# 9.7: Factoring Polynomials Completely

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

## Learning Objectives

- Factor out a common binomial.
- Factor by grouping.
- Factor a quadratic trinomial where \begin{align*}a \neq 1\end{align*}.
- Solve real world problems using polynomial equations.

## Introduction

We say that a polynomial is **factored completely** when we factor as much as we can and we can’t factor 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.

Here are some examples

**Example 1**

*Factor the following polynomials completely.*

a) \begin{align*}6x^2 - 30x + 24\end{align*}

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

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

**Solution**

a) \begin{align*}6x^2 - 30x + 24\end{align*}

Factor the common monomial. In this case 6 can be factored 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*} as a product of two binomials \begin{align*}(x \pm \underline{\;\;\;\;\;\;\;} )(x \pm \underline{\;\;\;\;\;\;\;})\end{align*}.

The two numbers that multiply to 6 and add to -5 are -2 and -3. Let's substitute them into the two parenthesis. The 6 is outside because it is factored out.

\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't factor anything else.

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

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

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

We recognize \begin{align*}x^2 - 4\end{align*} as a difference of squares. We factor as \begin{align*}2(x^2 - 4) = 2(x + 2) (x - 2).\end{align*}

If we look at each factor we see that we can't factor anything else.

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

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

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

We recognize as a perfect square and factor as \begin{align*}x (x + 3)^2.\end{align*}

If we look at each factor we see that we can't factor anything else.

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

**Example 2**

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

Factor the common monomial. In this case, factor -2 rather than 2. It is always easier to factor the negative number so that the leading 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 this result.

\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 this answers.

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

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

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

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

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*} as a perfect square and we factor it as \begin{align*} x(x^2 - 4)^2\end{align*}.

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

We factor and get: \begin{align*}x[(x + 2)^2 (x - 2)]^2 = x(x + 2)^2 (x - 2)^2.\end{align*}

We use square brackets “[” and “]” in this expression because x is multiplied by the expression \begin{align*}(x + 2)^2 (x - 2)\end{align*}. When we have “nested” grouping symbols we use brackets “[” and “]” to show the levels of nesting.

If we look at each factor now we see that we can't factor anything else.

The 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 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*} appears in both term of the polynomial. This common term can be factored by writing it in front of a parenthesis. Inside the parenthesis, we write all the terms that are left over when we divide them by the common factor.

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

This expression is now completely factored.

Let’s look at some more examples.

**Example 3**

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

## Factor 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 **factor by grouping**.

The next example illustrates how this process works.

**Example 4**

*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 a that is common to the last two terms. Factor 2 from the first two terms and factor a from the last two terms.

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

Now we notice that the binomial \begin{align*}(x + y)\end{align*} is common to both terms. We factor the common binomial and get.

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

Our polynomial is now factored completely.

**Example 5**

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

**Solution**

We factor \begin{align*}3x\end{align*} from the first two terms and factor 4 from the last two terms.

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

Now factor \begin{align*}(x + 2)\end{align*} from both terms.

\begin{align*}(x + 2) (3x + 4).\end{align*}

Now the polynomial is factored completely.

## Factor Quadratic Trinomials Where a ≠ 1

Factoring by grouping is a very useful method for factoring quadratic trinomials where \begin{align*}a \neq 1\end{align*}. A quadratic polynomial such as this one.

\begin{align*}ax^2 + bx + c\end{align*}

This does not factor as \begin{align*}(x \pm m) (x \pm n)\end{align*}, so it is not as simple as looking for two numbers that multiply to give \begin{align*}c\end{align*} and add to give \begin{align*}b\end{align*}. In this case, we must take into account the coefficient that appears in the first term.

To factor a quadratic polynomial where \begin{align*}a \neq 1\end{align*}, we follow the following steps.

- We find the product \begin{align*}ac\end{align*}.
- We look for two numbers that multiply to give \begin{align*}ac\end{align*} and add to give \begin{align*}b\end{align*}.
- 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 6**

*Factor the following quadratic trinomials by grouping.*

a) \begin{align*}3x^2 + 8x + 4\end{align*}

b) \begin{align*}6x^2 - 11x + 4\end{align*}

c) \begin{align*} 5x^2 - 6x + 1\end{align*}

**Solution**

Let’s follow the steps outlined above.

a) \begin{align*}3x^2 + 8x + 4\end{align*}

**Step 1** \begin{align*}ac = 3 \cdot 4 = 12\end{align*}

**Step 2** The number 12 can be written as a product of two numbers in any of these ways:

\begin{align*}& 12 = 1 \cdot 12 && \text{and} && 1 + 12 = 13 \\ & 12 = 2 \cdot 6 &&\text{and} && 2 + 6 = 8 && \ \text{This is the correct choice}.\\ & 12 = 3 \cdot 4 &&\text{and} && 3 + 4 = 7 \end{align*}

**Step 3** Re-write the middle term as: \begin{align*}8x = 2x + 6x\end{align*}, so the problem becomes the following.

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

**Step 4**: Factor an \begin{align*}x\end{align*} from the first two terms and 2 from the last two terms.

\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*}.

To check if this is correct we multiply \begin{align*}(3x + 2) (x + 2)\end{align*}.

\begin{align*}& \quad \quad \qquad \ 3x + \ 2\\ & \underline{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; \ x \ + \ 2}\\ & \quad \quad \qquad 6x + \ 4\\ & \underline{3x^2 \ + \ \ \ 2x\;\;\;\;\;\;\;\;\;}\\ & 3x^2 \ + \ \ 8x \ + \ 4\end{align*}

**The answer checks out.**

b) \begin{align*}6x^2 - 11x + 4\end{align*}

**Step 1** \begin{align*}ac = 6 \cdot 4 = 24\end{align*}

**Step 2** The number 24 can be written as a product of two numbers in any of these ways.

\begin{align*}& 24 = 1 \cdot 24 && \text{and} && 1 + 24 = 25 \\ & 24 = -1 \cdot (-24) &&\text{and} && -1 + (-24) = -25 \\ & 24 = 2 \cdot 12 &&\text{and} && 2 + 12 = 14 \\ & 24 = -2 \cdot (-12) &&\text{and} && -2 + (-12) = -14 \\ & 24 = 3 \cdot 8 &&\text{and} && 3 + 8 = 11 \\ & 24 = -3 \cdot (-8) &&\text{and} && -3 + (-8) = -11 \qquad \leftarrow \qquad \text{This is the correct choice.}\\ & 24 = 4 \cdot 6 &&\text{and} && 4 + 6 = 10 \\ & 24 = -4 \cdot (-6) &&\text{and} && -4 + (-6) = -10 \end{align*}

**Step 3** Re-write 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*}

**Step 4** 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*}.

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

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

c) \begin{align*}5x^2 -6x + 1\end{align*}

**Step 1** \begin{align*}ac = 5 \cdot 1 = 5\end{align*}

**Step 2** The number 5 can be written as a product of two numbers in any of these ways.

\begin{align*}& 5 = 1 \cdot 5 &&\text{and} && 1 + 5 = 6 \\ & 5 = -1 \cdot (-5) &&\text{and} && -1 + (-5) = -6 \qquad \leftarrow \qquad \text{This is the correct choice}\end{align*}

**Step 3** Rewrite the middle term as \begin{align*}-6x = -x - 5x.\end{align*} The problem becomes

\begin{align*}5x^2 - 6x+ 1 = 5x^2 - x- 5x+ 1\end{align*}

**Step 4** Factor by grouping: factor an \begin{align*}x\end{align*} from the first two terms and a factor of -1 from the last two terms

\begin{align*}x(5x - 1) - 1(5x - 1)\end{align*}

Now factor the common binomial \begin{align*}(5x - 1).\end{align*}

\begin{align*}(5x - 1) (x - 1).\end{align*}

Our answer is \begin{align*}(5x - 1) (x - 1)\end{align*}.

## Solve 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 7 Pythagorean Theorem**

*One leg of a right triangle is 3 feet longer than the other leg. The hypotenuse is 15 feet. Find the dimensions of the right triangle*.

**Solution**

Let \begin{align*}x =\end{align*} the length of one leg of the triangle, then the other leg will measure \begin{align*}x + 3\end{align*}.

Let’s draw a diagram.

Use the Pythagorean Theorem \begin{align*}(\text{leg}_1)^2 + (\text{leg}_2)^2 = (\text{hypotenuse})^2\end{align*} or \begin{align*}a^2 + b^2 = c^2.\end{align*}

Here \begin{align*}a\end{align*} and \begin{align*}b\end{align*} are the lengths of the legs and \begin{align*}c\end{align*} is the length of the hypotenuse.

Let’s substitute the values from the diagram.

\begin{align*}a^2 + b^2 & =c^2 \\ x^2 + (x + 3)^2 & = 15^2\end{align*}

In order to solve, we need to get the polynomial in standard form. We must first distribute, collect like terms and **re-write** in the form polynomial \begin{align*}= 0\end{align*}.

\begin{align*}x^2 + x^2 + 6x + 9 & = 225 \\ 2x^2 + 6x + 9 & = 225 \\ 2x^2 + 6x - 216 & = 0\end{align*}

**Factor** the common monomial \begin{align*}2(x + 3x - 108) = 0\end{align*}.

To factor the trinomial inside the parenthesis we need to numbers that multiply to -108 and add to 3. It would take a long time to go through all the options so let’s try some of the bigger factors.

\begin{align*}&-108 = -12 \cdot && \text{and} && -12 + 9 = -3 \\ &-108 = 12 \cdot (-9) && \text{and} && 12 + (-9) = 3 \qquad \leftarrow \qquad \text{This is the correct choice}.\end{align*}

We factor as: \begin{align*}2(x - 9) (x + 12) = 0\end{align*}.

**Set each term equal to zero and solve**

\begin{align*}x - 9 = 0 &&&& x + 12 = 0 \\ && \text{or} \\ x = 9 &&&& x = -12\end{align*}

It makes no sense to have a negative answer for the length of a side of the triangle, so the answer must be the following.

**Answer** \begin{align*}x = 9\end{align*} for one leg, and \begin{align*}x + 3 = 12\end{align*} for the other leg.

**Check** \begin{align*}9^2 +12^2 = 81 + 144 = 225 = 15^2\end{align*} so the answer checks.

**Example 8 Number Problems**

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

Solution

Let \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*}

In order to solve we must write the polynomial in standard form. Distribute, collect like terms and re-write in the form polynomial \begin{align*}= 0\end{align*}.

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

**Factor** by finding two numbers that multiply to -60 and add to 4. List some numbers that multiply to -60:

\begin{align*}& -60 = -4 \cdot 15 && \text{and} && -4 + 15 = 11 \\ & -60 = 4 \cdot (-15) && \text{and} && 4 + (-15) = -11 \\ & -60 = -5 \cdot 12 &&\text{and} && -5 + 12 = 7 \\ & -60 = 5 \cdot (-12) &&\text{and} && 5 + (-12) = -7 \\ & -60 = -6 \cdot 10 &&\text{and} && -6 + 10 = 4 \qquad \leftarrow \qquad \text{This is the correct choice}\\ & -60 = 6 \cdot (-10) &&\text{and} && 6 + (-10) = -4 \end{align*}

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

**Set each term equal to zero and 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 the following.

**Answer** \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.

**Example 9 Area of a rectangle**

*A rectangle has sides of* \begin{align*}x + 5\end{align*} *and* \begin{align*}x - 3\end{align*}. *What value of* \begin{align*}x\end{align*} *gives and area of 48*?

**Solution:**

Make a sketch of this situation.

Area of the rectangle \begin{align*}= \text{length} \times \text{width}\end{align*}

\begin{align*}(x + 5)(x - 3) = 48\end{align*}

In order to solve, we must write the polynomial in standard form. Distribute, collect like terms and **rewrite** in the form polynomial \begin{align*}= 0\end{align*}.

\begin{align*}x^2 + 2x - 15 & = 48\\ x^2 + 2x - 63 & = 0\end{align*}

**Factor** by finding two numbers that multiply to -63 and add to 2. List some numbers that multiply to -63.

\begin{align*}& -63 = -7 \cdot 9 &&\text{and} && -7 + 9 = 2 \qquad \leftarrow \qquad \text{This is the correct choice}\\ &-63 = 7 \cdot (-9) &&\text{and} && 7 + (-9) = -2 \end{align*}

The expression factors as \begin{align*}(x + 9)(x - 7) = 0\end{align*}.

**Set each term equal to zero and solve.**

\begin{align*}x + 9 = 0 &&&& x - 7 = 0\\ && \text{or}\\ x = -9 &&&& x = 7\end{align*}

Since we are looking for positive numbers the answer must be \begin{align*}x = 7\end{align*}.

**Answer** The width is \begin{align*}x - 3 = 4\end{align*} and the length is \begin{align*}x + 5 = 12\end{align*}.

**Check** \begin{align*}4 \cdot 12 = 48\end{align*} so the answer checks out.

## Review Questions

Factor completely.

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

Factor the following quadratic binomials by grouping.

- \begin{align*}4x^2 + 25x - 21\end{align*}
- \begin{align*}6x^2 + 7x + 1\end{align*}
- \begin{align*}4x^2 + 8x - 5\end{align*}
- \begin{align*}3x^2 + 16x + 21\end{align*}

Solve the following application problems:

- One leg of a right triangle is 7 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 and area of 108?
- The product of two positive numbers is 120. Find the two numbers if one numbers is 7 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?

## Review Answers

- \begin{align*}2 (x + 3) (x + 5)\end{align*}
- \begin{align*}-x (x - 7) (x - 10)\end{align*}
- \begin{align*}2(x - 4) (x + 4) (x^2 + 16)\end{align*}
- \begin{align*}3x (2x + 1)^2\end{align*}
- \begin{align*}(2x - 3) (3x + 5)\end{align*}
- \begin{align*}(x - 7) (5x + 1)\end{align*}
- \begin{align*}(9x - 1) (x - 1)\end{align*}
- \begin{align*}(x + 8) (4x - 5)\end{align*}
- \begin{align*}(4x - 3) (x + 7)\end{align*}
- \begin{align*}(6x + 1) (x + 1)\end{align*}
- \begin{align*}(2x - 1) (2x + 5)\end{align*}
- \begin{align*}(x + 3) (3x + 7)\end{align*}
- \begin{align*}\text{Leg} \ 1 = 5, \text{Leg} \ 2 = 12\end{align*}
- \begin{align*}x = 10\end{align*}
- Numbers are 8 and 15.
- You can frame a 2 foot \begin{align*}\times\end{align*} 2 foot picture.

## Texas Instruments Resources

*In the CK-12 Texas Instruments Algebra I FlexBook, there are graphing calculator activities designed to supplement the objectives for some of the lessons in this chapter. See http://www.ck12.org/flexr/chapter/9619.*

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