Lesson Objectives
• Demonstrate an understanding of how to factor a polynomial
• Learn about the zero-product property
• Learn how to solve polynomial equations by factoring

## How to Solve Equations by Factoring

As we move through our Algebra 2 course, we will develop methods for solving various polynomial equations. One such method involves factoring and the zero-product property. The zero-product property (a.k.a. zero-factor property) tells us:
if: a • b = 0
then one of the following possibilities must be true:
1) a = 0
2) b = 0
3) a = 0 and b = 0
In other words, the zero-product property is telling us if the result of multiplying two numbers together is zero, then either one of the numbers is zero, or they are both zero. We can use this property to solve equations that can be factored and set equal to zero. In most cases, we will be using this method with quadratic equations. A quadratic equation is one that contains a squared variable term and no term with any higher degree. A quadratic equation in standard form:
ax2 + bx + c = 0, a ≠ 0
(a, b, and c are real numbers)
To solve a quadratic equation by factoring, we can use the following steps.
• Write the quadratic equation in standard form
• ax2 + bx + c = 0
• Factor the left side of the equation (non-zero side)
• Use the zero-product property and set each factor with a variable equal to zero
• Solve each equation to find the solution or solutions for the quadratic equation
Let's look at a few examples.
Example 1: Solve each equation by factoring. $$7x^2 - 29=3 - 52x$$ Step 1) Write the quadratic equation in standard form: $$7x^2 + 52x - 32=0$$ Step 2) Factor the left side of the equation: $$(7x - 4)(x + 8)=0$$ Step 3) Use the zero-product property and set each factor with a variable equal to zero: $$7x - 4=0$$ $$x + 8=0$$ Step 4) Solve each equation to find the solution for the quadratic equation: $$7x - 4=0$$ $$x=\frac{4}{7}$$ $$x + 8=0$$ $$x=-8$$ The solutions for our quadratic equation are 4/7 and -8.
Example 2: Solve each equation by factoring. $$(x + 3)(x - 6)=(2x + 2)(x - 6)$$ Step 1) Write the quadratic equation in standard form:
We need to simplify each side first: $$x^2 - 3x - 18=2x^2 - 10x - 12$$ Now we can write our equation in standard form: $$-x^2 + 7x - 6=0$$ To make the leading term positive, multiply both sides by -1: $$x^2 - 7x + 6=0$$ Step 2) Factor the left side of the equation: $$(x - 6)(x - 1)=0$$ Step 3) Use the zero-product property and set each factor with a variable equal to zero: $$x - 6=0$$ $$x=6$$ $$x - 1=0$$ $$x=1$$ The solutions for our quadratic equation are 6 and 1.
Example 3: Solve each equation by factoring. $$2x^3 + x^2 - 98x - 49=0$$ This is not a quadratic equation since we have a variable raised to the 3rd power. Because the left side can be factored, we can use the same general strategy. Let's factor the left side using grouping. $$x^2(2x + 1) - 49(2x + 1)=0$$ $$(x^2 - 49)(2x + 1)=0$$ Notice our first binomial is the difference of two squares. $$(x + 7)(x - 7)(2x + 1)=0$$ We can set each factor equal to zero and solve the resulting equations. $$x + 7=0$$ $$x=-7$$ $$x - 7=0$$ $$x=7$$ $$2x + 1=0$$ $$x=-\frac{1}{2}$$ The solutions for our equation are -7, 7, and -1/2.

#### Skills Check:

Example #1

Solve each equation by factoring. $$21x^{2}- 86x + 12=x$$

A
$$x=-\frac{1}{5}, -8$$
B
$$x=-\frac{1}{7}, -2$$
C
$$x=\frac{1}{7}, -7$$
D
$$x=-\frac{2}{11}, 4$$
E
$$x=\frac{1}{7}, 4$$

Example #2

Solve each equation by factoring. $$40x^{2}- 20=-12x + 8x^{2}$$

A
$$x=-\frac{5}{8}, 1$$
B
$$x=\frac{5}{8}, -1$$
C
$$x=-\frac{2}{5}, -7$$
D
$$x=-\frac{8}{5}, 3$$
E
$$x=-\frac{13}{5}, 7$$

Example #3

Solve each equation by factoring. $$12x^{3}- 5x^{2} = 108x - 45$$

A
$$x=-5, 5, \frac{8}{9}$$
B
$$x=-12, \frac{1}{3}, \frac{2}{3}$$
C
$$x=-5, 4, 13$$
D
$$x=-3, 3, \frac{5}{12}$$
E
$$x=-13, \frac{1}{3}, \frac{5}{12}$$         