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# Solving Problems by Factoring

## Factor and use the zero product rule

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Practice Solving Problems by Factoring
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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 $h = -5t^2 + 9t + 2$ , 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.

### 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 $x^2-9x+18=0$ 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.

$x^2-9x+18 &= 0\\(x-6)(x-3) &= 0$

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 $x-6 = 0$ OR $x-3 = 0$ . Therefore, $x = 6$ or $x = 3$ . There will always be the same number of solutions as factors.

$6^2-9(6)+18 &=0 \quad or \quad 3^2-9(3)+18=0\\36-54+18 &=0 \qquad \quad \quad 9-27+18=0$

#### Example B

Solve $6x^2+x-4=11$ 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.

$& \ 6x^2+x-4 = \bcancel{11}\\& \underline{\; \; \; \; \; \; \; \; \; \; \; \; \; -11 = - \bcancel{11} \; \;}\\& 6x^2+x-15=0$

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 $x-$ term and factor.

$6x^2+x-15 &= 0\\6x^2-9x+10x-15 &= 0\\3x(2x-3)+5(2x-3) &= 0\\(2x-3)(3x+5) &= 0$

Lastly, set each factor equal to zero and solve.

$2x-3 &=0 \qquad \ 3x+5 = 0\\2x &=3 \quad or \quad \quad 3x=-5\\x &=\frac{3}{2} \qquad \qquad x=-\frac{5}{3}$

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

#### Example C

Solve $10x^2-25x=0$ 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.

$10x^2-25x &= 0\\5x(2x-5) &= 0$

Set the two factors equal to zero and solve.

$5x &=0 \qquad 2x-5=0\\x &=0 \quad or \quad \ \ 2x=5\\& \qquad \qquad \qquad x=\frac{5}{2}$

Check:

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

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

Let's factor and solve for t . $-5t^2 + 9t + 2$

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.

$-5t^2 + 10t - t + 2$ = $(-5t^2 + 10t)+ (-t + 2)$ = $5t(-t + 2)+ (-t + 2)$ = $(5t + 1)(-t + 2)$

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. $4x^2-12x+9=0$

2. $x^2-5x=6$

3. $8x-20x^2=0$

4. $12x^2+13x+7=12-4x$

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

$4x^2-12x+9 &= 0\\4x^2-6x-6x+9 &= 0\\2x(2x-3)-3(2x-3) &= 0\\(2x-3)(2x-3) &=0$

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 $x$ .

$2x-3 &= 0\\2x &= 3\\x &= \frac{3}{2}$

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.

$x^2-5x &= 6\\x^2-5x-6 &= 0$

Because there is no number in front of $x^2$ , we need to find the factors of -6 that add up to -5.

$(x-6)(x+1)=0$

Solving each factor for $x$ , we get that $x = 6$ or $x = -1$ .

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

$8x-20x^2 &= 0\\4x(2-5x) &= 0$

Solve each factor for $x$ .

$4x &=0 \qquad 2-5x=0\\x &=0 \quad or \qquad \ 2=5x\\& \qquad \qquad \qquad \frac{2}{5}=x$

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.

$12x^2+13x+7 &= 12-4x\\12x^2+17x-5 &= 0$

$ac = -60$ . The factors of -60 that add up to 17 are 20 and -3. Expand the $x-$ term and factor.

$12x^2+17x-5 &= 0\\12 x^2+20x-3x-5 &= 0\\4x(3x+5)-1(3x+5) &=0\\(3x+5)(4x-1) &= 0$

Solve each factor for $x$ .

$3x+5 &=0 \qquad 4x-1 = 0 \\3x &= -5 \quad or \quad 4x=1\\x &= -\frac{5}{3} \qquad \quad x = \frac{1}{4}$

### 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. $x^2+8x-9=0$
2. $x^2+6x=0$
3. $2x^2-5x=12$
4. $12x^2+7x-10=0$
5. $x^2=9$
6. $30x+25=-9x^2$
7. $2x^2+x-5=0$
8. $16x=32x^2$
9. $3x^2+28x=-32$
10. $36x^2-48=1$
11. $6x^2+x=4$
12. $5x^2+12x+4=0$

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

1. $8x^2+8x-5=10-6x$
2. $-18x^2=48x+14$
3. $36x^2-24=96x-39$
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?

### Vocabulary Language: English

Double Root

Double Root

A solution that is repeated twice.
solution

solution

A solution to an equation or inequality should result in a true statement when substituted for the variable in the equation or inequality.