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Algebra-2

You will factor second- and third-degree algebraic expressions.

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After completing this tutorial, you will be able to complete the following:

- Factor algebraic expressions of second- and third-degree polynomials by finding a common factor for all terms.
- Factor algebraic expressions of second-degree polynomials by finding the difference between two squares.
- Factor algebraic expressions of second-degree polynomials by recognizing perfect squares of binomials.
- Factor algebraic expressions of second- degree polynomials by using quadratic trinomials.
- Factor algebraic expressions of second- and third-degree polynomials using more than one factoring method.

The Process of Factoring

To change an algebraic expression from a product to a sum, we use the distributive property. For example, an expression such as x (x + 4) tells us to multiply x by (x + 4). When we do this, we get the sum x² + 4x. When we do that in reverse, by writing x² + 4x as the product of the two factors x and (x + 4), we are factoring.

For example:

Factoring the algebraic expression 2y + 6 = 2(y + 3), so the factors of 2y + 6 are 2 and (y + 3)

Methods of Factorization

There are many methods of factoring algebraic expressions. Those included in this Activity Object are as follows:

· Common factor method: In this method, find the factors that divide all the terms in the expression.

Example:

Factor the expression 12x² + 8x.

1. Notice that 4 divides into both 12x² and 8x.

12x² + 8x = 4(3x² + 2x)

2. Notice that x divides into both 3x² and 2x.

So, the answer is:

4x(3x + 2)

· Difference of two squares method: When the sum of two numbers is multiplied by their difference, the product is the difference of their squares.

(x + a)(x - a) = (x² - xa + ax - a²)

= x² - a²

Inversely, the difference of two squares can be factored.

x² - a² = (x + a)(x - a)

Example:

Factor the expression 4x² - 9.

1. Notice that 4 and 9 are squares.

4x² - 9 = (2x)² - 3²

2. Multiply the sum of the two numbers by their difference.

So, the answer is:

(2x +3)(2x - 3)

· Perfect square method: Some quadratics are "perfect square trinomials," which can be expressed as squares of binomials.

(x + a)² = (x + a)(x + a)

= x² + xa + ax + a²

= x² + 2ax + a²

For example,

Factor the expression x² + 10x + 25

1. Notice that the first term is a perfect square (x²) and the last term is a perfect square (25). The middle term is 2 times the square root of the first and last term (10x).

2. When an algebraic expression has this form, it can be factored into a binomial expression that is squared.

So, the answer is:

· Quadratic trinomial method: This method involves trinomials whose highest power is two.

Example:

Factor the expression x² - 14x - 15

1. Write down two sets of parentheses to indicate the product.

( )

2. Since the first term in the trinomial is the product of the first terms of the binomials, enter x as the first term of each binomial.

(x)

3. The product of the last terms of the binomials must equal -15, and their sum must equal -14, and one of the binomials' terms has to be negative.

Four different pairs of factors have a product that equals -15:

(3)(-5) = -15

(-3)(5) = -15

However, only one of those pairs has a sum of -14:

Therefore, the second terms in the binomial are -15 and 1 because these are the only two factors whose product is -15 (the last term of the trinomial) and whose sum is -14 (the coefficient of the middle term in the trinomial).

So, the answer is:

(x - 15)(x + 1)

The following key vocabulary terms will be used throughout this Activity Object:

Approximate Time | 20 Minutes |

Pre-requisite Concepts | Students should be familiar with algebraic expressions, the multiplication of algebraic expressions, and polynomials. |

Course | Algebra-2 |

Type of Tutorial | Concept Development |

Key Vocabulary | factoring algebraic expressions, greatest common factor, perfect squares |