We are asked to factor each quadratic equation and use the Zero Product Property to find the roots. The four quadratic equations are: a. $0 = x^2 - 7x + 12$ b. $0 = 6x^2 - 23x + 20$ c. $0 = x^2 - 9$ d. $0 = x^2 + 12x + 36$

1. Problem Description

We are asked to factor each quadratic equation and use the Zero Product Property to find the roots. The four quadratic equations are:

a. $0 = x^2 - 7x + 12$

b. $0 = 6x^2 - 23x + 20$

c. $0 = x^2 - 9$

d. $0 = x^2 + 12x + 36$

2. Solution Steps

a. $0 = x^2 - 7x + 12$

We need to find two numbers that multiply to 12 and add to -

7. These numbers are -3 and -

4. $0 = (x - 3)(x - 4)$

Using the Zero Product Property, we set each factor equal to zero:

$x - 3 = 0$ or $x - 4 = 0$

$x = 3$ or $x = 4$

b. $0 = 6x^2 - 23x + 20$

We need to factor this quadratic. We look for two numbers that multiply to $6 \times 20 = 120$ and add to -

2

3. These numbers are -8 and -

1

5. $0 = 6x^2 - 15x - 8x + 20$

$0 = 3x(2x - 5) - 4(2x - 5)$

$0 = (3x - 4)(2x - 5)$

Using the Zero Product Property, we set each factor equal to zero:

$3x - 4 = 0$ or $2x - 5 = 0$

$3x = 4$ or $2x = 5$

$x = \frac{4}{3}$ or $x = \frac{5}{2}$

c. $0 = x^2 - 9$

This is a difference of squares.

$0 = (x - 3)(x + 3)$

Using the Zero Product Property, we set each factor equal to zero:

$x - 3 = 0$ or $x + 3 = 0$

$x = 3$ or $x = -3$

d. $0 = x^2 + 12x + 36$

This is a perfect square trinomial.

$0 = (x + 6)(x + 6)$

$0 = (x + 6)^2$

Using the Zero Product Property, we set the factor equal to zero:

$x + 6 = 0$

$x = -6$

3. Final Answer

a. $x = 3, 4$

b. $x = \frac{4}{3}, \frac{5}{2}$

c. $x = 3, -3$

d. $x = -6$