Equations and Factoring

The Zero Product Theorems

 Zero Product Theorem  For Numbers If a and b are numbers then            ab = 0  implies that           a = 0 or b = 0

We have a similar theorem for polynomials:

 Zero Product Theorem For Functions If f(x) and g(x) are functions and            f(x) g(x)  =  0  then either      1.  f(x) = 0    or     2.  g(x) = 0

Example

Solve

(x - 4) (x + 3)  =  0

We have either

x - 4  =  0     or    x + 3  =  0

Hence the two solutions to (x - 4)(x + 3)  =  0 are

x = 4     or     x = -3

Exercises

Solve the following.

1. (x - 2) (x + 3)  =  0

2. (x + 4) (x - 5)  =  0

3. (3x + 4) (2 - x)  =  0

4. (5 + x) (2x + 3)  =  0

5. x (3x - 5)  =  0

Factoring and the Zero Product Rule

If we are given a trinomial = 0 then we first factor then we use the zero product rule.

Example

Solve

x2 - 3x + 2  =  0

We first factor:

(x - 2) (x - 1)  =  0

Then we use the zero product rule to get

x - 2  =  0     or     x - 1  =  0

Hence

x = 2     or     x = 1

Exercises

Solve the following.

1. x2 + 4x - 45 = 0

2. x2 - 2x + 1 = 0

3. 10x2 - 47x - 15 = 0

4. 6x2 + 17x + 10 = 0

Step by Step Process for Solving Using the Zero Product Method For Solving Quadratics

1. Put everything on the left hand side of the equation.

2. FOIL and combine like terms.

3. Multiply both sides of the equation by -1 of the leading coefficient is negative.

4. Factor

5. Use the Zero Product Theorem For Polynomials

6. Solve each of the factors

Example

Solve

21 = (x - 9) (x + 11)

Solution

1. 21 - (x - 9) (x + 11)  =  0

2. 21 - (x2 + 11x - 9x - 99)  =  0

21 - x2 - 2x + 99  =  0

120 - x2  - 2x  =  0

3. x2 + 2x - 120  =  0

4. (x - 10)(x + 12)  =  0

5. x - 10  =  0     or     x + 12  =  0

6. x = 10     or     x = -12

Exercises

Solve.

1. (2x - 5)(x + 4)  =  (3x + 12)

2. (7x - 6)(5x + 4)  =  2x(5x - 1) + 57

3. 2x4  =  35x2  - 3x3

Application

If air resistance is neglected, then the distance s in feet that an object falls in t seconds is given by

s  =  16t2

A bolt falls from a bridge 144 feet high.  How long does it take to hit the ground?

Solution

We set

144  =  16t2

16t2 - 144  =  0

16(t2 - 9)  =  0

16(t + 3)(t - 3)  =  0

t = -3     or     t = 3

Since a negative value for t is meaningless in the problem, we can conclude that it takes three seconds to hit the ground.

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