Factoring Polynomial Expressions

Previously, you learned how to factor the following types of quadratic expressions.

Type Example

General trinomial

Perfect square trinomial

Difference of two squares

Common monomial factor

2 x 2 – 5x – 12 = (2 x +3)(x – 4)

x 2 + 10x + 25 = ( x + 5) 2

4x 2 – 9 = (2 x + 3) (2 x – 3)

6 x 2 + 15x = 3x(2 x + 5)

In this lesson you will learn how to factor other types of polynomials.

Steps to Factoring

1. Always factor out what is common to all terms.

2. If four terms, try grouping two and two and factoring out what is common.

3. If three terms and a quadratic form (x2 + x + c or x4 + x2 + c or x6 + x3 + c. . .), try to reverse the distributive property.

4. If two terms, check to see if you have a special product

2 2 ( )( )a b a b a b

In the previous slide you discovered how to factor the difference of two cubes. This factorization and the factorization of the sum of two cubes are given below.

SPECIAL FACTORING PATTERNS

Sum of Two Cubes

Difference of Two Cubes

a 3 – b 3 = (a – b)(a 2 + a b + b 2)

a 3 + b 3 = (a + b)(a 2 – a b + b 2)

Example

x 3 + 8 = (x + 2)(x 2 – 2 x + 4)

8x 3 – 1 = (2 x – 1)(4 x 2 + 2 x + 1)

Factoring Polynomial Expressions

Factoring the Sum or Difference of Cubes

Factor each polynomial.

x 3 + 27

SOLUTION

x 3 + 27 = x 3 + 3 3

= (x + 3)(x 2 – 3x + 9)

Sum of two cubes.

Factoring the Sum or Difference of Cubes

Factor each polynomial.

x 3 + 27

SOLUTION

x 3 + 27 = x 3 + 3 3 Sum of two cubes.

= (x + 3)(x 2 – 3x + 9)

16 u 5 – 250 u 2

16 u 5 – 250 u 2 = 2 u 2(8 u 3 – 125)

= 2 u 2[(2 u) 3 – 5 3]

= 2 u 2(2 u – 5)(4 u 2 + 10 u +25)

Factor common monomial.

Difference of two cubes

Factoring by Grouping

For some polynomials, you can factor by grouping pairs of terms that have a common monomial factor.

Factor the polynomial x 3 – 2 x 2 – 9x + 18.

SOLUTION

x 3 – 2 x 2 – 9x + 18 = x 2(x – 2) – 9(x – 2) Factor by grouping.

= (x 2 – 9)(x – 2)

Difference of squares.= (x + 3) (x – 3)(x – 2)

Factoring Polynomials in Quadratic Form

An expression of the form a u 2 + b u + c where u is any expression in x is said to be in quadratic form. These can sometimes be factored using techniques you already know.

Factor each polynomial

SOLUTION

81x 4 – 16

81x 4 – 16 = (9x 2) 2 – 4 2

= (9x 2 + 4)(9x 2 – 4)

= (9x 2 + 4)(3x +2)(3x – 2)

Factoring Polynomials in Quadratic Form

An expression of the form a u 2 + b u + c where u is any expression in x is said to be in quadratic form. These can sometimes be factored using techniques you already know.

Factor each polynomial

SOLUTION

81x 4 – 16

81x 4 – 16 = (9x 2) 2 – 4 2

= (9x 2 + 4)(9x 2 – 4)

= (9x 2 + 4)(3x +2)(3x – 2)

4x 6 – 20x 4 + 24x 2

SOLUTION

4x 6 – 20x 4 + 24x 2 = 4x 2(x 4 – 5x 2 + 6)

= 4x 2(x 2 – 2)(x 2 – 3)

Solving Polynomial Equations by Factoring

Solve 2 x 5 + 24x = 14x 3 by using the zero product property.

SOLUTION

2 x 5 + 24x = 14x 3 Write original equation.

2 x 5 – 14x 3 + 24x = 0 Rewrite in standard form.

2 x (x 2 – 3)(x 2 – 4) = 0

Factor common monomial.2 x (x 4 – 7x 2 + 12) = 0

Factor trinomial.

2 x (x 2 – 3)(x + 2)(x – 2) = 0 Factor difference of squares.

Zero product propertyx = 0, x = 3, x = – 3, x = –2, or x = 2

The solutions are 0, 3, – 3, –2, and 2. Check these in the original equation.

Solving a Polynomial Equation in Real Life

Archeology In 1980 archeologists at the ruins of Caesara discovered a huge hydraulic concrete block with a volume of 330 cubic yards. The block’s dimensions are x yards high by 13x – 11 yards long by 13x – 15 yards wide. What is the height?

Verbal Model

Labels

SOLUTION

Volume = Height • Length • Width

Volume = 330

Height = x

Length = 13x – 11

Width = 13x – 15

(cubic yards)

(yards)

(yards)

(yards)

330 = x (13x – 11)(13x – 5)Algebraic Model

Solving a Polynomial Equation in Real Life

Archeology In 1980 archeologists at the ruins of Caesara discovered a huge hydraulic concrete block with a volume of 330 cubic yards. The block’s dimensions are x yards high by 13x – 11 yards long by 13x – 15 yards wide. What is the height?

SOLUTION

330 = x (13x – 11)(13x – 5)

0 = 169x 3 – 338 x 2 + 165 x – 330

0 = 169x 2(x – 2) + 165(x – 2)

0 = (169x 2 + 165)(x – 2)

The only real solution is x = 2, so 13x – 11 = 15 and 13x – 15 = 11. The block is 2 yards high. The dimensions are 2 yards by 15 yards by 11 yards.

Write in standard form.

Factor by grouping.