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Factoring Completely

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What if you had a polynomial like 3x^2 - 27 with multiple factors? How could you factor it completely? After completing this Concept, you'll be able to factor out common monomials and binomials from polynomials like this one.

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CK-12 Foundation: 0912S Factoring Polynomials Completely

The WTAMU Virtual Math Lab has a detailed page on factoring polynomials here: http://www.wtamu.edu/academic/anns/mps/math/mathlab/col_algebra/col_alg_tut7_factor.htm . This page contains many videos showing example problems being solved.

Guidance

We say that a polynomial is factored completely when we can’t factor it any more. Here are some suggestions that you should follow to make sure that you factor completely:

  • Factor all common monomials first.
  • Identify special products such as difference of squares or the square of a binomial. Factor according to their formulas.
  • If there are no special products, factor using the methods we learned in the previous sections.
  • Look at each factor and see if any of these can be factored further.

Example A

Factor the following polynomials completely.

a) 6x^2-30x+36

b) 2x^2-8

c) x^3+6x^2+9x

Solution

a) Factor out the common monomial. In this case 6 can be divided from each term:

6(x^2-5x+6)

There are no special products. We factor x^2-5x+6 as a product of two binomials: (x \ )(x \ )

The two numbers that multiply to 6 and add to -5 are -2 and -3, so:

6(x^2-5x+6)=6(x-2)(x-3)

If we look at each factor we see that we can factor no more.

The answer is 6(x-2)(x-3) .

b) Factor out common monomials: 2x^2-8=2(x^2-4)

We recognize x^2-4 as a difference of squares. We factor it as (x+2)(x-2) .

If we look at each factor we see that we can factor no more.

The answer is 2(x+2)(x-2) .

c) Factor out common monomials: x^3+6x^2+9x=x(x^2+6x+9)

We recognize x^2+6x+9 as a perfect square and factor it as (x+3)^2 .

If we look at each factor we see that we can factor no more.

The answer is x(x+3)^2 .

Example B

Factor the following polynomials completely:

a) -2x^4+162

b) x^5-8x^3+16x

Solution

a) Factor out the common monomial. In this case, factor out -2 rather than 2. (It’s always easier to factor out the negative number so that the highest degree term is positive.)

-2x^4+162=-2(x^4-81)

We recognize expression in parenthesis as a difference of squares. We factor and get:

-2(x^2-9)(x^2+9)

If we look at each factor we see that the first parenthesis is a difference of squares. We factor and get:

-2(x+3)(x-3)(x^2+9)

If we look at each factor now we see that we can factor no more.

The answer is -2(x+3)(x-3)(x^2+9) .

b) Factor out the common monomial: x^5-8x^3+14x=x(x^4-8x^2+16)

We recognize x^4-8x^2+16 as a perfect square and we factor it as x(x^2-4)^2 .

We look at each term and recognize that the term in parentheses is a difference of squares.

We factor it and get ((x+2)(x-2))^2 , which we can rewrite as (x+2)^2(x-2)^2 .

If we look at each factor now we see that we can factor no more.

The final answer is x(x+2)^2(x-2)^2 .

Factor out a Common Binomial

The first step in the factoring process is often factoring out the common monomials from a polynomial. But sometimes polynomials have common terms that are binomials. For example, consider the following expression:

x(3x+2)-5(3x+2)

Since the term (3x+2) appears in both terms of the polynomial, we can factor it out. We write that term in front of a set of parentheses containing the terms that are left over:

(3x+2)(x-5)

This expression is now completely factored.

Let’s look at some more examples.

Example C

Factor out the common binomials.

a) 3x(x-1)+4(x-1)

b) x(4x+5)+(4x+5)

Solution

a) 3x(x-1)+4(x-1) has a common binomial of (x-1) .

When we factor out the common binomial we get (x-1)(3x+4) .

b) x(4x+5)+(4x+5) has a common binomial of (4x+5) .

When we factor out the common binomial we get (4x+5)(x+1) .

Watch this video for help with the Examples above.

CK-12 Foundation: Factoring Polynomials Completely

Vocabulary

  • We say that a polynomial is factored completely when we factor as much as we can and we are unable to factor any more.

Guided Practice

Factor completely: 24x^3-28x^2+8x .

Solution:

First, notice that each term has 4x as a factor. Start by factoring out 4x :

24x^3-28x^2+8x=4x(6x^2-7x+2)

Next, factor the trinomial in the parenthesis. Since a\neq 1 find a\cdot c :  6\cdot 2=12 . Find the factors of 12 that add up to -7. Since 12 is positive and -7 is negative, the two factors should be negative:

 12&=-1\cdot -12 && and && -1+-12=-13\\12&=-2\cdot -6 && and && -2+-6=-8\\12&=-3\cdot -4 && and && -3+-4=-7

Rewrite the trinomial using -7x=-3x-4x , and then factor by grouping:

6x^2-7x+2 &= 6x^2-3x-4x+2\\ &=3x(2x-1)-2(2x-1)\\ &=(3x-2)(2x-1)

The final factored answer is:

4x(3x-2)(2x-1)

Explore More

Factor completely.

  1. 2x^2+16x+30
  2. 5x^2-70x+245
  3. -x^3+17x^2-70x
  4. 2x^4-512
  5. 25x^4-20x^3+4x^2
  6. 12x^3+12x^2+3x
  7. 12c^2-75
  8. 6x^2-600
  9. -5t^2-20t-20
  10. 6x^2+18x-24
  11. -n^2+10n-21
  12. 2a^2-14a-16

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