Download Algebra Quadratic Equations and the Zero Product Property

Algebra
Objective: To solve equations in factored form
Quadratic Equations and the Zero Product Property
A polynomial is in factored form if it is written as the product of two or more factors.
The following polynomial equations are not in factored form:
x2 – 7x = 0
x2 + 7x + 10 = 0
x3 – 3x2 – 24x = 0
The following polynomial equations are in factored form:
x(x – 7) = 0
(x + 5)(x + 2) = 0
x(x – 8)(x + 5) = 0
A value of x that makes any of the factors zero is a solution of the polynomial equation. This is based on the following property:
Zero-Product Property
For all numbers x and y, if xy = 0 then x = 0 or y = 0.
This means that if the product of any two factors is zero, that at least one of the factors must be zero.
This section focuses on solving equations that are already in factored form, set equal to zero. Therefore, we will use the zero-product
property to determine all the solutions of the polynomial equations. We need to determine what value of the unknown will make that
factor equal to zero.
Example 2
Example 1
Solve: (x – 2)(x + 3) = 0
Solve: 2x(x – 7) = 0
Either: x – 2 = 0 or x + 3 = 0
2x = 0 or x – 7 = 0
So: x = 2 or x = -3
x = 0 or x = 7
Solution: {-3, 2}
Solution: {0, 7}
Example 3
Example 4
Solve: (x + 9)2 = 0
Solve: (2x + 1)(3x – 2)(x – 1) = 0
(x + 9)(x + 9) = 0
2x + 1 = 0 or 3x – 2 = 0 or x – 1 = 0
x + 9 = 0 or x + 9 = 0
2x = -1 or 3x = 2 or x = 1
x = -9
x= −
Solution: {-9}
2
1
or x =
or x = 1
3
2
 1 2 
Solution: − , ,1
 2 3 
Solve. Leave answers in set notation, { }.
1.
(y + 5)( y – 7) = 0
2.
(n + 1)(n + 9) = 0
3.
15n(n + 15) = 0
4.
2x(x – 20) = 0
5.
(2t – 3)(3t – 2) = 0
6.
(2w + 3)(3w + 2) = 0
7.
(2m – 3)(3m + 2) = 0
8.
(2n + 3)(3n – 2) = 0
9.
(4u + 7)(5u – 1) = 0
10.
3x(2x + 1)(2x + 5) = 0
11.
n(5n – 2)(2n + 5) = 0
12.
(y + 3)2 = 0
13.
4(y – 11)(y – 8) = 0
14.
8(i + 9)(i – 9) = 0
15.
p(6p – 5) = 0
18.
(y – 3)2(4y – 3) = 0
16.
2
(4c + 9) = 0
17.
2
(5c – 6) = 0