Solve quadratic equations by factoring using the zero product property.

Problem Overview:

The task involves solving quadratic equations using the Zero Product Property. The Zero Product Property states that if the product of two factors is zero, then at least one of the factors must be zero. Mathematically, if \( A \cdot B = 0 \), then either \( A = 0 \) or \( B = 0 \).

We are given 10 quadratic equations in factored form, and we need to solve each equation by setting each factor equal to zero.

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Step-by-Step Solution:

#### 1. \( (4d - 7)(8d - 16) = 0 \)

Set each factor equal to zero:
- \( 4d - 7 = 0 \)
\[
4d = 7 \implies d = \frac{7}{4}
\]

- \( 8d - 16 = 0 \)
\[
8d = 16 \implies d = 2
\]

So, the solutions are:
\[
d = \frac{7}{4}, \, d = 2
\]

#### 2. \( (5p - 6)(3p + 15) = 0 \)

Set each factor equal to zero:
- \( 5p - 6 = 0 \)
\[
5p = 6 \implies p = \frac{6}{5}
\]

- \( 3p + 15 = 0 \)
\[
3p = -15 \implies p = -5
\]

So, the solutions are:
\[
p = \frac{6}{5}, \, p = -5
\]

#### 3. \( (9x + 2)(7x + 9) = 0 \)

Set each factor equal to zero:
- \( 9x + 2 = 0 \)
\[
9x = -2 \implies x = -\frac{2}{9}
\]

- \( 7x + 9 = 0 \)
\[
7x = -9 \implies x = -\frac{9}{7}
\]

So, the solutions are:
\[
x = -\frac{2}{9}, \, x = -\frac{9}{7}
\]

#### 4. \( (24t + 60)(12t - 24) = 0 \)

Set each factor equal to zero:
- \( 24t + 60 = 0 \)
\[
24t = -60 \implies t = -\frac{60}{24} = -\frac{5}{2}
\]

- \( 12t - 24 = 0 \)
\[
12t = 24 \implies t = 2
\]

So, the solutions are:
\[
t = -\frac{5}{2}, \, t = 2
\]

#### 5. \( (11h - 55)(2h + 10) = 0 \)

Set each factor equal to zero:
- \( 11h - 55 = 0 \)
\[
11h = 55 \implies h = 5
\]

- \( 2h + 10 = 0 \)
\[
2h = -10 \implies h = -5
\]

So, the solutions are:
\[
h = 5, \, h = -5
\]

#### 6. \( (2g - 16)(6g - 36) = 0 \)

Set each factor equal to zero:
- \( 2g - 16 = 0 \)
\[
2g = 16 \implies g = 8
\]

- \( 6g - 36 = 0 \)
\[
6g = 36 \implies g = 6
\]

So, the solutions are:
\[
g = 8, \, g = 6
\]

#### 7. \( (5v - 5)(8v + 7) = 0 \)

Set each factor equal to zero:
- \( 5v - 5 = 0 \)
\[
5v = 5 \implies v = 1
\]

- \( 8v + 7 = 0 \)
\[
8v = -7 \implies v = -\frac{7}{8}
\]

So, the solutions are:
\[
v = 1, \, v = -\frac{7}{8}
\]

#### 8. \( (6r + 3)(10r - 1) = 0 \)

Set each factor equal to zero:
- \( 6r + 3 = 0 \)
\[
6r = -3 \implies r = -\frac{3}{6} = -\frac{1}{2}
\]

- \( 10r - 1 = 0 \)
\[
10r = 1 \implies r = \frac{1}{10}
\]

So, the solutions are:
\[
r = -\frac{1}{2}, \, r = \frac{1}{10}
\]

#### 9. \( (3z + 10)(4z + 5) = 0 \)

Set each factor equal to zero:
- \( 3z + 10 = 0 \)
\[
3z = -10 \implies z = -\frac{10}{3}
\]

- \( 4z + 5 = 0 \)
\[
4z = -5 \implies z = -\frac{5}{4}
\]

So, the solutions are:
\[
z = -\frac{10}{3}, \, z = -\frac{5}{4}
\]

#### 10. \( (u - 4)(u - 8) = 0 \)

Set each factor equal to zero:
- \( u - 4 = 0 \)
\[
u = 4
\]

- \( u - 8 = 0 \)
\[
u = 8
\]

So, the solutions are:
\[
u = 4, \, u = 8
\]

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Final Answers:

\[
\boxed{
\begin{aligned}
1. & \quad d = \frac{7}{4}, \, d = 2 \\
2. & \quad p = \frac{6}{5}, \, p = -5 \\
3. & \quad x = -\frac{2}{9}, \, x = -\frac{9}{7} \\
4. & \quad t = -\frac{5}{2}, \, t = 2 \\
5. & \quad h = 5, \, h = -5 \\
6. & \quad g = 8, \, g = 6 \\
7. & \quad v = 1, \, v = -\frac{7}{8} \\
8. & \quad r = -\frac{1}{2}, \, r = \frac{1}{10} \\
9. & \quad z = -\frac{10}{3}, \, z = -\frac{5}{4} \\
10. & \quad u = 4, \, u = 8
\end{aligned}
}
\]