# Solving Problems by Factoring

## Factor and use the zero product rule

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The height of a ball that is thrown straight up in the air from a height of 2 meters above the ground with a velocity of 9 meters per second is given by the quadratic equation \begin{align*}h = -5t^2 + 9t + 2\end{align*}, where t is the time in seconds. How long does it take the ball to hit the ground?

To solve word problems involving quadratic equations, you will need to know how to either factor the quadratic equation and set it equal to zero or else use the quadratic formula, in order to solve for \begin{align*}x.\end{align*}

Let's solve the following quadratics by factoring.

1. \begin{align*}x^2-9x+18=0\end{align*}

The only difference between this problem and previous ones is the addition of the \begin{align*}=\end{align*} sign. Now that this is present, we need to solve for \begin{align*}x\end{align*}. We can still factor the way we always have. Because \begin{align*}a = 1\end{align*}, determine the two factors of 18 that add up to -9.

\begin{align*}x^2-9x+18 &= 0\\ (x-6)(x-3) &= 0\end{align*}

Now, we have two factors that, when multiplied, equal zero. Recall that when two numbers are multiplied together and one of them is zero, the product is always zero.

Zero-Product Property: If \begin{align*}ab = 0\end{align*}, then \begin{align*}a = 0\end{align*} or \begin{align*}b = 0\end{align*}.

This means that \begin{align*}x-6 = 0\end{align*} OR \begin{align*}x-3 = 0\end{align*}. Therefore, \begin{align*}x = 6\end{align*} or \begin{align*} x = 3\end{align*}. There will always be the same number of solutions as factors.

\begin{align*} 6^2-9(6)+18 &=0 \quad or \quad 3^2-9(3)+18=0\\ 36-54+18 &=0 \qquad \quad \quad 9-27+18=0 \end{align*}

1. \begin{align*}6x^2+x-4=11\end{align*}

At first glance, this might not look factorable to you. However, before we factor, we must combine like terms. Also, the Zero-Product Property tells us that in order to solve for the factors, one side of the equation must be zero.

\begin{align*}& \ 6x^2+x-4 = \bcancel{11}\\ & \underline{\; \; \; \; \; \; \; \; \; \; \; \; \; -11 = - \bcancel{11} \; \;}\\ & 6x^2+x-15=0\end{align*}

Now, factor. The product of \begin{align*}ac\end{align*} is -90. What are the two factors of -90 that add up to 1? 10 and -9. Expand the \begin{align*}x-\end{align*}term and factor.

\begin{align*}6x^2+x-15 &= 0\\ 6x^2-9x+10x-15 &= 0\\ 3x(2x-3)+5(2x-3) &= 0\\ (2x-3)(3x+5) &= 0\end{align*}

Lastly, set each factor equal to zero and solve.

\begin{align*}2x-3 &=0 \qquad \ 3x+5 = 0\\ 2x &=3 \quad or \quad \quad 3x=-5\\ x &=\frac{3}{2} \qquad \qquad x=-\frac{5}{3}\end{align*}

\begin{align*}6 \left(\frac{3}{2}\right)^2 +\frac{3}{2}-4 &= 11 \qquad \ \ 6 \left(- \frac{5}{3}\right)^2 -\frac{5}{3}-4 = 11 \\ 6 \cdot \frac{9}{4}+\frac{3}{2}-4 &=11 \quad or \quad \ \ 6 \cdot \frac{25}{9}-\frac{5}{3}-4 =11 \\ \frac{27}{2}+\frac{3}{2}-4 &=11 \qquad \qquad \quad \frac{50}{3}-\frac{5}{3}-4=11\\ 15-4 &=11 \qquad \qquad \qquad \quad 15-4=11\end{align*}

1. \begin{align*}10x^2-25x=0\end{align*}

Here is an example of a quadratic equation without a constant term. The only thing we can do is take out the GCF.

\begin{align*}10x^2-25x &= 0\\ 5x(2x-5) &= 0\end{align*}

Set the two factors equal to zero and solve.

\begin{align*}5x &=0 \qquad 2x-5=0\\ x &=0 \quad or \quad \ \ 2x=5\\ & \qquad \qquad \qquad x=\frac{5}{2}\end{align*}

Check:

\begin{align*}& 10(0)^2-25(0) = 0 \qquad \qquad 10\left(\frac{5}{2}\right)^2- 25 \left(\frac{5}{2}\right)=0\\ & \qquad \qquad \quad \ \ 0 = 0 \qquad or \quad \quad 10 \cdot \frac{25}{4}-\frac{125}{2}=0 \\ & \qquad \qquad \qquad \qquad \qquad \qquad \qquad \frac{125}{2} - \frac{125}{2} =0\end{align*}

### Examples

#### Example 1

Earlier, you were asked to find the time it takes the ball to hit the ground.

When the ball hits the ground, the height h is 0. So the equation becomes \begin{align*}0 = -5t^2 + 9t + 2\end{align*}

Let's factor and solve for t. \begin{align*}-5t^2 + 9t + 2\end{align*}

We need to find the factors of \begin{align*}-10\end{align*} that add up to 9. Testing the possibilities, we find 10 and -1 to be the correct combination.

\begin{align*}-5t^2 + 10t - t + 2\end{align*} = \begin{align*}(-5t^2 + 10t)+ (-t + 2)\end{align*} = \begin{align*}5t(-t + 2)+ (-t + 2)\end{align*} = \begin{align*}(5t + 1)(-t + 2)\end{align*}

Now set this factorization equal to zero and solve.

\begin{align*}(5t + 1)(-t + 2)=0\end{align*}

Because t represents the time, it must be positive. Only \begin{align*}(-t + 2)=0\end{align*} results in a positive value.

\begin{align*}t = 2\end{align*}, therefore it takes the ball 2 seconds to reach the ground.

Solve the following equations by factoring.

#### Example 2

\begin{align*}4x^2-12x+9=0\end{align*}

\begin{align*}ac = 36\end{align*}. The factors of 36 that also add up to -12 are -6 and -6. Expand the \begin{align*}x-\end{align*}term and factor.

\begin{align*}4x^2-12x+9 &= 0\\ 4x^2-6x-6x+9 &= 0\\ 2x(2x-3)-3(2x-3) &= 0\\ (2x-3)(2x-3) &=0\end{align*}

The factors are the same. When factoring a perfect square trinomial, the factors will always be the same. In this instance, the solutions for \begin{align*}x\end{align*} will also be the same. Solve for \begin{align*}x\end{align*}.

\begin{align*}2x-3 &= 0\\ 2x &= 3\\ x &= \frac{3}{2}\end{align*}

When the two factors are the same, we call the solution for \begin{align*}x\end{align*} a double root because it is the solution twice.

#### Example 3

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

Here, we need to get everything on the same side of the equals sign in order to factor.

\begin{align*}x^2-5x &= 6\\ x^2-5x-6 &= 0\end{align*}

Because there is no number in front of \begin{align*}x^2\end{align*}, we need to find the factors of -6 that add up to -5.

\begin{align*}(x-6)(x+1)=0\end{align*}

Solving each factor for \begin{align*}x\end{align*}, we get that \begin{align*}x = 6\end{align*} or \begin{align*}x = -1\end{align*}.

#### Example 4

\begin{align*}8x-20x^2=0\end{align*}

Here there is no constant term. Find the GCF to factor.

\begin{align*}8x-20x^2 &= 0\\ 4x(2-5x) &= 0\end{align*}

Solve each factor for \begin{align*}x\end{align*}.

\begin{align*}4x &=0 \qquad 2-5x=0\\ x &=0 \quad or \qquad \ 2=5x\\ & \qquad \qquad \qquad \frac{2}{5}=x\end{align*}

#### Example 5

\begin{align*}12x^2+13x+7=12-4x\end{align*}

This problem is slightly more complicated. Combine all like terms onto the same side of the equals sign so that one side is zero.

\begin{align*}12x^2+13x+7 &= 12-4x\\ 12x^2+17x-5 &= 0\end{align*}

\begin{align*}ac = -60\end{align*}. The factors of -60 that add up to 17 are 20 and -3. Expand the \begin{align*}x-\end{align*}term and factor.

\begin{align*}12x^2+17x-5 &= 0\\ 12 x^2+20x-3x-5 &= 0\\ 4x(3x+5)-1(3x+5) &=0\\ (3x+5)(4x-1) &= 0\end{align*}

Solve each factor for \begin{align*}x\end{align*}.

\begin{align*}3x+5 &=0 \qquad 4x-1 = 0 \\ 3x &= -5 \quad or \quad 4x=1\\ x &= -\frac{5}{3} \qquad \quad x = \frac{1}{4}\end{align*}

### Review

Solve the following quadratic equations by factoring, if possible.

1. \begin{align*}x^2+8x-9=0\end{align*}
2. \begin{align*}x^2+6x=0\end{align*}
3. \begin{align*}2x^2-5x=12\end{align*}
4. \begin{align*}12x^2+7x-10=0\end{align*}
5. \begin{align*}x^2=9\end{align*}
6. \begin{align*}30x+25=-9x^2\end{align*}
7. \begin{align*}2x^2+x-5=0\end{align*}
8. \begin{align*}16x=32x^2\end{align*}
9. \begin{align*}3x^2+28x=-32\end{align*}
10. \begin{align*}36x^2-48=1\end{align*}
11. \begin{align*}6x^2+x=4\end{align*}
12. \begin{align*}5x^2+12x+4=0\end{align*}

Challenge Solve these quadratic equations by factoring. They are all factorable.

1. \begin{align*}8x^2+8x-5=10-6x\end{align*}
2. \begin{align*}-18x^2=48x+14\end{align*}
3. \begin{align*}36x^2-24=96x-39\end{align*}
4. Real Life Application George is helping his dad build a fence for the backyard. The total area of their backyard is 1600 square feet. The width of the house is half the length of the yard, plus 7 feet. How much fencing does George’s dad need to buy?

To see the Review answers, open this PDF file and look for section 5.4.

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### Vocabulary Language: English

TermDefinition
Double Root A solution that is repeated twice.
solution A solution to an equation or inequality should result in a true statement when substituted for the variable in the equation or inequality.