ The answer checks out.

Example 6

Factor $x^2 - 17x + 16$.

Solution

We are looking for an answer that is a product of two binomials in parentheses: $(x\;\;\;\;)(x\;\;\;\;)$

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

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

In general, whenever $b$ is negative and $a$ and $c$ are positive, the two binomial factors will have minus signs instead of plus signs.

## 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 binomials in parentheses: $(x\;\;\;\;)(x\;\;\;\;\;)$

Once again, we must take the negative sign into account. The number -15 can be written as the product of the following numbers:

$& -15 = -1 \cdot 15 && \text{and} && -1 + 15 = 14\\& -15 = 1 \cdot (-15) && \text{and} && 1 + (-15) = -14\\& -15 = -3 \cdot 5 && \text{and} && -3 + 5 = 2 \qquad \qquad This \ is \ the \ correct \ choice.\\& -15 = 3 \cdot (-5) && \text{and} && 3 + (-5) = -2$

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

We can check to see if this is correct by multiplying:

$& \quad \quad \ \ x - \ 3\\& \underline{\;\;\;\;\;\;\;\;\; x + \;5\;}\\& \quad \quad 5x - 15\\& \underline{x^2 - 3x\;\;\;\;\;\;\;\;}\\& x^2 + 2x - 15$ The answer checks out.

Example 8

Factor $x^2 - 10x - 24$.

Solution

We are looking for an answer that is a product of two binomials in parentheses: $(x\;\;\;\;)(x\;\;\;\;)$

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

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

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 binomials in parentheses: $(x\;\;\;\;)(x\;\;\;\;)$

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

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

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 and then apply the methods you learned so far in this section

Example 10

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

Solution

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

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

We’re looking for a product of two binomials in parentheses: $-(x\;\;\;\;)(x\;\;\;\;)$

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

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

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

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

## Lesson Summary

• A quadratic of the form $x^2 + bx + c$ factors as a product of two binomials in parentheses: $(x + m)(x + n)$
• 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 out -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. $x^2 - 17x - 60$
25. $x^2 + 5x - 36$

Feb 23, 2012

Feb 12, 2015