Unit 7 · Lesson 2 of 5
Multiplying and dividing rational expressions
You'll be able to: multiply and divide fractions that have polynomials on top and bottom, and give the answer in lowest terms.
Step 1 of 5Warm-up
Two quick questions from earlier lessons. Answer from memory first, then check.
Simplify \(\dfrac{x^2 - 9}{x^2 + 5x + 6}\).
Factor the top and the bottom: \(\dfrac{(x-3)(x+3)}{(x+2)(x+3)}\).
Cancel the common factor \((x+3)\). Answer: \(\dfrac{x-3}{x+2}\).
\(\dfrac{3}{4} \div \dfrac{9}{10}\)
Dividing means multiplying by the reciprocal: \(\dfrac{3}{4} \cdot \dfrac{10}{9}\).
Cancel first: 3 goes into 9 (leaves 3), and 2 goes into 4 and 10 (leaves 2 and 5). So \(\dfrac{1}{2} \cdot \dfrac{5}{3} = \dfrac{5}{6}\).
Step 2 of 5The idea
You already multiply number fractions this way: cancel what you can, then multiply top times top and bottom times bottom. A rational expression follows the same rules.
The one new step: you can only cancel factors (things being multiplied). So you factor everything first. Then the pieces that cancel are easy to see.
Multiplying. Factor every top and bottom. Cancel any factor that is on a top and on a bottom (it can be in different fractions). Multiply what is left.
Dividing. Keep the first fraction, flip the second one (its reciprocal), and multiply: \[\dfrac{A}{B} \div \dfrac{C}{D} = \dfrac{A}{B} \cdot \dfrac{D}{C}\]
Answer form. Leaving the answer factored, like \(\dfrac{2x(x-3)}{x-2}\), is fine and is usually how answer choices look.
A useful leftover from last lesson: \(\dfrac{a-b}{b-a} = -1\). If you see \(5 - x\) and \(x - 5\), write \(5 - x = -(x-5)\) and they cancel, leaving a minus sign.
Why this shows up on the placement test: "multiply and simplify" problems check factoring and fraction rules in one question, so they are a favorite.
Step 3 of 5Worked examples
Example 1: \(\dfrac{x^2 - 9}{4x} \cdot \dfrac{8x^2}{x^2 + x - 6}\)
- Factor everything: \(x^2 - 9 = (x-3)(x+3)\) and \(x^2 + x - 6 = (x+3)(x-2)\). (Difference of squares, then two numbers that multiply to \(-6\) and add to \(1\).)
- Rewrite: \(\dfrac{(x-3)(x+3)}{4x} \cdot \dfrac{8x^2}{(x+3)(x-2)}\).
- Cancel \((x+3)\), which is on a top and on a bottom.
- Cancel the numbers and \(x\)'s: \(\dfrac{8x^2}{4x} = 2x\).
- What is left: \(\dfrac{2x(x-3)}{x-2}\).
Example 2: \(\dfrac{x^2 - 25}{3x + 6} \div \dfrac{x^2 + 5x}{x + 2}\)
- Flip the second fraction and multiply: \(\dfrac{x^2 - 25}{3x + 6} \cdot \dfrac{x + 2}{x^2 + 5x}\).
- Factor: \(\dfrac{(x-5)(x+5)}{3(x+2)} \cdot \dfrac{x+2}{x(x+5)}\). (\(3x+6\) has GCF 3; \(x^2+5x\) has GCF \(x\).)
- Cancel \((x+5)\) and \((x+2)\).
- What is left: \(\dfrac{x-5}{3x}\).
Step 4 of 5Practice
Type your answer and press Check. If you miss twice, open the worked solution. Then mark how it went. Use parentheses around a top or bottom that has more than one term, like (x+2)/(x-3).
Multiply and simplify: \(\dfrac{6x^2}{5y} \cdot \dfrac{10y^3}{3x}\)
Cancel first: \(6 \div 3 = 2\), \(10 \div 5 = 2\), \(\dfrac{x^2}{x} = x\), \(\dfrac{y^3}{y} = y^2\).
Multiply what is left: \(2 \cdot 2 \cdot x \cdot y^2\). Answer: \(4xy^2\).
Multiply and simplify: \(\dfrac{x+3}{x-2} \cdot \dfrac{x^2 - 4}{2x + 6}\)
Factor: \(x^2 - 4 = (x-2)(x+2)\) and \(2x + 6 = 2(x+3)\).
\(\dfrac{x+3}{x-2} \cdot \dfrac{(x-2)(x+2)}{2(x+3)}\). Cancel \((x+3)\) and \((x-2)\).
Answer: \(\dfrac{x+2}{2}\).
Multiply and simplify: \(\dfrac{x^2 + 5x + 6}{x^2 - 9} \cdot \dfrac{x - 3}{x + 4}\)
Factor: \(x^2 + 5x + 6 = (x+2)(x+3)\) and \(x^2 - 9 = (x-3)(x+3)\).
\(\dfrac{(x+2)(x+3)}{(x-3)(x+3)} \cdot \dfrac{x-3}{x+4}\). Cancel \((x+3)\) and \((x-3)\).
Answer: \(\dfrac{x+2}{x+4}\).
Divide and simplify: \(\dfrac{4x}{x+1} \div \dfrac{8x^2}{x^2 - 1}\)
Flip the second fraction: \(\dfrac{4x}{x+1} \cdot \dfrac{x^2 - 1}{8x^2}\).
Factor \(x^2 - 1 = (x-1)(x+1)\) and cancel \((x+1)\): \(\dfrac{4x(x-1)}{8x^2}\).
\(\dfrac{4x}{8x^2} = \dfrac{1}{2x}\). Answer: \(\dfrac{x-1}{2x}\).
Divide and simplify: \(\dfrac{x^2 - x - 12}{x^2 + 2x} \div \dfrac{x^2 - 16}{x^2 + 6x + 8}\)
Flip the second fraction, then factor everything:
\(\dfrac{(x-4)(x+3)}{x(x+2)} \cdot \dfrac{(x+2)(x+4)}{(x-4)(x+4)}\).
Cancel \((x-4)\), \((x+2)\), and \((x+4)\). Answer: \(\dfrac{x+3}{x}\).
Multiply and simplify: \(\dfrac{5 - x}{x + 3} \cdot \dfrac{x^2 + 3x}{x^2 - 25}\)
Factor: \(5 - x = -(x-5)\), \(x^2 + 3x = x(x+3)\), \(x^2 - 25 = (x-5)(x+5)\).
\(\dfrac{-(x-5)}{x+3} \cdot \dfrac{x(x+3)}{(x-5)(x+5)}\). Cancel \((x-5)\) and \((x+3)\); the minus sign stays.
Answer: \(-\dfrac{x}{x+5}\).
Divide and simplify: \(\dfrac{x^2 - y^2}{3x} \div \dfrac{x + y}{6x^2}\)
Flip the second fraction: \(\dfrac{x^2 - y^2}{3x} \cdot \dfrac{6x^2}{x + y}\).
Factor \(x^2 - y^2 = (x-y)(x+y)\) and cancel \((x+y)\). Also \(\dfrac{6x^2}{3x} = 2x\).
Answer: \(2x(x-y)\), which is the same as \(2x^2 - 2xy\).
Which expression is equivalent to \(\dfrac{x^2 - 4}{x^2 + 4x + 3} \div \dfrac{x - 2}{x + 1}\)?
- \(\dfrac{x+3}{x+2}\)
- \(\dfrac{x-2}{x+3}\)
- \(\dfrac{x+2}{x+1}\)
- \(\dfrac{x+2}{x+3}\)
Flip the second fraction and factor: \(\dfrac{(x-2)(x+2)}{(x+1)(x+3)} \cdot \dfrac{x+1}{x-2}\).
Cancel \((x-2)\) and \((x+1)\). What is left is \(\dfrac{x+2}{x+3}\). Choice D.
(A) is the right answer upside down. (B) and (C) come from cancelling the wrong factors.
Step 5 of 5Wrap-up
You can now multiply and divide rational expressions: flip the second fraction if you are dividing, factor everything, cancel matching factors, and multiply what is left.
Read or watch next: OpenStax Intermediate Algebra 2e, 7.1 Multiply and Divide Rational Expressions (free textbook, with more practice at the end of the section).
Also good: Khan Academy: Multiplying and dividing rational expressions (short videos). Quick rules: Rational expressions cheat sheet.
Something unclear? Ask your teacher bot a follow-up question, for example: "Why do I flip only the second fraction when I divide?" It knows this course and can explain things another way.