Algebra 2 Practice Worksheet - Free Printable

Problem Analysis:

The worksheet contains two types of problems:
1. Finding all zeros of given polynomial functions.
2. Stating the excluded values for given rational expressions.

Let's solve each part step by step.

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Part 1: Finding all zeros

#### Problem 1:
$$
f(x) = (x - 5)(2x + 1)
$$
To find the zeros, set \( f(x) = 0 \):
$$
(x - 5)(2x + 1) = 0
$$
Using the zero-product property, either:
$$
x - 5 = 0 \quad \text{or} \quad 2x + 1 = 0
$$
Solve each equation:
- From \( x - 5 = 0 \): \( x = 5 \)
- From \( 2x + 1 = 0 \): \( 2x = -1 \implies x = -\frac{1}{2} \)

Thus, the zeros are:
$$
\boxed{5, -\frac{1}{2}}
$$

#### Problem 2:
$$
f(x) = (x - 4)(x - 3)
$$
Set \( f(x) = 0 \):
$$
(x - 4)(x - 3) = 0
$$
Using the zero-product property:
- From \( x - 4 = 0 \): \( x = 4 \)
- From \( x - 3 = 0 \): \( x = 3 \)

Thus, the zeros are:
$$
\boxed{4, 3}
$$

#### Problem 3:
$$
f(x) = (2x - 2)(x + 5)
$$
Set \( f(x) = 0 \):
$$
(2x - 2)(x + 5) = 0
$$
Using the zero-product property:
- From \( 2x - 2 = 0 \): \( 2x = 2 \implies x = 1 \)
- From \( x + 5 = 0 \): \( x = -5 \)

Thus, the zeros are:
$$
\boxed{1, -5}
$$

#### Problem 4:
$$
f(x) = (3x - 1)(x + 9)(4x - 1)
$$
Set \( f(x) = 0 \):
$$
(3x - 1)(x + 9)(4x - 1) = 0
$$
Using the zero-product property:
- From \( 3x - 1 = 0 \): \( 3x = 1 \implies x = \frac{1}{3} \)
- From \( x + 9 = 0 \): \( x = -9 \)
- From \( 4x - 1 = 0 \): \( 4x = 1 \implies x = \frac{1}{4} \)

Thus, the zeros are:
$$
\boxed{\frac{1}{3}, -9, \frac{1}{4}}
$$

#### Problem 5:
$$
f(x) = (x - 8)(4x + 2)(3x - 8)
$$
Set \( f(x) = 0 \):
$$
(x - 8)(4x + 2)(3x - 8) = 0
$$
Using the zero-product property:
- From \( x - 8 = 0 \): \( x = 8 \)
- From \( 4x + 2 = 0 \): \( 4x = -2 \implies x = -\frac{1}{2} \)
- From \( 3x - 8 = 0 \): \( 3x = 8 \implies x = \frac{8}{3} \)

Thus, the zeros are:
$$
\boxed{8, -\frac{1}{2}, \frac{8}{3}}
$$

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Part 2: Stating the excluded values

#### Problem 6:
$$
\frac{30x^3}{15x}
$$
Excluded values occur when the denominator is zero. Set the denominator equal to zero:
$$
15x = 0 \implies x = 0
$$
Thus, the excluded value is:
$$
\boxed{0}
$$

#### Problem 7:
$$
\frac{x^2 + 8x + 7}{x + 1}
$$
Excluded values occur when the denominator is zero. Set the denominator equal to zero:
$$
x + 1 = 0 \implies x = -1
$$
Thus, the excluded value is:
$$
\boxed{-1}
$$

#### Problem 8:
$$
\frac{60x^2}{40x}
$$
Excluded values occur when the denominator is zero. Set the denominator equal to zero:
$$
40x = 0 \implies x = 0
$$
Thus, the excluded value is:
$$
\boxed{0}
$$

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Final Answers:

1. Zeros: \( \boxed{5, -\frac{1}{2}} \)
2. Zeros: \( \boxed{4, 3} \)
3. Zeros: \( \boxed{1, -5} \)
4. Zeros: \( \boxed{\frac{1}{3}, -9, \frac{1}{4}} \)
5. Zeros: \( \boxed{8, -\frac{1}{2}, \frac{8}{3}} \)
6. Excluded value: \( \boxed{0} \)
7. Excluded value: \( \boxed{-1} \)
8. Excluded value: \( \boxed{0} \)