Multiplying Rational Expressions Practice Worksheet A from Mathcation.com, featuring eight problems to simplify rational expressions.

Let’s solve each problem one by one. We’re multiplying rational expressions — that means we multiply the numerators together and the denominators together, then simplify if possible.

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Problem 1:
> $\frac{8x}{3} \cdot \frac{5x}{3}$

Multiply numerators: $8x \cdot 5x = 40x^2$
Multiply denominators: $3 \cdot 3 = 9$
→ Answer: $\frac{40x^2}{9}$
No simplification needed.

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Problem 2:
> $\frac{4}{9x} \cdot \frac{2x^2}{5}$

Multiply numerators: $4 \cdot 2x^2 = 8x^2$
Multiply denominators: $9x \cdot 5 = 45x$
→ $\frac{8x^2}{45x}$

Now simplify: cancel one x from top and bottom → $\frac{8x}{45}$

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Problem 3:
> $\frac{5}{2x} \cdot \frac{4}{9x}$

Numerators: $5 \cdot 4 = 20$
Denominators: $2x \cdot 9x = 18x^2$
→ $\frac{20}{18x^2}$

Simplify fraction: divide numerator and denominator by 2 → $\frac{10}{9x^2}$

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Problem 4:
> $\frac{5x+5}{3} \cdot \frac{4x-2}{7}$

First, factor where possible:

$5x + 5 = 5(x + 1)$
$4x - 2 = 2(2x - 1)$

So expression becomes:
$\frac{5(x+1)}{3} \cdot \frac{2(2x-1)}{7}$

Multiply numerators: $5(x+1) \cdot 2(2x-1) = 10(x+1)(2x-1)$
Multiply denominators: $3 \cdot 7 = 21$

→ $\frac{10(x+1)(2x-1)}{21}$

We can leave it like this or expand the numerator, but since no common factors with denominator, this is simplified.

Alternatively, expanded: $10(2x^2 + 2x - x -1) = 10(2x^2 + x -1) = 20x^2 + 10x -10$, so $\frac{20x^2 + 10x -10}{21}$ — but factored form is usually preferred unless told otherwise.

We’ll keep it as $\frac{10(x+1)(2x-1)}{21}$

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Problem 5:
> $\frac{2x-8}{5} \cdot \frac{4}{3x}$

Factor numerator: $2x - 8 = 2(x - 4)$

So: $\frac{2(x-4)}{5} \cdot \frac{4}{3x}$

Multiply numerators: $2(x-4) \cdot 4 = 8(x-4)$
Multiply denominators: $5 \cdot 3x = 15x$

→ $\frac{8(x-4)}{15x}$

No further simplification.

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Problem 6:
> $\frac{4x}{9} \cdot \frac{5}{3x+2}$

Multiply straight across:

Numerator: $4x \cdot 5 = 20x$
Denominator: $9 \cdot (3x+2) = 9(3x+2)$

→ $\frac{20x}{9(3x+2)}$

Can’t simplify further.

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Problem 7:
> $\frac{4}{5x^2} \cdot \frac{5x}{3x^2 + 2x}$

First, factor denominator of second fraction: $3x^2 + 2x = x(3x + 2)$

So now: $\frac{4}{5x^2} \cdot \frac{5x}{x(3x+2)}$

Multiply numerators: $4 \cdot 5x = 20x$
Multiply denominators: $5x^2 \cdot x(3x+2) = 5x^3(3x+2)$

Wait — better to cancel before multiplying!

Look at: $\frac{4}{5x^2} \cdot \frac{5x}{x(3x+2)}$

Cancel 5 in numerator and denominator:
→ $\frac{4}{x^2} \cdot \frac{x}{x(3x+2)}$

Now cancel one x from numerator and denominator:
→ $\frac{4}{x} \cdot \frac{1}{x(3x+2)}$? Wait, let's do step by step.

Actually:

After canceling 5:
$\frac{4}{x^2} \cdot \frac{x}{x(3x+2)}$

Now, the x in numerator cancels with one x in denominator of first fraction? Let’s write all together:

Numerator: $4 \cdot x = 4x$
Denominator: $x^2 \cdot x(3x+2) = x^3(3x+2)$ → too messy.

Better: rewrite as:

$\frac{4 \cdot 5x}{5x^2 \cdot x(3x+2)} = \frac{20x}{5x^3(3x+2)}$

Now simplify: 20/5 = 4, and x/x^3 = 1/x^2

→ $\frac{4}{x^2(3x+2)}$

Yes! That’s simpler.

Alternatively, cancel step-by-step:

Start: $\frac{4}{5x^2} \cdot \frac{5x}{x(3x+2)}$

Cancel 5: $\frac{4}{x^2} \cdot \frac{x}{x(3x+2)}$

Now, the x in numerator and one x in denominator (from x(3x+2)) cancel:

→ $\frac{4}{x^2} \cdot \frac{1}{(3x+2)}$ → wait, no:

Actually, after canceling 5, we have:

$\frac{4}{x^2} \cdot \frac{x}{x(3x+2)}$

The x in numerator and the x in the denominator of the second fraction cancel → leaves:

$\frac{4}{x^2} \cdot \frac{1}{(3x+2)}$ → but that would be $\frac{4}{x^2(3x+2)}$

Wait — actually, the second fraction’s denominator is x(3x+2), and we have an x in its numerator, so they cancel to 1, leaving:

$\frac{4}{x^2} \cdot \frac{1}{3x+2} = \frac{4}{x^2(3x+2)}$

Yes.

But earlier I thought there was another x — let me double-check original:

Original: $\frac{4}{5x^2} \cdot \frac{5x}{3x^2 + 2x}$

And $3x^2 + 2x = x(3x + 2)$, correct.

So: $\frac{4 \cdot 5x}{5x^2 \cdot x(3x+2)} = \frac{20x}{5x^3(3x+2)} = \frac{4}{x^2(3x+2)}$

Perfect.

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Problem 8:
> $\frac{4x^2 + 5}{5x} \cdot \frac{5x^2 - 4}{3x + 9}$

Check for factoring:

$4x^2 + 5$ — doesn’t factor nicely (discriminant negative if quadratic, but it’s sum of squares basically)

$5x^2 - 4$ — difference of squares? Not quite, since 5 isn't square. Actually, $5x^2 - 4 = (\sqrt{5}x - 2)(\sqrt{5}x + 2)$ — not helpful for rational expressions.

$3x + 9 = 3(x + 3)$

So expression: $\frac{4x^2 + 5}{5x} \cdot \frac{5x^2 - 4}{3(x + 3)}$

Multiply numerators: $(4x^2 + 5)(5x^2 - 4)$
Multiply denominators: $5x \cdot 3(x + 3) = 15x(x + 3)$

No common factors between numerator and denominator, so answer is:

$\frac{(4x^2 + 5)(5x^2 - 4)}{15x(x + 3)}$

We could expand numerator, but it’s not necessary.

Expanded: $4x^2 \cdot 5x^2 = 20x^4$, $4x^2 \cdot (-4) = -16x^2$, $5 \cdot 5x^2 = 25x^2$, $5 \cdot (-4) = -20$

So: $20x^4 + 9x^2 - 20$

Thus: $\frac{20x^4 + 9x^2 - 20}{15x(x + 3)}$

Either form is acceptable, but factored numerator might be preferred if it factors — but it doesn’t easily.

We’ll go with the factored form unless specified.

But since the problem says “simplify”, and no cancellation possible, either is fine. Let’s use expanded numerator for clarity.

Actually, let’s check if numerator factors:

$20x^4 + 9x^2 - 20$

Let $u = x^2$, then $20u^2 + 9u - 20$

Discriminant: $81 + 1600 = 1681 = 41^2$

So roots: $u = \frac{-9 \pm 41}{40}$ → $u = \frac{32}{40} = \frac{4}{5}$ or $u = \frac{-50}{40} = -\frac{5}{4}$

So factors: $20(u - 4/5)(u + 5/4)$ → messy.

Better to leave as $(4x^2 + 5)(5x^2 - 4)$

So final answer: $\frac{(4x^2 + 5)(5x^2 - 4)}{15x(x + 3)}$

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Now, compiling all answers:

1) $\frac{40x^2}{9}$
2) $\frac{8x}{45}$
3) $\frac{10}{9x^2}$
4) $\frac{10(x+1)(2x-1)}{21}$
5) $\frac{8(x-4)}{15x}$
6) $\frac{20x}{9(3x+2)}$
7) $\frac{4}{x^2(3x+2)}$
8) $\frac{(4x^2 + 5)(5x^2 - 4)}{15x(x + 3)}$

All checked and simplified.

Final Answer:
1) $\frac{40x^2}{9}$
2) $\frac{8x}{45}$
3) $\frac{10}{9x^2}$
4) $\frac{10(x+1)(2x-1)}{21}$
5) $\frac{8(x-4)}{15x}$
6) $\frac{20x}{9(3x+2)}$
7) $\frac{4}{x^2(3x+2)}$
8) $\frac{(4x^2 + 5)(5x^2 - 4)}{15x(x + 3)}$