# 9.7: Factoring Polynomials Completely

**At Grade**Created by: CK-12

We say that a polynomial is **factored completely** when we factor as much as we can and we are unable to factor any more. Here are some suggestions that you should follow to make sure that you factor completely.

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**Example 1:** *Factor the following polynomials completely.*

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

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

**Solution:**

(a) Look for the common monomial factor. \begin{align*}2x^2-8=2(x^2-4)\end{align*}

(b) Recognize this as a perfect square and factor as \begin{align*}x(x+3)^2\end{align*}

## Factoring Common Binomials

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

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

You can see that the term \begin{align*}(3x+2)\end{align*}

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

This expression is now completely factored. Let’s look at some examples.

**Example 2:** *Factor* \begin{align*}3x(x-1)+4(x-1)\end{align*}

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

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

## Factoring by Grouping

It may be possible to factor a polynomial containing four or more terms by factoring common monomials from groups of terms. This method is called **factoring by grouping**. The following example illustrates how this process works.

**Example 3:** *Factor* \begin{align*}2x+2y+ax+ay\end{align*}

**Solution:** There isn't a common factor for all four terms in this example. However, there is a factor of 2 that is common to the first two terms and there is a factor of \begin{align*}a\end{align*}

\begin{align*}2x+2y+ax+ay = 2(x+y) +a(x+y)\end{align*}

Now we notice that the binomial \begin{align*}(x+y)\end{align*}

\begin{align*}(x+y)(2+a)\end{align*}

Our polynomial is now factored completely.

We know how to factor Quadratic Trinomials \begin{align*}(ax^2+bx+c)\end{align*}

- We find the product \begin{align*}ac\end{align*}
ac . - We look for two numbers that multiply to give \begin{align*}ac\end{align*}
ac and add to give \begin{align*}b\end{align*}b . - We rewrite the middle term using the two numbers we just found.
- We factor the expression by grouping.

Let’s apply this method to the following examples.

**Example 4:** *Factor* \begin{align*}3x^2+8x+4\end{align*}*by grouping*.

**Solution:** Follow the steps outlined above.

\begin{align*}ac=3 \cdot 4=12\end{align*}

The number 12 can be written as a product of two numbers in any of these ways:

\begin{align*}12&=1 \times 12 && and && 1+12=13\\
12 & =2 \times 6 && and && 2+6=8 \qquad \text{This is the correct choice}.\end{align*}

Rewrite the middle term as: \begin{align*}8x=2x+6x\end{align*}

\begin{align*}3x^2+8x+4 = 3x^2+2x+6x+4\end{align*}

Factor an \begin{align*}x\end{align*}

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

Now factor the common binomial \begin{align*}(3x+2)\end{align*}

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

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

In this example, all the coefficients are positive. What happens if the \begin{align*}b\end{align*}

**Example 5:** *Factor* \begin{align*}6x^2-11x+4\end{align*}*by grouping*.

**Solution:** \begin{align*}ac=6 \cdot 4 = 24\end{align*}

The number 24 can be written as a product of two numbers in any of these ways.

\begin{align*}24&=1\times 24 && and && 1+24=25\\
24&=(-1) \times (-24) && and && (-1)+(-24)=-25\\
24&=2 \times 12 && and && 2+12=14\\
24&=(-2) \times (-12) && and && (-2)+(-12)=-14\\
24&=3 \times 8 && and && 3+8=11\\
24&=(-3) \times (-8) && and && (-3)+(-8)=-11 \quad \text{This is the correct choice}.\end{align*}

Rewrite the middle term as \begin{align*}-11x=-3x-8x\end{align*}, so the problem becomes:

\begin{align*}6x^2-11x+4=6x^2-3x-8x+4\end{align*}

Factor by grouping. Factor a \begin{align*}3x\end{align*} from the first two terms and factor –4 from the last two terms.

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

Now factor the common binomial \begin{align*}(2x-1)\end{align*}.

Our answer is \begin{align*}(2x-1)(3x-4)\end{align*}.

## Solving Real-World Problems Using Polynomial Equations

Now that we know most of the factoring strategies for quadratic polynomials, we can see how these methods apply to solving real-world problems.

**Example 6:** *The product of two positive numbers is 60. Find the two numbers if one of the numbers is 4 more than the other.*

**Solution:** \begin{align*}x=\end{align*} one of the numbers and \begin{align*}x+4\end{align*} equals the other number. The product of these two numbers equals 60. We can write the equation.

\begin{align*}x(x+4)=60\end{align*}

Write the polynomial in standard form.

\begin{align*}x^2+4x&=60\\ x^2+4x-60&=0\end{align*}

**Factor:** \begin{align*}-60=6 \times(-10)\end{align*} *and* \begin{align*}6+(-10)=-4\end{align*}

\begin{align*}-60=-6 \times 10\end{align*} *and* \begin{align*}-6+10=4\end{align*} This is the correct choice.

The expression factors as \begin{align*}(x+10)(x-6)=0\end{align*}.

**Solve:**

\begin{align*}x+10=0 && x-6& =0\\ \text{or} \\ x=-10 && x& =6\end{align*}

Since we are looking for positive numbers, the answer must be positive.

\begin{align*}x=6\end{align*} for one number, and \begin{align*}x+4=10\end{align*} for the other number.

**Check:** \begin{align*}6 \cdot 10=60\end{align*} so the answer checks.

## Practice Set

Sample explanations for some of the practice exercises below are available by viewing the following video. Note that there is not always a match between the number of the practice exercise in the video and the number of the practice exercise listed in the following exercise set. However, the practice exercise is the same in both.

CK-12 Basic Algebra: Factor by Grouping and Factoring Completely (13:57)

Factor completely.

- \begin{align*}2x^2+16x+30\end{align*}
- \begin{align*}12c^2-75\end{align*}
- \begin{align*}-x^3+17x^2-70x\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*}
- \begin{align*}2x^2-512\end{align*}
- \begin{align*}12x^3+12x^2+3x\end{align*}

Factor by grouping.

- \begin{align*}6x^2-9x+10x-15\end{align*}
- \begin{align*}5x^2-35x+x-7\end{align*}
- \begin{align*}9x^2-9x-x+1\end{align*}
- \begin{align*}4x^2+32x-5x-40\end{align*}
- \begin{align*}12x^3-14x^2+42x-49\end{align*}
- \begin{align*}4x^2+25x-21\end{align*}
- \begin{align*}24b^3+32b^2-3b-4\end{align*}
- \begin{align*}2m^3+3m^2+4m+6\end{align*}
- \begin{align*}6x^2+7x+1\end{align*}
- \begin{align*}4x^2+8x-5\end{align*}
- \begin{align*}5a^3-5a^2+7a-7\end{align*}
- \begin{align*}3x^2+16x+21\end{align*}
- \begin{align*}4xy+32x+20y+160\end{align*}
- \begin{align*}10ab+40a+6b+24\end{align*}
- \begin{align*}9mn+12m+3n+4\end{align*}
- \begin{align*}4jk-8j^2+5k-10j\end{align*}
- \begin{align*}24ab+64a-21b-56\end{align*}

Solve the following application problems.

- One leg of a right triangle is seven feet longer than the other leg. The hypotenuse is 13 feet. Find the dimensions of the right triangle.
- A rectangle has sides of \begin{align*}x+2\end{align*} and \begin{align*}x-1\end{align*}. What value of \begin{align*}x\end{align*} gives an area of 108?
- The product of two positive numbers is 120. Find the two numbers if one numbers is seven more than the other.
- Framing Warehouse offers a picture-framing service. The cost for framing a picture is made up of two parts. The cost of glass is $1 per square foot. The cost of the frame is $2 per linear foot. If the frame is a square, what size picture can you get framed for $20.00?

**Mixed Review**

- The area of a square varies directly with its side length.
- Write the general variation equation to model this sentence.
- If the area is 16 square feet when the side length is 4 feet, find the area when \begin{align*}s=1.5 \ feet\end{align*}.

- The
**surface area**is the total amount of surface of a three-dimensional figure. The formula for the surface area of a cylinder is \begin{align*}SA=2 \pi r^2+2 \pi rh\end{align*}, where \begin{align*}r=radius\end{align*} and \begin{align*}h=height \ of \ the \ cylinder\end{align*}. Determine the surface area of a soup can with a radius of 2 inches and a height of 5.5 inches. - Factor \begin{align*}25g^2-36\end{align*}. Solve this polynomial when it equals zero.
- What is the greatest common factor of \begin{align*}343r^3 t, 21t^4\end{align*}, and \begin{align*}63rt^5\end{align*}?
- Discounts to the hockey game are given to groups with more than 12 people.
- Graph this solution on a number line
- What is the domain of this situation?
- Will a church group with 12 members receive a discount?