Multiplying and Dividing Rational Expressions Worksheets - Math Monks - Free Printable

Let’s solve each problem one by one. We’ll multiply or divide the rational expressions, simplify where possible, and show our work clearly.

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Problem ①:
(2x + 4) · (3x)/(3x + 6)

Step 1: Factor anything we can.
2x + 4 = 2(x + 2)
3x + 6 = 3(x + 2)

So now it becomes:
2(x + 2) · (3x)/[3(x + 2)]

Step 2: Cancel common factors.
The (x + 2) in numerator and denominator cancel.
Also, 3 in numerator and denominator cancel.

Left with: 2 · x = 2x

✔ Final Answer for ①: 2x

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Problem ②:
(3x + 6) · (2x)/(2x + 4)

Factor:
3x + 6 = 3(x + 2)
2x + 4 = 2(x + 2)

Now:
3(x + 2) · (2x)/[2(x + 2)]

Cancel (x + 2), and cancel 2:

Left with: 3 · x = 3x

✔ Final Answer for ②: 3x

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Problem ③:
(x² - 4)/(x² - 1) · (x + 1)/(x + 2)

Factor everything:

x² - 4 = (x - 2)(x + 2)
x² - 1 = (x - 1)(x + 1)

So expression becomes:
[(x - 2)(x + 2)] / [(x - 1)(x + 1)] · (x + 1)/(x + 2)

Now cancel:

- (x + 2) top and bottom → gone
- (x + 1) top and bottom → gone

Left with: (x - 2)/(x - 1)

✔ Final Answer for ③: (x - 2)/(x - 1)

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Problem ④:
(x² - 1)/(x - 3) · (x² - 9)/(x + 1)

Factor:

x² - 1 = (x - 1)(x + 1)
x² - 9 = (x - 3)(x + 3)

Expression becomes:
[(x - 1)(x + 1)]/(x - 3) · [(x - 3)(x + 3)]/(x + 1)

Cancel:

- (x + 1) top and bottom → gone
- (x - 3) top and bottom → gone

Left with: (x - 1)(x + 3)

Multiply out if needed? Usually leave factored unless told otherwise. But let’s expand to be safe:

= x² + 3x - x - 3 = x² + 2x - 3

But actually, since no instruction to expand, either is fine — but often simplified form means expanded polynomial. Let’s check standard practice: usually they want simplified rational expression, which here is a polynomial. So yes, expand.

Wait — actually, after canceling, we have (x - 1)(x + 3). That’s already simplified. But sometimes teachers prefer expanded. Since original problems had polynomials, maybe expand.

Actually, looking at other answers — like #1 and #2 gave monomials, #3 gave fraction. Here, after canceling, it's a product of two binomials — so we can write as (x - 1)(x + 3) OR expand.

I think both are acceptable, but to match style of #1 and #2 (which gave single terms), perhaps expand.

Let me double-check cancellation:

Original:
Numerator: (x² - 1)(x² - 9) = (x-1)(x+1)(x-3)(x+3)
Denominator: (x - 3)(x + 1)

After canceling (x+1) and (x-3), left with (x-1)(x+3)

Yes.

If we expand: x*x + x*3 -1*x -1*3 = x² + 3x - x - 3 = x² + 2x - 3

I’ll go with expanded form since it’s a polynomial result.

✔ Final Answer for ④: x² + 2x - 3

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Problem ⑤:
(x² - x - 12)/(3x - 9) ÷ (x - 4)/12

Dividing fractions → flip second and multiply.

First, factor:

x² - x - 12 = (x - 4)(x + 3)
3x - 9 = 3(x - 3)

So first fraction: (x - 4)(x + 3)/[3(x - 3)]

Second fraction flipped: 12/(x - 4)

Now multiply:

[(x - 4)(x + 3)/[3(x - 3)]] · [12/(x - 4)]

Cancel (x - 4) top and bottom.

Now: (x + 3)/[3(x - 3)] · 12

Which is: 12(x + 3)/[3(x - 3)]

Simplify 12/3 = 4

So: 4(x + 3)/(x - 3)

We can leave like this, or expand numerator: 4x + 12 over (x - 3)

Usually leave factored unless specified.

✔ Final Answer for ⑤: 4(x + 3)/(x - 3) or (4x + 12)/(x - 3) — I’ll pick the factored version as simpler.

Actually, let’s see what’s more simplified — both are fine. I’ll write as \frac{4(x + 3)}{x - 3}

But since others used inline, maybe write as 4(x+3)/(x-3)

Okay.

✔ Final Answer for ⑤: 4(x + 3)/(x - 3)

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Problem ⑥:
x²/(6 - 3x) ÷ x/(8 - 4x)

Flip second fraction: multiply by (8 - 4x)/x

So: [x²/(6 - 3x)] · [(8 - 4x)/x]

Factor denominators and numerators:

6 - 3x = 3(2 - x) → note: 2 - x = -(x - 2), but let’s keep for now
8 - 4x = 4(2 - x)

So:

[x² / 3(2 - x)] · [4(2 - x)/x]

Now cancel:

- (2 - x) top and bottom → gone
- x² and x → leaves x in numerator

So: x · 4 / 3 = 4x/3

Wait — let’s do step by step:

Numerator: x² * 4(2 - x)
Denominator: 3(2 - x) * x

Cancel (2 - x): gone
Cancel x from x² and x: left with x in num

So: (x * 4) / 3 = 4x/3

Yes.

Note: 2 - x is same as -(x - 2), but since it appears in both num and den, cancels directly without sign issue.

✔ Final Answer for ⑥: 4x/3

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Problem ⑦:
(x - 4)/(x² - 5x + 6) ÷ (x - 3)/(x² - 6x + 9)

Flip second fraction: multiply by (x² - 6x + 9)/(x - 3)

Factor all:

x² - 5x + 6 = (x - 2)(x - 3)
x² - 6x + 9 = (x - 3)²

So:

[(x - 4)/((x - 2)(x - 3))] · [(x - 3)²/(x - 3)]

Wait — second fraction is (x² - 6x + 9)/(x - 3) = (x - 3)² / (x - 3) = (x - 3) [after canceling one (x-3)]

Actually, better to write full:

= [(x - 4)/((x - 2)(x - 3))] · [(x - 3)(x - 3)/(x - 3)]

No — (x² - 6x + 9) = (x - 3)^2, and divided by (x - 3), so overall second part is (x - 3)^2 / (x - 3) = (x - 3)

But in multiplication, we have:

First fraction: (x - 4) / [(x - 2)(x - 3)]
Times second flipped: (x - 3)^2 / (x - 3) → wait no:

Original division: A ÷ B = A × (1/B)

B is (x - 3)/(x² - 6x + 9) = (x - 3)/(x - 3)^2

So 1/B = (x - 3)^2 / (x - 3) = (x - 3)

Easier:

A = (x - 4)/(x² - 5x + 6) = (x - 4)/[(x - 2)(x - 3)]
B = (x - 3)/(x² - 6x + 9) = (x - 3)/(x - 3)^2

So A ÷ B = A × (reciprocal of B) = A × [(x - 3)^2 / (x - 3)] = A × (x - 3)

Because (x - 3)^2 / (x - 3) = x - 3

So: [(x - 4)/((x - 2)(x - 3))] × (x - 3)

Now cancel (x - 3):

Left with: (x - 4)/(x - 2)

✔ Final Answer for ⑦: (x - 4)/(x - 2)

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Problem ⑧:
(x² - 9)/(6x) ÷ (x² - 9)/(18x³)

Flip second fraction: multiply by (18x³)/(x² - 9)

So: [(x² - 9)/(6x)] · [18x³/(x² - 9)]

Now cancel (x² - 9) top and bottom → gone

Left with: (1)/(6x) · 18x³ = 18x³ / (6x)

Simplify: 18/6 = 3, x³/x = x²

So: 3x²

✔ Final Answer for ⑧: 3x²

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Final Answers Summary:

① 2x
② 3x
③ (x - 2)/(x - 1)
④ x² + 2x - 3
⑤ 4(x + 3)/(x - 3)
⑥ 4x/3
⑦ (x - 4)/(x - 2)
⑧ 3x²

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Final Answer:
① 2x
② 3x
③ \frac{x - 2}{x - 1}
④ x^2 + 2x - 3
⑤ \frac{4(x + 3)}{x - 3}
⑥ \frac{4x}{3}
⑦ \frac{x - 4}{x - 2}
⑧ 3x^2