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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?

Watch This

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

Khan Academy: Solving Quadratic Equations by Factoring.avi

Guidance

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.

x29x+18(x6)(x3)=0=0

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:

629(6)+183654+18=0or329(3)+18=0=0927+18=0

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.

 6x2+x4=1111=116x2+x15=0

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.

6x2+x156x29x+10x153x(2x3)+5(2x3)(2x3)(3x+5)=0=0=0=0

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.

10x225x5x(2x5)=0=0

Set the two factors equal to zero and solve.

5xx=02x5=0=0or  2x=5x=52

Check:

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.

(5t+1)(t+2)=0

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

Answers

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

4x212x+94x26x6x+92x(2x3)3(2x3)(2x3)(2x3)=0=0=0=0

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.

2x32xx=0=3=32

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.

x25xx25x6=6=0

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

(x6)(x+1)=0

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*}

Vocabulary

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

Practice

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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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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