Lesson Objectives

- Demonstrate an understanding of how to Factor Trinomials
- Learn how to write a Quadratic Equation in Standard Form
- Learn about the Zero-Product Property
- Learn how to solve a Quadratic Equation using Factoring

## How to Solve Quadratic Equations Using Factoring

Over the course of the last few lessons, we have learned to factor quadratic expressions. A quadratic expression contains a squared variable and no term with a higher degree. We will expand on this knowledge and learn how to solve a quadratic equation using factoring. A quadratic equation is an equation that contains a squared variable and no other term with a higher degree. Generally, we think about a quadratic equation in standard form:

ax

a ≠ 0 (since we must have a variable squared)

a, b, and c are any real numbers (a can't be zero)

Some examples of a quadratic equation are:

4x

5x

xy = 0

x could be 0, y could be a non-zero number

y could be 0, x could be a non-zero number

x and y could both be zero

We can apply this to more advanced examples. Suppose we saw the following:

(x - 2)(x + 3) = 0

In this case, we have a quantity (x - 2) multiplied by another quantity (x + 3). The result of this multiplication is zero. This means we can use our zero-product property. To do this, we set each factor equal to zero and solve:

(x - 2) = 0

(x + 3) = 0

x - 2 = 0

x = 2

x + 3 = 0

x = -3

Essentially, x could be 2 or x could be -3. In either scenario, the equation would be true:

Let's check x = 2:

(x - 2)(x + 3) = 0

(2 - 2)(2 + 3) = 0

0(5) = 0

0 = 0

Let's check x = -3:

(-3 - 2)(-3 + 3) = 0

(-5)(0) = 0

0 = 0

Example 1: Solve each quadratic equation using factoring.

x

Step 1) Write the quadratic equation in standard form. We want to subtract 18 away from each side of the equation:

x

Step 2) Factor the left side:

x

(x + 6)(x - 3) = 0

Step 3) Use the zero-product property and set each factor with a variable equal to zero:

x + 6 = 0

x - 3 = 0

x = -6

x = 3

We can say that x = -6, 3

This means x can be -6 or x can be 3. Either will work as a solution.

Step 4) Check the result:

Plug in a -6 for x:

(-6)

36 - 18 - 18 = 0

36 - 36 = 0

0 = 0

Plug in a 3 for x:

(3)

9 + 9 - 18 = 0

18 - 18 = 0

0 = 0

Example 2: Solve each quadratic equation using factoring.

3x

Step 1) Write the quadratic equation in standard form. We want to add 14x to both sides of the equation:

3x

Step 2) Factor the left side:

3x

(3x - 1)(x + 5) = 0

Step 3) Use the zero-product property and set each factor with a variable equal to zero:

3x - 1 = 0

x + 5 = 0

x = 1/3

x = -5

We can say that x = -5, 1/3

This means x can be -5 or x can be 1/3. Either will work as a solution.

Step 4) Check the result:

Plug in a -5 for x:

3(-5)

3(25) - 70 - 5 = 0

75 - 75 = 0

0 = 0

Plug in a 1/3 for x:

3(1/3)

3(1/9) + (14/3) - 5 = 0

1/3 + 14/3 - 5 = 0

15/3 - 5 = 0

5 - 5 = 0

0 = 0

ax

^{2}+ bx + c = 0a ≠ 0 (since we must have a variable squared)

a, b, and c are any real numbers (a can't be zero)

Some examples of a quadratic equation are:

4x

^{2}+ 7x - 15 = 05x

^{2}+ 18x + 9 = 0### Zero-Product Property

Up to this point, we have not attempted to solve an equation in which the exponent on a variable was not 1. For these types of problems, obtaining a solution can be a bit more work than what we have seen so far. When a quadratic equation is in standard form and the left side can be factored, we can solve the quadratic equation using factoring. This works based on the zero-product property (also known as the zero-factor property). The zero product property tells us if the product of two numbers is zero, then at least one of them must be zero:xy = 0

x could be 0, y could be a non-zero number

y could be 0, x could be a non-zero number

x and y could both be zero

We can apply this to more advanced examples. Suppose we saw the following:

(x - 2)(x + 3) = 0

In this case, we have a quantity (x - 2) multiplied by another quantity (x + 3). The result of this multiplication is zero. This means we can use our zero-product property. To do this, we set each factor equal to zero and solve:

(x - 2) = 0

(x + 3) = 0

x - 2 = 0

x = 2

x + 3 = 0

x = -3

Essentially, x could be 2 or x could be -3. In either scenario, the equation would be true:

Let's check x = 2:

(x - 2)(x + 3) = 0

(2 - 2)(2 + 3) = 0

0(5) = 0

0 = 0

Let's check x = -3:

(-3 - 2)(-3 + 3) = 0

(-5)(0) = 0

0 = 0

### Solving a Quadratic Equation using Factoring

- Place the quadratic equation in standard form
- Factor the left side
- Use the zero-product property and set each factor with a variable equal to zero
- Check the result

Example 1: Solve each quadratic equation using factoring.

x

^{2}+ 3x = 18Step 1) Write the quadratic equation in standard form. We want to subtract 18 away from each side of the equation:

x

^{2}+ 3x - 18 = 0Step 2) Factor the left side:

x

^{2}+ 3x - 18 » (x + 6)(x - 3)(x + 6)(x - 3) = 0

Step 3) Use the zero-product property and set each factor with a variable equal to zero:

x + 6 = 0

x - 3 = 0

x = -6

x = 3

We can say that x = -6, 3

This means x can be -6 or x can be 3. Either will work as a solution.

Step 4) Check the result:

Plug in a -6 for x:

(-6)

^{2}+ 3(-6) - 18 = 036 - 18 - 18 = 0

36 - 36 = 0

0 = 0

Plug in a 3 for x:

(3)

^{2}+ 3(3) - 18 = 09 + 9 - 18 = 0

18 - 18 = 0

0 = 0

Example 2: Solve each quadratic equation using factoring.

3x

^{2}- 5 = -14xStep 1) Write the quadratic equation in standard form. We want to add 14x to both sides of the equation:

3x

^{2}+ 14x - 5 = 0Step 2) Factor the left side:

3x

^{2}+ 14x - 5 » (3x - 1)(x + 5)(3x - 1)(x + 5) = 0

Step 3) Use the zero-product property and set each factor with a variable equal to zero:

3x - 1 = 0

x + 5 = 0

x = 1/3

x = -5

We can say that x = -5, 1/3

This means x can be -5 or x can be 1/3. Either will work as a solution.

Step 4) Check the result:

Plug in a -5 for x:

3(-5)

^{2}+ 14(-5) - 5 = 03(25) - 70 - 5 = 0

75 - 75 = 0

0 = 0

Plug in a 1/3 for x:

3(1/3)

^{2}+ 14(1/3) - 5 = 03(1/9) + (14/3) - 5 = 0

1/3 + 14/3 - 5 = 0

15/3 - 5 = 0

5 - 5 = 0

0 = 0

#### Skills Check:

Example #1

Solve each equation. $$x^{2}+ 2x - 8=0$$

Please choose the best answer.

A

$$x=5, -1$$

B

$$x=-4, 2$$

C

$$x=4, -2$$

D

$$x=5, -5$$

E

$$x=\frac{1}{2}, -3$$

Example #2

Solve each equation. $$x^{2}- 5x=6$$

Please choose the best answer.

A

$$x=-1, 6$$

B

$$x=-4, -7$$

C

$$x=1, 4$$

D

$$x=4, 2$$

E

$$x=-\frac{2}{5}, 6$$

Example #3

Solve each equation. $$3x^{2}+ 26x + 17=1$$

Please choose the best answer.

A

$$x=-\frac{1}{3}, 4$$

B

$$x=\frac{2}{3}, 8$$

C

$$x=-\frac{2}{3}, -8$$

D

$$x=-\frac{6}{5}, -2$$

E

$$x=-4, \frac{1}{3}$$

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