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# 9.7: Factoring Polynomials Completely

Difficulty Level: At Grade Created by: CK-12

## Learning Objectives

• Factor out a common binomial.
• Factor by grouping.
• Factor a quadratic trinomial where a1\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) 6x230x+24\begin{align*}6x^2 - 30x + 24\end{align*}

b) 2x28\begin{align*}2x^2 - 8\end{align*}

c) x3+6x2+9x\begin{align*}x^3 + 6x^2 + 9x\end{align*}

Solution

a) 6x230x+24\begin{align*}6x^2 - 30x + 24\end{align*}

Factor the common monomial. In this case 6 can be factored from each term.

6(x25x+6)\begin{align*}6(x^2 - 5x + 6)\end{align*}

There are no special products. We factor x25x+6\begin{align*}x^2 - 5x + 6\end{align*} as a product of two binomials (x±)(x±)\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.

6(x25x+6)=6(x2)(x3)\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 6(x2)(x3)\begin{align*}6(x - 2) (x - 3)\end{align*}

b) 2x28\begin{align*}2x^2 - 8\end{align*}

Factor common monomials 2x28=2(x24).\begin{align*}2x^2 - 8 = 2(x^2 - 4).\end{align*}

We recognize x24\begin{align*}x^2 - 4\end{align*} as a difference of squares. We factor as 2(x24)=2(x+2)(x2).\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 2(x+2)(x2).\begin{align*}2(x + 2) (x - 2).\end{align*}

c) x3+6x2+9x\begin{align*}x^3 + 6x^2 + 9x\end{align*}

Factor common monomials x3+6x2+9x=x(x2+6x+9).\begin{align*}x^3 + 6x^2 + 9x = x(x^2 + 6x + 9).\end{align*}

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

Example 2

Factor the following polynomials completely.

a) 2x4+162\begin{align*}-2x^4 + 162\end{align*}

b) x58x3+16x\begin{align*}x^5 - 8x^3 + 16x\end{align*}

Solution

a) 2x4+162\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.

2x4+162=2(x481)\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.

2(x29)(x2+9)\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.

2(x+3)(x3)(x2+9)\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 2(x+3)(x3)(x2+9)\begin{align*}-2(x + 3) (x - 3) (x^2 + 9)\end{align*}

b) x58x3+16x\begin{align*}x^5 - 8x^3 + 16x\end{align*}

Factor out the common monomial x58x3+14x=x(x48x2+16)\begin{align*}x^5 - 8x^3 + 14x = x(x^4 - 8x^2 + 16)\end{align*}.

We recognize x48x2+16\begin{align*}x^4 - 8x^2 + 16\end{align*} as a perfect square and we factor it as x(x24)2\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: x[(x+2)2(x2)]2=x(x+2)2(x2)2.\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 (x+2)2(x2)\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: x(x+2)2(x2)2.\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.

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

You can see that the term (3x+2)\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.

(3x+2)(x5)\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) 3x(x1)+4(x1)\begin{align*}3x(x - 1) + 4(x - 1)\end{align*}

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

Solution

a) 3x(x1)+4(x1)\begin{align*}3x(x - 1) + 4(x - 1)\end{align*} has a common binomial of (x1)\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.

1. We find the product \begin{align*}ac\end{align*}.
2. We look for two numbers that multiply to give \begin{align*}ac\end{align*} and add to give \begin{align*}b\end{align*}.
3. We rewrite the middle term using the two numbers we just found.
4. 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*}

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.

1. \begin{align*}2x^2 + 16x + 30\end{align*}
2. \begin{align*}-x^3 + 17x^2 - 70x\end{align*}
3. \begin{align*}2x^2 - 512\end{align*}
4. \begin{align*}12x^3 + 12x^2 + 3x\end{align*}

Factor by grouping.

1. \begin{align*}6x^2 - 9x + 10x - 15\end{align*}
2. \begin{align*}5x^2 - 35x + x - 7\end{align*}
3. \begin{align*}9x^2 - 9x - x + 1\end{align*}
4. \begin{align*}4x^2 + 32x - 5x- 40\end{align*}

Factor the following quadratic binomials by grouping.

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

Solve the following application problems:

1. 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.
2. 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?
3. The product of two positive numbers is 120. Find the two numbers if one numbers is 7 more than the other.
4. 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?

1. \begin{align*}2 (x + 3) (x + 5)\end{align*}
2. \begin{align*}-x (x - 7) (x - 10)\end{align*}
3. \begin{align*}2(x - 4) (x + 4) (x^2 + 16)\end{align*}
4. \begin{align*}3x (2x + 1)^2\end{align*}
5. \begin{align*}(2x - 3) (3x + 5)\end{align*}
6. \begin{align*}(x - 7) (5x + 1)\end{align*}
7. \begin{align*}(9x - 1) (x - 1)\end{align*}
8. \begin{align*}(x + 8) (4x - 5)\end{align*}
9. \begin{align*}(4x - 3) (x + 7)\end{align*}
10. \begin{align*}(6x + 1) (x + 1)\end{align*}
11. \begin{align*}(2x - 1) (2x + 5)\end{align*}
12. \begin{align*}(x + 3) (3x + 7)\end{align*}
13. \begin{align*}\text{Leg} \ 1 = 5, \text{Leg} \ 2 = 12\end{align*}
14. \begin{align*}x = 10\end{align*}
15. Numbers are 8 and 15.
16. 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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