What if you had a polynomial like \begin{align*}3x^2 - 27\end{align*}

### Watch This

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) \begin{align*}6x^2-30x+36\end{align*}

b) \begin{align*}2x^2-8\end{align*}

c) \begin{align*}x^3+6x^2+9x\end{align*}

**Solution**

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

\begin{align*}6(x^2-5x+6)\end{align*}

There are no special products. We factor \begin{align*}x^2-5x+6\end{align*}

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

\begin{align*}6(x^2-5x+6)=6(x-2)(x-3)\end{align*}

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

The answer is \begin{align*}6(x-2)(x-3)\end{align*}

b) Factor out common monomials: \begin{align*}2x^2-8=2(x^2-4)\end{align*}

We recognize \begin{align*}x^2-4\end{align*}

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

The answer is \begin{align*}2(x+2)(x-2)\end{align*}

c) Factor out common monomials: \begin{align*}x^3+6x^2+9x=x(x^2+6x+9)\end{align*}

We recognize \begin{align*}x^2+6x+9\end{align*}

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

The answer is \begin{align*}x(x+3)^2\end{align*}

#### Example B

Factor the following polynomials completely:

a) \begin{align*}-2x^4+162\end{align*}

b) \begin{align*}x^5-8x^3+16x\end{align*}

**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.)

\begin{align*}-2x^4+162=-2(x^4-81)\end{align*}

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

\begin{align*}-2(x^2-9)(x^2+9)\end{align*}

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

\begin{align*}-2(x+3)(x-3)(x^2+9)\end{align*}

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

The answer is \begin{align*}-2(x+3)(x-3)(x^2+9)\end{align*}

b) Factor out the common monomial: \begin{align*}x^5-8x^3+14x=x(x^4-8x^2+16)\end{align*}

We recognize \begin{align*}x^4-8x^2+16\end{align*}

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

We factor it and get \begin{align*}((x+2)(x-2))^2\end{align*}

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

The final answer is \begin{align*}x(x+2)^2(x-2)^2\end{align*}.

**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:

\begin{align*}x(3x+2)-5(3x+2)\end{align*}

Since the term \begin{align*}(3x+2)\end{align*} 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:

\begin{align*}(3x+2)(x-5)\end{align*}

This expression is now completely factored.

Let’s look at some more examples.

#### Example C

*Factor out the common binomials.*

a) \begin{align*}3x(x-1)+4(x-1)\end{align*}

b) \begin{align*}x(4x+5)+(4x+5)\end{align*}

**Solution**

a) \begin{align*}3x(x-1)+4(x-1)\end{align*} has a common binomial of \begin{align*}(x-1)\end{align*}.

When we factor out the common binomial we get \begin{align*}(x-1)(3x+4)\end{align*}.

b) \begin{align*}x(4x+5)+(4x+5)\end{align*} has a common binomial of \begin{align*}(4x+5)\end{align*}.

When we factor out the common binomial we get \begin{align*}(4x+5)(x+1)\end{align*}.

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: \begin{align*}24x^3-28x^2+8x\end{align*}.*

**Solution:**

First, notice that each term has \begin{align*}4x\end{align*} as a factor. Start by factoring out \begin{align*}4x\end{align*}:

\begin{align*}24x^3-28x^2+8x=4x(6x^2-7x+2)\end{align*}

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

\begin{align*} 12&=-1\cdot -12 && and && -1+-12=-13\\ 12&=-2\cdot -6 && and && -2+-6=-8\\ 12&=-3\cdot -4 && and && -3+-4=-7\end{align*}

Rewrite the trinomial using \begin{align*}-7x=-3x-4x\end{align*}, and then factor by grouping:

\begin{align*}6x^2-7x+2 &= 6x^2-3x-4x+2\\ &=3x(2x-1)-2(2x-1)\\ &=(3x-2)(2x-1)\end{align*}

The final factored answer is:

\begin{align*}4x(3x-2)(2x-1)\end{align*}

### Explore More

Factor completely.

- \begin{align*}2x^2+16x+30\end{align*}
- \begin{align*}5x^2-70x+245\end{align*}
- \begin{align*}-x^3+17x^2-70x\end{align*}
- \begin{align*}2x^4-512\end{align*}
- \begin{align*}25x^4-20x^3+4x^2\end{align*}
- \begin{align*}12x^3+12x^2+3x\end{align*}
- \begin{align*}12c^2-75\end{align*}
- \begin{align*}6x^2-600\end{align*}
- \begin{align*}-5t^2-20t-20\end{align*}
- \begin{align*}6x^2+18x-24\end{align*}
- \begin{align*}-n^2+10n-21\end{align*}
- \begin{align*}2a^2-14a-16\end{align*}