<meta http-equiv="refresh" content="1; url=/nojavascript/"> Factorization of Quadratic Expressions ( Read ) | Algebra | CK-12 Foundation
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What if you had a quadratic expression like $x^2 + 9x + 14$ in which all the coefficients were positive? How could you factor that expression? After completing this Concept, you'll be able to factor quadratic expressions like this one with positive coefficient values.

### Guidance

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 A

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 B

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 C

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 $(x + 2)(x + 6)$ .

Watch this video for help with the Examples above.

### Vocabulary

• 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.

### Guided Practice

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 $(x + 6)(x + 6)$ .

### Practice

1. $x^2 + 10x + 9$
2. $x^2 + 15x + 50$
3. $x^2 + 10x + 21$
4. $x^2 + 16x + 48$
5. $x^2+14x+45$
6. $x^2+15x+50$
7. $x^2+22x+40$
8. $x^2+15x+56$
9. $x^2+2x+1$
10. $x^2+10x+24$
11. $x^2+17x+72$
12. $x^2+25x+150$