<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
Dismiss
Skip Navigation

Factorization of Quadratic Expressions

Factor quadratics with positive coefficients

Atoms Practice
Estimated8 minsto complete
%
Progress
Practice Factorization of Quadratic Expressions
 
 
 
MEMORY METER
This indicates how strong in your memory this concept is
Practice
Progress
Estimated8 minsto complete
%
Practice Now
Turn In
Factorization of Quadratic Expressions

Factorization of Quadratic Expressions

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

\begin{align*}ax^2 + bx + c\end{align*}ax2+bx+c

where \begin{align*}a, b,\end{align*}a,b, and \begin{align*}c\end{align*}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 \begin{align*}a, b,\end{align*}a,b, and \begin{align*}c\end{align*}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 \begin{align*}c = 0\end{align*}c=0. For example, for the quadratic \begin{align*}ax^2 + bx\end{align*}ax2+bx, the common factor is \begin{align*}x\end{align*}x and this expression is factored as \begin{align*}x(ax + b)\end{align*}x(ax+b). Now we’ll see how to factor quadratics where \begin{align*}c\end{align*}c 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*}a=1,b is positive and \begin{align*}c\end{align*}c is positive. The quadratic trinomials will take the form

\begin{align*}x^2 + bx +c\end{align*}x2+bx+c

You know from multiplying binomials that when you multiply two factors \begin{align*}(x + m)(x + n)\end{align*}(x+m)(x+n), 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*}(x+m)(x+n)=x2+nx+mx+mn

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*}(x+m)(x+n)=x2+(n+m)x+mn

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*}We see that x2+(n+m)x+mnis the same form asx2+bx+c

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

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

The factors of \begin{align*}x^2 + bx + c\end{align*}x2+bx+c are always two binomials

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

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

Factoring

1. Factor \begin{align*}x^2 + 5x + 6\end{align*}x2+5x+6.

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

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

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

\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*}6=166=23andand1+6=72+3=5This is the correct choice.

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

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

\begin{align*}& \quad \quad x + 2\\ & \underline{\;\;\;\;\;\;\;\;\;\;x + 3}\\ & \quad \quad \ 3x + 6\\ & \underline{x^2 + 2x\;\;\;\;\;\;}\\ & x^2 + 5x + 6\end{align*}x+2x+3 3x+6x2+2xx2+5x+6

The answer checks out.

2. Factor \begin{align*}x^2 + 7x + 12\end{align*}x2+7x+12.

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

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*}12=11212=2612=34andandand1+12=132+6=83+4=7This is the correct choice.

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

3. Factor \begin{align*}x^2 + 8x + 12\end{align*}x2+8x+12.

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

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*}12=11212=2612=34andandand1+12=132+6=8This is the correct choice.3+4=7

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

Example

Example 1

Factor \begin{align*}x^2 + 12x + 36\end{align*}x2+12x+36.

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

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*}36=13636=21836=31236=4936=66andandandandand1+36=372+18=203+12=154+9=136+6=12This is the correct choice.

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

Review 

Factor the following quadratic polynomials.

  1. \begin{align*}x^2 + 10x + 9\end{align*}x2+10x+9
  2. \begin{align*}x^2 + 15x + 50\end{align*}x2+15x+50
  3. \begin{align*}x^2 + 10x + 21\end{align*}x2+10x+21
  4. \begin{align*}x^2 + 16x + 48\end{align*}x2+16x+48
  5. \begin{align*}x^2+14x+45\end{align*}
  6. \begin{align*}x^2+15x+50\end{align*}
  7. \begin{align*}x^2+22x+40\end{align*}
  8. \begin{align*}x^2+15x+56\end{align*}
  9. \begin{align*}x^2+2x+1\end{align*}
  10. \begin{align*}x^2+10x+24\end{align*}
  11. \begin{align*}x^2+17x+72\end{align*}
  12. \begin{align*}x^2+25x+150\end{align*}

Review (Answers)

To view the Review answers, open this PDF file and look for section 9.8. 

Notes/Highlights Having trouble? Report an issue.

Color Highlighted Text Notes
Please to create your own Highlights / Notes
Show More

Vocabulary

TermDefinition
Greatest Common Factor The greatest common factor of two numbers is the greatest number that both of the original numbers can be divided by evenly.
linear factors Linear factors are expressions of the form ax+b where a and b are real numbers.
Polynomial A polynomial is an expression with at least one algebraic term, but which does not indicate division by a variable or contain variables with fractional exponents.
Quadratic Formula The quadratic formula states that for any quadratic equation in the form ax^2+bx+c=0, x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}.
Quadratic Polynomials A quadratic polynomial is a polynomial of the 2nd degree, in other words, a polynomial with an x^2 term.
Trinomial A trinomial is a mathematical expression with three terms.

Image Attributions

Explore More

Sign in to explore more, including practice questions and solutions for Factorization of Quadratic Expressions.
Please wait...
Please wait...