powerpoint

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!