1. Lesson 7.4 Solving polynomial equations in factored form
EQ: How do you solve an equation in factored form?
Objective: To use zero product property to solve factored equations
Zero Product property: If a • b = 0, then either a = 0 or b = 0.
Examples: What value of x would give a product of zero.
(x – 2)5 = 0
x = 2
x = –3
(x + 3)7 = 0
(7 – x)2 = 0
x = 7
x = 2
(3x – 6)5 = 0

2. Examples: Determine the values for x that makes each equation true.
Either x + 3 = 0 or x – 2 = 0
1. (x + 3) (x – 2) = 0
–3 –3
x = –3
x = 2
Check –3 and 2
(–3 + 3)(–3 – 2) = 0
(0)(–5) = 0
(2 + 3) (2 – 2) = 0
Solutions –3 and 2
(5)(0) = 0
The graph of y = (x + 3)(x – 2) would be a parabola that crosses the x-axis at (–3,0) and (2,0) Two roots/solutions

3. Ex: Solve for x
2x + 4 = or x – 3 = 0
(2x + 4)(x – 3) = 0
– 4 – 4
x = 3
2x = –4
x = –2
Solutions: –2 and 3
EX:
is equal to (x + 5)(x + 5) = 0
(x + 5) is repeated. There will be only one answer.
x + 5 = 0
–5 –5
The parabola touches the x-axis once at the point (–5,0)
x = –5

4. Factor and solve the equation.
Factoring is the opposite of the distributive property where the expression is factored.
Factor x(x + 3) = 0
Zero Product prop x = 0 or x + 3 = 0
Roots are 0 and -3
x = 0 or x = -3
The solutions to polynomial equations are called roots.

5. Factor k(5k + 4) = 0
Zero product prop k = 0 or 5k + 4 = 0