Factoring Techniques

Factoring Techniques#

Pull out Common Factors#

Common Factors

The act of pulling out common factors from a sum can be thought of as applying the distributive property of multiplication in reverse.

A B + A C = A (B + C)

For example, $21 + 14 = 7 \cdot 3 + 7 \cdot 2 = 7 (3 + 2)$ .

Video Resource

Factor Polynomials Using the Greatest Common Factor Method (Links to an external site)
A review of identifying and pulling out the greatest common factors when factoring a polynomial.

Video Resource

Factoring Algebraic Expressions with Rational Exponents (Links to an external site)
A review of identifying and pulling out the greatest common factors when factoring expressions with rational exponents.

Example 1#

Factoring

Factor $10 x^{5} + 15 x^{4}$ by pulling out common factors.

Example 2#

Factoring

Factor $4 x^{3} (2 x^{3} + 1)^{5} + 30 x^{6} (2 x^{3} + 1)^{4}$ by first pulling out common factors.

Example 3#

Factoring

Factor $4 x^{3} (2 x^{3} + 1)^{1 / 2} + 3 x^{6} (2 x^{3} + 1)^{- 1 / 2}$ by first pulling out common factors.

Difference of Squares#

Applying the FOIL Method to the Product $(A + B) (A - B)$ .

When applying the FOIL technique to a product of the form $(A + B) (A - B)$ , we get the following result.

\begin{aligned} (A + B) (A - B) & = A \cdot A - A \cdot B + B \cdot A - B \cdot B \\ = A^{2} - A B + A B - B^{2} \\ = A^{2} - B^{2} \end{aligned}

We refer to $A^{2} - B^{2}$ as a difference of squares, which can always be factored in the following manner.

A^{2} - B^{2} = (A + B) (A - B)

Example 4#

Applying difference of squares

Factor $x^{2} - 25$ .

Example 5#

Applying difference of squares

Factor $9 x^{3} - 4 x^{5}$ .

AC Grouping Method#

Steps of the AC Grouping Method.

The AC grouping method is a technique for factoring certain quadratic expressions of the form

a x^{2} + b x + c .

The process is broken down into the following five steps.

Find two integers, $r$ and $s$ , that multiply to $a \times c$ and sum to $b$ .
Replace $b x$ with $r x$ + $s x$ .
Group the first two terms and the last two terms of $a x^{2} + r x + s x + c$ .
Pull out common factors from each group.
Pull out common factor.

Example 6#

Applying AC grouping

Factor $6 x^{2} + 7 x - 5$ using the AC grouping method.

Step 1: Find two integers that multiply to

6 (- 5) = - 30

and sum to

7

.

Since the product is negative, the two numbers must have opposite signs. And since the sum is positive, the larger number in absolute value, must be positive.

Product equals $- 30$	Sum equals $7$ ?
$- 1 \times 30 = - 30$	$- 1 + 30 = 29$ NO
$- 2 \times 15 = - 30$	$- 2 + 15 = 13$ NO
$- 3 \times 10 = - 30$	$- 3 + 10 = 7$ YES

Example 7#

Applying AC grouping

Factor $x^{2} - 13 x + 36$ using the AC grouping method.

Step 1: Find two integers that multiply to

1 \times 36 = 36

and sum to

- 13

.

Since the product is positive, the two numbers must have the same sign. And since the sum is negative, both numbers must be negative.

Product equals $36$	Sum equals $- 13$ ?
$- 1 \times - 36$	NO
$- 2 \times - 18$	NO
$- 3 \times - 12$	NO
$- 4 \times - 9$	YES

A Special Case of the AC Grouping Method#

Important Observation

If the coefficient of $x^{2}$ is one (i.e., $a = 1$ in $a x^{2} + b x + c$ ), then Step 1 of the $A C$ grouping method is to find two numbers that multiply to $c$ and sum to $b$ . If those numbers exist, call them $r$ and $s$ , then the factorization can be written as

x^{2} + b x + c = (x + r) (x + s)

where $r + s = b$ and $r s = c$ .

Example 8#

AC grouping

Factor $x^{2} - 4 x - 12$ .

Step 1: Find two integers that multiply to

- 12

and sum to

- 4

.

Since the product is negative, the two numbers must have opposite signs. And since the sum is negative, the larger number in absolute value, must be negative.

Product equals $- 12$	Sum equals $- 4$ ?
$1 \times - 12$	NO
$2 \times - 6$	YES

Example 9#

AC grouping and pulling out common factors

Factor $7 x^{4} + 35 x^{3} + 42 x^{2}$ .

Step 3: Factor

x^{2} + 5 x + 6

by finding two integers that multiply to

6

and sum to

5

.

Product equals $6$	Sum equals $5$ ?
$1 \times 6$	NO
$2 \times 3$	YES

Therefore,

\begin{aligned} 7 x^{4} + 35 x^{3} + 42 x^{2} & = 7 x^{2} (x^{2} + 5 x + 6) & Step 2 \\ = 7 x^{2} (x + 2) (x + 3) & AC grouping \end{aligned}

Factoring Techniques

Contents

Factoring Techniques#

Pull out Common Factors#

Example 1#

Example 2#

Example 3#

Difference of Squares#

Example 4#

Example 5#

AC Grouping Method#

Example 6#

Example 7#

A Special Case of the AC Grouping Method#

Example 8#

Example 9#