Algebra 2 Practice Worksheet | Worksheets Worksheets - Free Printable

Let's solve each problem step by step.

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Part 1: Find all zeros

To find the zeros of a function $ f(x) $, we set $ f(x) = 0 $ and solve for $ x $. Since each function is given in factored form, we can use the Zero Product Property: if $ a \cdot b = 0 $, then $ a = 0 $ or $ b = 0 $.

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#### 1. $ f(x) = (x - 5)(2x + 1) $

Set each factor equal to zero:

- $ x - 5 = 0 $ → $ x = 5 $
- $ 2x + 1 = 0 $ → $ 2x = -1 $ → $ x = -\frac{1}{2} $

✔ Zeros: $ x = 5 $, $ x = -\frac{1}{2} $

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#### 2. $ f(x) = (x - 4)(x - 3) $

- $ x - 4 = 0 $ → $ x = 4 $
- $ x - 3 = 0 $ → $ x = 3 $

✔ Zeros: $ x = 4 $, $ x = 3 $

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#### 3. $ f(x) = (2x - 2)(x + 5) $

- $ 2x - 2 = 0 $ → $ 2x = 2 $ → $ x = 1 $
- $ x + 5 = 0 $ → $ x = -5 $

✔ Zeros: $ x = 1 $, $ x = -5 $

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#### 4. $ f(x) = (3x - 1)(x + 9)(4x - 1) $

Set each factor to zero:

- $ 3x - 1 = 0 $ → $ 3x = 1 $ → $ x = \frac{1}{3} $
- $ x + 9 = 0 $ → $ x = -9 $
- $ 4x - 1 = 0 $ → $ 4x = 1 $ → $ x = \frac{1}{4} $

✔ Zeros: $ x = \frac{1}{3} $, $ x = -9 $, $ x = \frac{1}{4} $

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#### 5. $ f(x) = (x - 8)(4x + 2)(3x - 8) $

Set each factor to zero:

- $ x - 8 = 0 $ → $ x = 8 $
- $ 4x + 2 = 0 $ → $ 4x = -2 $ → $ x = -\frac{1}{2} $
- $ 3x - 8 = 0 $ → $ 3x = 8 $ → $ x = \frac{8}{3} $

✔ Zeros: $ x = 8 $, $ x = -\frac{1}{2} $, $ x = \frac{8}{3} $

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Part 2: State the excluded values

Excluded values are values of $ x $ that make the denominator zero, since division by zero is undefined.

We simplify expressions first, but we must identify values that make the original denominator zero — even if they cancel.

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#### 6. $ \frac{30x^3}{15x} $

Simplify:
$$
\frac{30x^3}{15x} = \frac{30}{15} \cdot \frac{x^3}{x} = 2x^2 \quad \text{(for } x \neq 0\text{)}
$$

But the original expression has a denominator $ 15x $, so it's undefined when $ x = 0 $.

✔ Excluded value: $ x = 0 $

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#### 7. $ \frac{x^2 + 8x + 7}{x + 1} $

Factor numerator:
$$
x^2 + 8x + 7 = (x + 7)(x + 1)
$$

So:
$$
\frac{(x + 7)(x + 1)}{x + 1}
$$

We can cancel $ x + 1 $ only if $ x \neq -1 $ (because at $ x = -1 $, denominator is zero).

So the expression simplifies to $ x + 7 $, but $ x = -1 $ is excluded.

✔ Excluded value: $ x = -1 $

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#### 8. $ \frac{60x^2}{40x} $

Simplify:
$$
\frac{60x^2}{40x} = \frac{60}{40} \cdot \frac{x^2}{x} = \frac{3}{2}x \quad \text{(for } x \neq 0\text{)}
$$

Denominator is $ 40x $, which is zero when $ x = 0 $.

✔ Excluded value: $ x = 0 $

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✔ Final Answers:

#### Find all zeros:
1. $ x = 5, -\frac{1}{2} $
2. $ x = 4, 3 $
3. $ x = 1, -5 $
4. $ x = \frac{1}{3}, -9, \frac{1}{4} $
5. $ x = 8, -\frac{1}{2}, \frac{8}{3} $

#### State the excluded values:
6. $ x = 0 $
7. $ x = -1 $
8. $ x = 0 $

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