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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 \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 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 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) 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*} as a product of two binomials: \begin{align*}(x \ )(x \ )\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*} as a difference of squares. We factor it as \begin{align*}(x+2)(x-2)\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*} as a perfect square and factor it as \begin{align*}(x+3)^2\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 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) 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*} 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 parentheses is a difference of squares.

We factor it and get \begin{align*}((x+2)(x-2))^2\end{align*}, which we can rewrite as \begin{align*}(x+2)^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 3

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*}.

Factor by Grouping

Sometimes, we can 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 is no factor common to all the terms. However, the first two terms have a common factor of 2 and the last two terms have a common factor of \begin{align*}a\end{align*}. Factor 2 from the first two terms and factor \begin{align*}a\end{align*} from the last two terms:

\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*} is common to both terms. We factor the common binomial and get:

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

Example 5

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

Solution

We factor 3 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 of the form \begin{align*}ax^2+bx+c\end{align*}, where \begin{align*}a \neq 1\end{align*}.

A quadratic like this doesn’t factor as \begin{align*}(x \pm m)(x \pm n)\end{align*}, so it’s not as simple as looking for two numbers that multiply to \begin{align*}c\end{align*} and add up to \begin{align*}b\end{align*}. Instead, we also have to take into account the coefficient in the first term.

To factor a quadratic polynomial where \begin{align*}a \neq 1\end{align*}, we follow these steps:

  1. We find the product \begin{align*}ac\end{align*}.
  2. We look for two numbers that multiply to \begin{align*}ac\end{align*} and add up to \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 \qquad This \ is \ the \ correct \ choice.\\ 12 &= 3 \cdot 4 && \text{and} && 3 + 4 = 7\end{align*}

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

\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 a 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) \qquad This \ is \ the \ answer.\end{align*}

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

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

The solution 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 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: \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 a -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) \qquad This \ is \ the \ answer.\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 This \ is \ the \ correct \ choice.\end{align*}

Step 3: Re-write the middle term: \begin{align*}-6x = -x - 5x\end{align*}, so 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 \begin{align*}a - 1\end{align*} 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) \qquad This \ is \ the \ answer.\end{align*}

Solve Real-World Problems Using Polynomial Equations

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

Example 7

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

Solution

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

Use the Pythagorean Theorem: \begin{align*}a^2+b^2=c^2\end{align*}, where \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. When we substitute the values from the diagram, we get \begin{align*}x^2+(x+3)^2=15^2\end{align*}.

In order to solve this equation, we need to get the polynomial in standard form. We must first distribute, collect like terms and rewrite in the form “polynomial = 0.”

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

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

To factor the trinomial inside the parentheses, we need two numbers that multiply to -108 and add to 3. It would take a long time to go through all the options, so let’s start by trying some of the bigger factors:

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

We factor the expression 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}\\ & \underline{\underline{x=9}} &&&& \underline{\underline{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 \begin{align*}x = 9\end{align*}. That means the short leg is 9 feet and the long leg is 12 feet.

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

Example 8

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

Solution

Let \begin{align*}x =\end{align*} one of the numbers; then \begin{align*}x + 4\end{align*} is the other number.

The product of these two numbers is 60, so 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 rewrite:

\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 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}\\ & \underline{\underline{x=-10}} &&&& \underline{\underline{x=6}}\end{align*}

Since we are looking for positive numbers, the answer must be \begin{align*}x = 6\end{align*}. One number is 6, and the other number is 10.

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

Example 9

A rectangle has sides of length \begin{align*}x + 5\end{align*} and \begin{align*}x - 3\end{align*}. What is \begin{align*}x\end{align*} if the area of the rectangle is 48?

Solution

Make a sketch of this situation:

Using the formula Area = length \begin{align*}\times\end{align*} width, we have \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:

\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 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}\\ & \underline{\underline{x=-9}} &&&& \underline{\underline{x=7}}\end{align*}

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

Resources

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.

Review Questions

Factor completely.

  1. \begin{align*}2x^2+16x+30\end{align*}
  2. \begin{align*}5x^2-70x+245\end{align*}
  3. \begin{align*}-x^3+17x^2-70x\end{align*}
  4. \begin{align*}2x^4-512\end{align*}
  5. \begin{align*}25x^4-20x^3+4x^2\end{align*}
  6. \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*}
  5. \begin{align*}2a^2-6ab+3ab-9b^2\end{align*}
  6. \begin{align*}5x^2+15x-2xy-6y\end{align*}

Factor the following quadratic trinomials 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*}
  5. \begin{align*}6x^2-2x-4\end{align*}
  6. \begin{align*}8x^2-14x-15\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. 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 an 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. A rectangle has a 50-foot diagonal. What are the dimensions of the rectangle if it is 34 feet longer than it is wide?
  5. Two positive numbers have a sum of 8, and their product is equal to the larger number plus 10. What are the numbers?
  6. Two positive numbers have a sum of 8, and their product is equal to the smaller number plus 10. What are the numbers?
  7. Framing Warehouse offers a picture framing service. The cost for framing a picture is made up of two parts: glass costs $1 per square foot and the frame costs $2 per foot. If the frame has to be a square, what size picture can you get framed for $20?

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