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:
ax2 + bx + c = 0
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:
4x2 + 7x - 15 = 0
5x2 + 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
Let's look at a few examples.
Example 1: Solve each quadratic equation using factoring.
x2 + 3x  = 18
Step 1) Write the quadratic equation in standard form. We want to subtract 18 away from each side of the equation:
x2 + 3x - 18  = 0
Step 2) Factor the left side:
x2 + 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 = 0
36 - 18 - 18 = 0
36 - 36 = 0
0 = 0
Plug in a 3 for x:
(3)2 + 3(3) - 18 = 0
9 + 9 - 18 = 0
18 - 18 = 0
0 = 0
Example 2: Solve each quadratic equation using factoring.
3x2 - 5  = -14x
Step 1) Write the quadratic equation in standard form. We want to add 14x to both sides of the equation:
3x2 + 14x - 5  = 0
Step 2) Factor the left side:
3x2 + 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  = 0
3(25) - 70 - 5 = 0
75 - 75 = 0
0 = 0
Plug in a 1/3 for x:
3(1/3)2 + 14(1/3) - 5  = 0
3(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. $$x^{2}+ 2x - 8=0$$

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

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

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