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9.14: Solving Problems by Factoring

Difficulty Level: At Grade Created by: CK-12
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What if you had a triangle in which one leg was one unit shorter than the other and the hypotenuse was 5 units? How could you find the length of the two legs? After completing this Concept, you'll be able to solve real-world applications like this one that involve factoring polynomial equations.

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CK-12 Foundation: 0914S Solving Real-World Problems By Factoring

Guidance

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

Example A

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 (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 B

The product of two positive numbers is 60. Find the two numbers if one number 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 (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 C

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 (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.

Watch this video for help with the Examples above.

CK-12 Foundation: Solving Real-World Problems by Factoring

Guided Practice

Consider the rectangle in Example C with 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 now 20?

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)=20\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& =20\\ x^2+2x-35& =0\end{align*}

Factor by finding two numbers that multiply to -35 and add to 2. List some numbers that multiply to -35:

\begin{align*}-35 &= -7 \cdot 5 && \text{and} && -7 + 5 = -2 \\ -35 &= 7 \cdot (-5) && \text{and} && 7 + (-5) = 2\end{align*}

The expression factors as \begin{align*}(x+7)(x-5)=0\end{align*}.

Set each term equal to zero and solve:

\begin{align*}& x+7=0 &&&& x-5=0\\ & && \text{or}\\ & \underline{\underline{x=-7}} &&&& \underline{\underline{x=5}}\end{align*}

Since we are looking for positive numbers the answer must be \begin{align*}x = 5\end{align*}. So the width is \begin{align*}x - 3 = 2\end{align*} and the length is \begin{align*}x + 5 = 10\end{align*}.

Check: \begin{align*}2 \cdot 10 = 20\end{align*}, so the answer checks.

Explore More

Solve the following application problems:

  1. One leg of a right triangle is 1 feet longer than the other leg. The hypotenuse is 5. Find the dimensions of the right triangle.
  2. 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.
  3. 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?
  4. A rectangle has sides of \begin{align*}x - 1\end{align*} and \begin{align*}x + 1\end{align*}. What value of \begin{align*}x\end{align*} gives an area of 120?
  5. The product of two positive numbers is 120. Find the two numbers if one numbers is 7 more than the other.
  6. A rectangle has a 50-foot diagonal. What are the dimensions of the rectangle if it is 34 feet longer than it is wide?
  7. Two positive numbers have a sum of 8, and their product is equal to the larger number plus 10. What are the numbers?
  8. Two positive numbers have a sum of 8, and their product is equal to the smaller number plus 10. What are the numbers?
  9. 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?

Answers for Explore More Problems

To view the Explore More answers, open this PDF file and look for section 9.14. 

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Oct 01, 2012
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Apr 14, 2016
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