Final math test - 9th grade | m_martinez03 | Live - Free Printable

Here are the step-by-step solutions for each problem on the worksheet.

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

Step 1: Factor the numerators and denominators where possible.
* $6p - 6$ becomes $6(p - 1)$.
* $9p - 9$ becomes $9(p - 1)$.
* The expression is now: $\frac{6(p - 1)}{p} \cdot \frac{2p^2}{9(p - 1)}$

Step 2: Cancel out common terms.
* The term $(p - 1)$ appears in the top left and bottom right, so they cancel out.
* You have $p$ on the bottom and $p^2$ on the top. One $p$ cancels out, leaving just $p$ on the top.
* The numbers $6$ and $9$ can both be divided by $3$. $6$ becomes $2$, and $9$ becomes $3$.

Step 3: Multiply what is left.
* Numerator (top): $2 \cdot 2p = 4p$
* Denominator (bottom): $1 \cdot 3 = 3$

Result: $\frac{4p}{3}$
Correct Option: A

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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: Factor every quadratic expression.
* $k^2 + 12k + 32$ factors to $(k + 4)(k + 8)$.
* $k^2 + 13k + 40$ factors to $(k + 5)(k + 8)$.
* $k^2 + 5k$ factors to $k(k + 5)$.
* $k^2 + 6k + 8$ factors to $(k + 2)(k + 4)$.

Step 2: Rewrite the multiplication with factored forms.
$$ \frac{(k + 4)(k + 8)}{(k + 5)(k + 8)} \cdot \frac{k(k + 5)}{(k + 2)(k + 4)} $$

Step 3: Cancel matching terms from the top and bottom.
* $(k + 8)$ cancels.
* $(k + 5)$ cancels.
* $(k + 4)$ cancels.

Step 4: See what remains.
* Top: $k$
* Bottom: $k + 2$

Result: $\frac{k}{k + 2}$
Correct Option: D

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Problem 3: Simplify $\frac{2x + x^2}{x^2 + 5x + 6}$

Step 1: Factor the numerator and denominator.
* Numerator ($2x + x^2$): Factor out an $x$ to get $x(2 + x)$ or $x(x + 2)$.
* Denominator ($x^2 + 5x + 6$): Find two numbers that multiply to 6 and add to 5. Those are 2 and 3. So it factors to $(x + 2)(x + 3)$.

Step 2: Rewrite the fraction.
$$ \frac{x(x + 2)}{(x + 2)(x + 3)} $$

Step 3: Cancel the common term $(x + 2)$.
* Top: $x$
* Bottom: $x + 3$

Result: $\frac{x}{x + 3}$
Correct Option: B

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Problem 4: Product of $\frac{4x}{x - 1}$ and $\frac{x^2 - 1}{3x + 3}$

Step 1: Set up the multiplication.
$$ \frac{4x}{x - 1} \cdot \frac{x^2 - 1}{3x + 3} $$

Step 2: Factor the expressions.
* $x^2 - 1$ is a difference of squares: $(x - 1)(x + 1)$.
* $3x + 3$ has a common factor of 3: $3(x + 1)$.

Step 3: Rewrite and cancel.
$$ \frac{4x}{x - 1} \cdot \frac{(x - 1)(x + 1)}{3(x + 1)} $$
* The $(x - 1)$ on the bottom left cancels with the $(x - 1)$ on the top right.
* The $(x + 1)$ on the top right cancels with the $(x + 1)$ on the bottom right.

Step 4: Multiply the remaining parts.
* Top: $4x$
* Bottom: $3$

Result: $\frac{4x}{3}$
Correct Option: A

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Final Answer:
1. A
2. D
3. B
4. A