## Simplifying Expressions Using Common Factors

## Introduction

The concept of factoring can best be understood as being the opposite of multiplying. In algebra, a pair of binomials can be multiplied as follows:

A factoring problem is the exact opposite of this. The polynomial 14x^{3} + 21x is slightly different than the answer above. The terms 14x^{3} and 21x have a common factor of 7x. Dividing each term by the common factor results in 2x^{2} + 3 being left as the remaining (second) factor.

## Lesson

The word

*factoring*means to break down a number or expression into smaller pieces, or factors. These factors can be multiplied together to obtain the original number.

When algebraic expressions, the simplest kind of problems are those where each of the terms in the expression can be reduced by a common factor.

**Example 1****:** Find a common factor in each expression. Then simplify the expression.

*Scroll over the problem to see the answer.

The same concept can also be used on expressions with 3 or more terms. Choose a factor that can be divided into each term to in order to simplify the expression.

**Example 2****:** Find a common factor in each expression containing 3 or more terms. Then simplify the expression.

*Scroll over the problem to see the answer.

In future lessons, you will factor more complicated expressions such as

- 8x
^{3}y^{5}+ 4x^{2}y in which each term has a common factor of 4x^{2}y - x
^{2}+ 6x + 5 which can be done using another factoring method

The method outlined in this lesson works for problems whose terms contain a single variable and have a common factor. Other lessons cover the above (bulleted) methods of factoring.

## Try It

Find a common factor for each of the terms. Then factor the expression.

1) 3d^{2} + 9

2) 15e^{3} + 7e^{2}

3) 4f^{2 }+ 20f

4) 3g^{5} + 3g^{4} + 3g^{3}

5) 21h^{3} + 49h

6) i^{4} + 3i^{3} + 11i^{2}

7) 75j^{4} + 66j^{3} + 45j^{2}

Scroll down for answers:

Answers:

1) 3d^{2} + 9 = 3(d^{2} + 3)

2) 15e^{3} + 7e^{2} = e^{2}(15e + 7)

3) 4f^{2 }+ 20f = 4f(f + 5)

4) 3g^{5} + 3g^{4} + 3g^{3} = 3g^{3}(g^{2} + g + 1)

5) 21h^{3} + 49h = 7h(3h^{2} + 7)

6) i^{4} + 3i^{3} + 11i^{2} = i^{2}(i^{2} + 3i + 11)

7) 75j^{4} + 66j^{3} + 45j^{2} = 3j^{2}(25j^{2} + 22j + 15)

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