Unit 7 · Lesson 1 of 5
Simplifying rational expressions
You'll be able to: find the values \(x\) can't be, and simplify a fraction of polynomials like \(\dfrac{x^2 - 9}{x^2 + 5x + 6}\).
Step 1 of 5Warm-up
Two quick questions from earlier lessons. Both are building blocks for today.
Solve \(x^2 - x - 12 = 0\).
Two numbers that multiply to \(-12\) and add to \(-1\): \(-4\) and \(3\). So \((x - 4)(x + 3) = 0\).
Answer: \(x = 4\) or \(x = -3\).
Simplify \(\dfrac{42}{56}\) to lowest terms.
Write both as products: \(42 = 14 \cdot 3\) and \(56 = 14 \cdot 4\).
Cancel the common factor 14: \(\dfrac{42}{56} = \dfrac{3}{4}\).
Step 2 of 5The idea
A rational expression is a fraction with polynomials on top and bottom, like \(\dfrac{x + 1}{x - 3}\). You simplify it the same way you simplified \(\dfrac{42}{56}\) in the warm-up: write top and bottom as products, then cancel what they share.
First, though, notice the values \(x\) is not allowed to be. A denominator can never be 0. A value that makes it 0 is an excluded value. In \(\dfrac{x + 1}{x - 3}\), \(x\) can't be 3.
1. Excluded values: set the original denominator equal to 0 and solve. Do this before canceling.
2. Factor the top and the bottom completely.
3. Cancel common factors. Only whole factors (things multiplied), never terms (things added).
Opposites cancel to \(-1\): \(\dfrac{a - b}{b - a} = -1\), because \(b - a = -(a - b)\).
Why this shows up on the placement test: "simplify the rational expression" is a standard placement question, and the wrong answer choices are usually the results of canceling terms.
Step 3 of 5Worked examples
Example 1: Simplify \(\dfrac{x^2 - 9}{x^2 + 5x + 6}\)
- Factor the bottom: \(x^2 + 5x + 6 = (x + 2)(x + 3)\).
- Excluded values: \((x + 2)(x + 3) = 0\) gives \(x = -2\) or \(x = -3\). So \(x \ne -2\) and \(x \ne -3\).
- Factor the top: \(x^2 - 9 = (x + 3)(x - 3)\). (Difference of squares.)
- \(\dfrac{(x + 3)(x - 3)}{(x + 2)(x + 3)}\). Cancel the common factor \((x + 3)\).
- Answer: \(\dfrac{x - 3}{x + 2}\), with \(x \ne -2, -3\).
Example 2: Simplify \(\dfrac{10 - 5x}{x^2 - 4}\)
- Factor the bottom: \(x^2 - 4 = (x + 2)(x - 2)\). Excluded values: \(x \ne -2\) and \(x \ne 2\).
- Factor the top: \(10 - 5x = 5(2 - x)\).
- \(2 - x\) and \(x - 2\) are opposites: \(2 - x = -(x - 2)\). So the top is \(-5(x - 2)\).
- \(\dfrac{-5(x - 2)}{(x + 2)(x - 2)}\). Cancel \((x - 2)\).
- Answer: \(\dfrac{-5}{x + 2}\), with \(x \ne \pm 2\). You can also write it as \(-\dfrac{5}{x + 2}\).
Step 4 of 5Practice
Type your answer and press Check. Write fractions with parentheses, like (x-3)/(x+2). If you miss twice, open the worked solution. Then mark how it went.
What are the excluded values of \(\dfrac{x + 1}{x^2 - 2x - 15}\)?
Set the denominator equal to 0: \(x^2 - 2x - 15 = 0\).
Factor: \((x - 5)(x + 3) = 0\), so \(x = 5\) or \(x = -3\).
Answer: \(x\) can't be \(5\) or \(-3\).
Simplify \(\dfrac{x + 5}{x^2 + 5x}\).
Factor the bottom: \(x^2 + 5x = x(x + 5)\). Excluded values: \(x \ne 0\), \(x \ne -5\).
\(\dfrac{x + 5}{x(x + 5)}\). Cancel \((x + 5)\). A 1 is left on top.
Answer: \(\dfrac{1}{x}\).
Simplify \(\dfrac{3x + 12}{x^2 - 16}\).
Top: \(3x + 12 = 3(x + 4)\). Bottom: \(x^2 - 16 = (x + 4)(x - 4)\). Excluded values: \(x \ne \pm 4\).
\(\dfrac{3(x + 4)}{(x + 4)(x - 4)}\). Cancel \((x + 4)\).
Answer: \(\dfrac{3}{x - 4}\).
Simplify \(\dfrac{x^2 + 7x + 10}{x^2 + 3x - 10}\).
Top: multiply to 10, add to 7: \((x + 5)(x + 2)\). Bottom: multiply to \(-10\), add to 3: \((x + 5)(x - 2)\). Excluded values: \(x \ne -5\), \(x \ne 2\).
Cancel \((x + 5)\).
Answer: \(\dfrac{x + 2}{x - 2}\).
Simplify \(\dfrac{7 - x}{x - 7}\).
The top and bottom are opposites: \(7 - x = -(x - 7)\). Excluded value: \(x \ne 7\).
\(\dfrac{-(x - 7)}{x - 7}\). Cancel \((x - 7)\).
Answer: \(-1\).
Simplify \(\dfrac{2x^2 - 8x}{12 - 3x}\).
Top: \(2x^2 - 8x = 2x(x - 4)\). Bottom: \(12 - 3x = 3(4 - x) = -3(x - 4)\). Excluded value: \(x \ne 4\).
\(\dfrac{2x(x - 4)}{-3(x - 4)}\). Cancel \((x - 4)\).
Answer: \(\dfrac{2x}{-3} = -\dfrac{2x}{3}\).
Which is \(\dfrac{x^2 + 4x}{x^2 + 2x - 8}\) simplified correctly?
- \(\dfrac{x}{x - 2}\)
- \(\dfrac{4x}{2x - 8}\)
- \(\dfrac{1}{x - 2}\)
- \(\dfrac{x + 4}{x - 2}\)
Factor: top \(x(x + 4)\), bottom \((x + 4)(x - 2)\). The only common factor is \((x + 4)\).
(a) Cancel \((x + 4)\): \(\dfrac{x}{x - 2}\). This one.
(b) This "cancels" the \(x^2\) terms. Terms can't be canceled.
(c) This drops the \(x\) on top as well. After canceling \((x + 4)\), the top is \(x\), not 1.
(d) This cancels the \(x\) instead of the \((x + 4)\). There is no \(x\) factor on the bottom to cancel with.
Simplify \(\dfrac{2x^2 + 7x + 3}{4x^2 - 1}\).
Top (ac method): \(ac = 6\), numbers \(1\) and \(6\). \(2x^2 + x + 6x + 3 = x(2x + 1) + 3(2x + 1) = (2x + 1)(x + 3)\).
Bottom (difference of squares): \(4x^2 - 1 = (2x + 1)(2x - 1)\). Excluded values: \(x \ne \pm\dfrac{1}{2}\).
Cancel \((2x + 1)\). Answer: \(\dfrac{x + 3}{2x - 1}\).
Step 5 of 5Wrap-up
You can now simplify a rational expression: find the excluded values, factor the top and bottom, and cancel common factors, remembering that opposites like \(a - b\) and \(b - a\) cancel to \(-1\).
Read or watch next: OpenStax Intermediate Algebra 2e, 7.1 Multiply and Divide Rational Expressions (free textbook; simplifying is the first part of the section).
Also good: Khan Academy: Reducing rational expressions to lowest terms (short videos). Quick rules: Rational expressions cheat sheet.
Something unclear? Ask your teacher bot a follow-up question, for example: "Why do I find the excluded values before canceling and not after?" It knows this course and can explain things another way.