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5.4: Solving Quadratics using Factoring

Difficulty Level: Advanced Created by: CK-12
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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 h=5t2+9t+2, where t is the time in seconds. How long does it take the ball to hit the ground?

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Watch the first part of this video, until about 4:40.

Khan Academy: Solving Quadratic Equations by Factoring.avi


In this lesson we have not actually solved for x. 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. Let’s go through an example.

Example A

Solve x29x+18=0 by factoring.

Solution: The only difference between this problem and previous ones from the concepts before is the addition of the = sign. Now that this is present, we need to solve for x. We can still factor the way we always have. Because a=1, determine the two factors of 18 that add up to -9.


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 ab=0, then a=0 or b=0.

This means that x6=0 OR x3=0. Therefore, x=6 or x=3. There will always be the same number of solutions as factors.

Check your answer:


Example B

Solve 6x2+x4=11 by factoring.

Solution: 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.


Now, factor. The product of ac is -90. What are the two factors of -90 that add up to 1? 10 and -9. Expand the xterm and factor.


Lastly, set each factor equal to zero and solve.

2x32xx=0 3x+5=0=3or3x=5=32x=53

Check your work:

6(32)2+324694+324272+324154=11  6(53)2534=11=11or  6259534=11=11503534=11=11154=11

Example C

Solve 10x225x=0 by factoring.

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


Set the two factors equal to zero and solve.

5xx=02x5=0=0or  2x=5x=52


10(0)225(0)=010(52)225(52)=0  0=0or102541252=012521252=0

Intro Problem Revisit When the ball hits the ground, the height h is 0. So the equation becomes 0=5t2+9t+2.

Let's factor and solve for t. 5t2+9t+2

We need to find the factors of 10 that add up to 9. Testing the possibilities, we find 10 and -1 to be the correct combination.

5t2+10tt+2 = (5t2+10t)+(t+2) = 5t(t+2)+(t+2) = (5t+1)(t+2)

Now set this factorization equal to zero and solve.


Because t represents the time, it must be positive. Only (t+2)=0 results in a positive value.

t=2, therefore it takes the ball 2 seconds to reach the ground.

Guided Practice

Solve the following equations by factoring.

1. 4x212x+9=0

2. x25x=6

3. 8x20x2=0

4. 12x2+13x+7=124x


1. ac=36. The factors of 36 that also add up to -12 are -6 and -6. Expand the xterm and factor.


The factors are the same. When factoring a perfect square trinomial, the factors will always be the same. In this instance, the solutions for x will also be the same. Solve for x.


When the two factors are the same, we call the solution for x a double root because it is the solution twice.

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


Because there is no number in front of x2, we need to find the factors of -6 that add up to -5.


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*}.

3. 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*}

4. 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*}


The answer to an equation. With quadratic equations, solutions can also be called zeros or roots.
Double Root
A solution that is repeated twice.


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?

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Double Root

A solution that is repeated twice.


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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