The document discusses the zero factor theorem and provides examples of using it to solve quadratic equations. The zero factor theorem states that if p and q are algebraic expressions, then pq = 0 if and only if p = 0 or q = 0. This means a quadratic equation can be solved by factoring it into two linear factors and setting each factor equal to zero. Five examples are provided that show factoring quadratic equations, applying the zero factor theorem to set the factors equal to zero, and solving for the roots.

1. By: Ms. G. Martin

2. Zero Factor Theorem
If p and q are algebraic expressions, then
pq = 0 if and only if p = 0 or q = 0.
This theorem states that if ax2 + bx + c = 0 can
be written as a product of two first-degree
polynomials, then the solutions can be found by
setting each factor equal to zero.

3. Example 1. Solve 3x2 + 6x = 0
Steps Solution
1. Write the equation in general
form.
3x2 + 6x = 0
2. Factor the quadratic equation
on the left side of the equation.
3x(x + 2) = 0
3. Use the Zero Product Property 3x(x + 2) = 0
3x = 0 and x + 2 = 0
4. Solve for the roots. 3𝑥
3
=
0
3
and x = 0 - 2
x = 0 and x = - 2

4. Checking:
The roots of 3x2 + 6x = 0 are x = 0 and x = -2
When = 0
3x2 + 6x = 0
3(0)2 + 6(0) = 0
0 + 0 = 0
0 = 0
When x = -2
3x2 + 6x = 0
3(-2)2 + 6(-2) = 0
3(4) – 12 = 0
12 – 12 = 0
0 = 0

5. Example 2. Solve for the roots of 4x2 = 12x.
Steps Solution
1. Write in standard form. 4x2 – 12x = 0
2. Factor the quadratic
expression
4x (x – 3) = 0
3. Use the Zero Product Property 4x(x – 3) = 0
4x = 0 and x – 3 = 0
4. Solve for the roots 4𝑥
4
=
0
4
and x = 0 + 3
x = 0 and x = 3
Therefore, the roots of 4x2 = 12x are x = 0 and x = 3

6. Steps Solution
1. Write in general form. x2 – 5x – 14 = 0
2. Factor the quadratic.
Since it is a quadratic trinomial
square, look for the factor of c
which gives the result of b.
c = (-14) = -7 and 2
b = (-5) = -7 + 2
x2 = x . x
(x – 7) (x + 2) = 0
3. Apply the Zero Factor
Theorem.
(x – 7) = 0 and (x + 2) = 0
4. Solve for the roots. x = 7 and x = -2
Example 3. Solve for the solution set of x2 – 5x =
14

7. Checking:
Therefore, the solution sets are 7 and -2.
When x = 7
x2 – 5x = 14
72 – 5(7) = 14
49 – 35 = 14
14 = 14
When x = -2
x2 – 5x = 14
(-2)2 – 5(-2) = 14
4 + 10 = 14
14 = 14

8. Example 4. Find the roots of x2 = 10x – 25.
Steps Solution
1. Write in general term x2 – 10x + 25 = 0
2. Factor the equation c = (25) = -5 . -5
b = (-10) = (-5) + (-5)
x2= x . x
(x – 5) (x – 5) = 0
3. Apply the zero factor theorem.
Solve the equation.
(x – 5) (x – 5) = 0
x = 5 and x = 5
This equation has one solution, since x – 5 appears as a factor twice in the
solution. It is called double root or root of multiplicity 2 of the equation x2 = 10x –
25.

9. Example 5. Solve x(x + 2) – 17 = 18.
Steps Solution
1. Simplify.
Write in general form.
x(x + 2) – 17 = 18
x2 + 2x – 17 – 18 = 0
x2 + 2x – 35 = 0
2. Factor the equation c = (-35) = 7 . -5
b = (2) = (7) + (-5)
x2 = x . X
(x + 7 (x – 5) = 0
3. Apply the zero factor theorem (x + 7) = 0 and (x – 5) = 0
4. Solve for the solution x = – 7 and x = 5

10. THANK YOU!