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Solving Problems by Factoring

Factor and use the zero product rule

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Solving Quadratics using Factoring

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?

Factoring Quadratics

In this lesson we have not actually solved for \begin{align*}x\end{align*}. Now, we will apply factoring to solving a quadratic equation. It adds one additional step to the end of what you have already been doing. 

 

Solve the following problems by factoring

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

The only difference between this problem and previous ones from the concepts before 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.

Check your answer:

\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*}

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

 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*}

Check your work:

\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*}

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

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 how long does it take 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 than #2. 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?

Answers for Review Problems

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

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Vocabulary

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.

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