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# Factoring Completely

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# Factoring Completely

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.

### Watch This

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.

### 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. $2a^2-14a-16$