Quadratic Equations, the Zero Factor Theorem, and Factoring

Quadratic equation:
- any equation that can be written in the form π‘Žπ‘₯ 2 + 𝑏π‘₯ + 𝑐 = 0
o 𝑏 and/or 𝑐 can equal zero, but π‘Ž cannot
o if π‘Ž = 0 the equation is NOT quadratic
- the degree of a quadratic equation is always 2
- the following are all examples of quadratic equations
o 2π‘₯ 2 βˆ’ 3π‘₯ βˆ’ 5 = 0
o βˆ’π‘₯ 2 + 4π‘₯ = 0
1
3
o 2 π‘₯2 + 4 = 0
o βˆ’3π‘₯ 2 = 0
Zero Factor Theorem:
- the product of two or more factors is zero if and only if at least one of
the factors is zero
o π‘₯𝑦 = 0 if and only if π‘₯ = 0 or 𝑦 = 0
o (π‘₯ βˆ’ 2)(π‘₯ + 3) = 0 if and only if π‘₯ βˆ’ 2 = 0 or π‘₯ + 3 = 0
ο‚§ this means that π‘₯ = 2 or π‘₯ = βˆ’3, so there are two solutions
o this Theorem does NOT work for any other numbers but zero
ο‚§ if π‘₯ βˆ™ 𝑦 = 1, there are an infinite number of possible values
for the factors π‘₯ and 𝑦
When an equation is factorable, we set it equal to zero so we can use the
Zero Factor Theorem to solve it.
Steps for Solving an Equation by Factoring:
1. write the equation as a polynomial and set it equal to zero
2. factor the polynomial (review the Steps for Factoring if needed)
3. use Zero Factor Theorem to solve
Example 1: Solve the following equations for π‘₯ and enter exact answers
only (no decimal approximations). If there is more than one solution,
separate your answers with commas. If there are no real solutions, enter
NO SOLUTION.
a. 15π‘₯ 2 βˆ’ 2 = 13π‘₯
b. π‘₯(3π‘₯ + 10) = 8
Quadratic equations are not the only type of equation that can be solved by
factoring. Other polynomial equations such as 2π‘₯ 4 βˆ’ 168π‘₯ 2 + 486 = 0
(which we will see in a future lesson) that are not quadratic can still be
solved by factoring.
Example 2: Solve the following equations for π‘₯ and enter exact answers
only (no decimal approximations). If there is more than one solution,
separate your answers with commas. If there are no real solutions, enter
NO SOLUTION.
a. π‘₯ =
b. b
15βˆ’4π‘₯ 2
17
c. 10 + 17π‘₯ = 6π‘₯ 2
d. d
b. 26π‘₯ + 24 = 5π‘₯ 2
d. (π‘₯ βˆ’ 1)(π‘₯ βˆ’ 2) = 6
e. π‘₯ 2 βˆ’ 25 = 0
f. (3π‘₯ )2 βˆ’ 16 = 0
Again, this method of solving equations can be used to solve more than
just quadratic equations, as we will see in future lessons.
Answers to Examples:
2
2
3
4
1a. π‘₯ = βˆ’ 15 , 1 ; 1b. π‘₯ = βˆ’4, 3 ; 2a. π‘₯ = βˆ’5, 4 ; 2b. π‘₯ = 5 , 6 ;
1 10
2c. π‘₯ = βˆ’ 2 ,
3
4 4
; 2d. π‘₯ = βˆ’1, 4 ; 2e. = βˆ’5, 5 ; 2f. π‘₯ = βˆ’ 3 , 3 ;