Final math test - 9th grade worksheet - Free Printable

Let's solve each problem step by step.

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Problem 1: Multiply \( \frac{6p - 6}{p} \cdot \frac{2p^2}{9p - 9} \)

#### Step 1: Factorize the numerators and denominators
- \( 6p - 6 = 6(p - 1) \)
- \( 9p - 9 = 9(p - 1) \)

So the expression becomes:
\[
\frac{6(p - 1)}{p} \cdot \frac{2p^2}{9(p - 1)}
\]

#### Step 2: Simplify by canceling common factors
- The \( (p - 1) \) terms in the numerator and denominator cancel out.
- The \( p \) in the denominator of the first fraction cancels with one \( p \) from \( 2p^2 \) in the second fraction.

This leaves:
\[
\frac{6}{1} \cdot \frac{2p}{9} = \frac{6 \cdot 2p}{9} = \frac{12p}{9} = \frac{4p}{3}
\]

#### Final Answer:
\[
\boxed{\frac{4p}{3}}
\]

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Problem 2: Multiply \( \frac{k^2 + 12k + 32}{k^2 + 13k + 40} \cdot \frac{k^2 + 5k}{k^2 + 6k + 8} \)

#### Step 1: Factorize the numerators and denominators
- \( k^2 + 12k + 32 = (k + 4)(k + 8) \)
- \( k^2 + 13k + 40 = (k + 5)(k + 8) \)
- \( k^2 + 5k = k(k + 5) \)
- \( k^2 + 6k + 8 = (k + 2)(k + 4) \)

So the expression becomes:
\[
\frac{(k + 4)(k + 8)}{(k + 5)(k + 8)} \cdot \frac{k(k + 5)}{(k + 2)(k + 4)}
\]

#### Step 2: Simplify by canceling common factors
- The \( (k + 8) \) terms in the numerator and denominator cancel out.
- The \( (k + 5) \) terms in the numerator and denominator cancel out.
- The \( (k + 4) \) terms in the numerator and denominator cancel out.

This leaves:
\[
\frac{k}{k + 2}
\]

#### Final Answer:
\[
\boxed{\frac{k}{k + 2}}
\]

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Problem 3: For all values of \( x \) for which the expression is defined, \( \frac{2x + x^2}{x^2 + 5x + 6} \) is equivalent to

#### Step 1: Factorize the numerator and denominator
- Numerator: \( 2x + x^2 = x(x + 2) \)
- Denominator: \( x^2 + 5x + 6 = (x + 2)(x + 3) \)

So the expression becomes:
\[
\frac{x(x + 2)}{(x + 2)(x + 3)}
\]

#### Step 2: Simplify by canceling common factors
- The \( (x + 2) \) terms in the numerator and denominator cancel out (assuming \( x \neq -2 \)).

This leaves:
\[
\frac{x}{x + 3}
\]

#### Final Answer:
\[
\boxed{\frac{x}{x + 3}}
\]

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Problem 4: What is the product of \( \frac{4x}{x - 1} \) and \( \frac{x^2 - 1}{3x + 3} \) expressed in simplest form?

#### Step 1: Factorize the numerators and denominators
- \( x^2 - 1 = (x - 1)(x + 1) \)
- \( 3x + 3 = 3(x + 1) \)

So the expression becomes:
\[
\frac{4x}{x - 1} \cdot \frac{(x - 1)(x + 1)}{3(x + 1)}
\]

#### Step 2: Simplify by canceling common factors
- The \( (x - 1) \) terms in the numerator and denominator cancel out.
- The \( (x + 1) \) terms in the numerator and denominator cancel out.

This leaves:
\[
\frac{4x}{3}
\]

#### Final Answer:
\[
\boxed{\frac{4x}{3}}
\]

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Final Answers:

1. \( \boxed{\frac{4p}{3}} \)
2. \( \boxed{\frac{k}{k + 2}} \)
3. \( \boxed{\frac{x}{x + 3}} \)
4. \( \boxed{\frac{4x}{3}} \)