Using Factoring to Solve Problems

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

Real-World Application: Right Triangles

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.

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.

Real-World Application: Finding Unknown Numbers 

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

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.

Real-World Application: Area

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?

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.

Source: cK-12, https://www.ck12.org/algebra/applications-of-factoring/lesson/Solving-Problems-by-Factoring-ALG-I/
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