What if you had a quadratic expression like \begin{align*}x^2 + 9x + 14\end{align*} 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.

### Watch This

CK-12 Foundation: 0908S Factoring Quadratic Expressions

### Guidance

**Quadratic polynomials** are polynomials of the \begin{align*}2^{nd}\end{align*} degree. The standard form of a quadratic polynomial is written as

\begin{align*}ax^2 + bx + c\end{align*}

where \begin{align*}a, b,\end{align*} and \begin{align*}c\end{align*} 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 \begin{align*}a, b,\end{align*} and \begin{align*}c\end{align*}. (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 \begin{align*}c = 0\end{align*}. For example, for the quadratic \begin{align*}ax^2 + bx\end{align*}, the common factor is \begin{align*}x\end{align*} and this expression is factored as \begin{align*}x(ax + b)\end{align*}. Now we’ll see how to factor quadratics where \begin{align*}c\end{align*} is nonzero.

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

First, let’s consider the case where \begin{align*}a = 1, b\end{align*} is positive and \begin{align*}c\end{align*} is positive. The quadratic trinomials will take the form

\begin{align*}x^2 + bx +c\end{align*}

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

\begin{align*}(x + m)(x + n) = x^2 + nx + mx + mn\end{align*}

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

\begin{align*}(x + m)(x + n) = x^2 + (n + m) x +mn\end{align*}

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

\begin{align*}& \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\end{align*}

This means that we need to find two numbers \begin{align*}m\end{align*} and \begin{align*}n\end{align*} where

\begin{align*} n + m = b \qquad \qquad \text{and} \qquad \qquad mn = c\end{align*}

The factors of \begin{align*}x^2 + bx + c\end{align*} are always two binomials

\begin{align*}(x + m)(x + n)\end{align*}

such that \begin{align*}n + m = b\end{align*} and \begin{align*}mn = c\end{align*}.

#### Example A

*Factor* \begin{align*}x^2 + 5x + 6\end{align*}.

**Solution**

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

\begin{align*}(x\;\;\;\;)(x\;\;\;\;)\end{align*}

We want two numbers \begin{align*}m\end{align*} and \begin{align*}n\end{align*} 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:

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

So the answer is \begin{align*}(x + 2)(x + 3)\end{align*}.

We can check to see if this is correct by multiplying \begin{align*}(x + 2)(x + 3)\end{align*}:

\begin{align*}& \quad \quad x + 2\\ & \underline{\;\;\;\;\;\;\;\;\;\;x + 3}\\ & \quad \quad \ 3x + 6\\ & \underline{x^2 + 2x\;\;\;\;\;\;}\\ & x^2 + 5x + 6\end{align*}

The answer checks out.

#### Example B

*Factor* \begin{align*}x^2 + 7x + 12\end{align*}.

**Solution**

We are looking for an answer that is a product of two binomials in parentheses: \begin{align*}(x\;\;\;\;\;)(x\;\;\;\;)\end{align*}

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

\begin{align*}& 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.\end{align*}

The answer is \begin{align*}(x + 3)(x + 4)\end{align*}.

#### Example C

*Factor* \begin{align*}x^2 + 8x + 12\end{align*}.

**Solution**

We are looking for an answer that is a product of two binomials in parentheses: \begin{align*}(x\;\;\;\;\;)(x\;\;\;\;)\end{align*}

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

\begin{align*}& 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\end{align*}

The answer is \begin{align*}(x + 2)(x + 6)\end{align*}.

Watch this video for help with the Examples above.

CK-12 Foundation: Factoring Quadratic Expressions

### Vocabulary

- A quadratic of the form \begin{align*}x^2 + bx + c\end{align*} factors as a product of two binomials in parentheses: \begin{align*}(x + m)(x + n)\end{align*}

- If \begin{align*}b\end{align*} and \begin{align*}c\end{align*} are positive, then both \begin{align*}m\end{align*} and \begin{align*}n\end{align*} are positive.

### Guided Practice

*Factor* \begin{align*}x^2 + 12x + 36\end{align*}.

**Solution**

We are looking for an answer that is a product of two binomials in parentheses: \begin{align*}(x\;\;\;\;)(x\;\;\;\;)\end{align*}

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

\begin{align*}& 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.\end{align*}

The answer is \begin{align*}(x + 6)(x + 6)\end{align*}.

### Explore More

Factor the following quadratic polynomials.

- \begin{align*}x^2 + 10x + 9\end{align*}
- \begin{align*}x^2 + 15x + 50\end{align*}
- \begin{align*}x^2 + 10x + 21\end{align*}
- \begin{align*}x^2 + 16x + 48\end{align*}
- \begin{align*}x^2+14x+45\end{align*}
- \begin{align*}x^2+27x+50\end{align*}
- \begin{align*}x^2+22x+40\end{align*}
- \begin{align*}x^2+15x+56\end{align*}
- \begin{align*}x^2+2x+1\end{align*}
- \begin{align*}x^2+10x+24\end{align*}
- \begin{align*}x^2+17x+72\end{align*}
- \begin{align*}x^2+25x+150\end{align*}