Multiplication of Rational Algebraic Expressions Worksheets - Free Printable

Problem Description:

The task involves solving the product of two rational algebraic expressions (RAEs) with the same denominators. The steps to solve such problems are outlined in the image, and we will follow them systematically.

#### Key Steps for Multiplying Rational Algebraic Expressions:
1. Factor the numerator and denominator completely.
2. Cancel out all common factors.
3. Multiply the numerators and denominators (or leave the answer in factored form).

We will solve the given examples step by step.

---

Example 1:

$$
\frac{8}{10a^2} \times \frac{a^2}{10a^2}
$$

#### Step 1: Factor the numerator and denominator completely.
- The first fraction is already in its simplest form: $\frac{8}{10a^2}$.
- The second fraction is also in its simplest form: $\frac{a^2}{10a^2}$.

#### Step 2: Cancel out all common factors.
- In the first fraction, $8$ and $10a^2$, the greatest common factor (GCF) of $8$ and $10$ is $2$. So, we can simplify:
$$
\frac{8}{10a^2} = \frac{4}{5a^2}.
$$
- In the second fraction, $a^2$ in the numerator and $10a^2$ in the denominator share a common factor of $a^2$. So, we can simplify:
$$
\frac{a^2}{10a^2} = \frac{1}{10}.
$$

Now, the expression becomes:
$$
\frac{4}{5a^2} \times \frac{1}{10}.
$$

#### Step 3: Multiply the numerators and denominators.
- Multiply the numerators: $4 \times 1 = 4$.
- Multiply the denominators: $5a^2 \times 10 = 50a^2$.

So, the product is:
$$
\frac{4}{50a^2}.
$$

#### Simplify the result:
- The GCF of $4$ and $50$ is $2$. Simplify:
$$
\frac{4}{50a^2} = \frac{2}{25a^2}.
$$

Thus, the final answer for Example 1 is:
$$
\boxed{\frac{2}{25a^2}}
$$

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Example 2:

$$
\frac{x^2 - 9}{x^2 + 6x + 9} \times \frac{x^2 + x}{x^2 + 6x + 9}
$$

#### Step 1: Factor the numerator and denominator completely.
- Factor the numerator and denominator of the first fraction:
$$
x^2 - 9 = (x - 3)(x + 3) \quad \text{(difference of squares)}.
$$
$$
x^2 + 6x + 9 = (x + 3)^2 \quad \text{(perfect square trinomial)}.
$$
So, the first fraction becomes:
$$
\frac{x^2 - 9}{x^2 + 6x + 9} = \frac{(x - 3)(x + 3)}{(x + 3)^2}.
$$

- Factor the numerator and denominator of the second fraction:
$$
x^2 + x = x(x + 1).
$$
The denominator is already factored:
$$
x^2 + 6x + 9 = (x + 3)^2.
$$
So, the second fraction becomes:
$$
\frac{x^2 + x}{x^2 + 6x + 9} = \frac{x(x + 1)}{(x + 3)^2}.
$$

Now, the expression becomes:
$$
\frac{(x - 3)(x + 3)}{(x + 3)^2} \times \frac{x(x + 1)}{(x + 3)^2}.
$$

#### Step 2: Cancel out all common factors.
- In the first fraction, $(x + 3)$ in the numerator and one $(x + 3)$ in the denominator cancel out:
$$
\frac{(x - 3)(x + 3)}{(x + 3)^2} = \frac{x - 3}{x + 3}.
$$

- The second fraction remains:
$$
\frac{x(x + 1)}{(x + 3)^2}.
$$

Now, the expression becomes:
$$
\frac{x - 3}{x + 3} \times \frac{x(x + 1)}{(x + 3)^2}.
$$

Combine the fractions:
$$
\frac{(x - 3) \cdot x(x + 1)}{(x + 3) \cdot (x + 3)^2} = \frac{x(x - 3)(x + 1)}{(x + 3)^3}.
$$

#### Step 3: The expression is now in its simplest form:
$$
\frac{x(x - 3)(x + 1)}{(x + 3)^3}.
$$

Thus, the final answer for Example 2 is:
$$
\boxed{\frac{x(x - 3)(x + 1)}{(x + 3)^3}}
$$

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Final Answers:

1. $\boxed{\frac{2}{25a^2}}$
2. $\boxed{\frac{x(x - 3)(x + 1)}{(x + 3)^3}}$