Math101 · Independent Ontario learning library
Math101learn.math101.caDividing Rational Expressions
Dividing rational expressions uses the reciprocal of the divisor: $A/B\div C/D=A/B\cdot D/C$.
Division of algebraic fractions appears in compound rates and solving rational equations. Restriction tracking preserves equivalence and prevents hidden undefined values.
Intuition and core definition
Dividing rational expressions uses the reciprocal of the divisor: $A/B\div C/D=A/B\cdot D/C$. Every original denominator must be nonzero, and the divisor $C/D$ must itself be nonzero, so both $C$ and $D$ carry restrictions. Factoring reveals cancellable factors after the reciprocal step.
Notation, language, and conditions
A rational expression is a quotient of polynomials. Cancellation is valid only for nonzero common factors in a product. Domain restrictions are collected from the original expressions before any cancellation; a simplified formula does not redefine missing input values.
Why this idea matters
Dividing algebraic fractions combines reciprocal multiplication with a domain audit that also excludes zeros of the divisor's numerator.
A dependable method
- Factor all numerators and denominators and state original restrictions.
- Confirm the divisor is not zero.
- Change division to multiplication and invert only the divisor.
- Cancel common nonzero factors across the product.
- Multiply remaining factors and retain every original restriction.
Worked example
Representations and interpretation
A factor grid shows numerator factors above a division bar and denominator factors below. Flipping the divisor changes their positions; strike-through marks are justified only between matching multiplicative factors.
Reasoning about variations
If the divisor is zero at a value even though its written denominator is nonzero, division is still undefined. That is why zeros of the divisor’s numerator become restrictions after division.
Common mistakes
How to check your work
- Multiply the simplified quotient by the original divisor at a legal input and recover the dividend.
- Substitute a convenient legal value into both forms.
- Audit restrictions against every original denominator and the divisor numerator.
Practice
- Simplify $\frac{x^2-4}{x}\div\frac{x+2}{3x}$.
- What extra restriction comes from dividing by $\frac{x-5}{x+1}$?
- May $x+2$ cancel in $(x+2)/(x+2+y)$?
Answers and brief solutions
Show answers
- $3(x-2)$ $\frac{(x-2)(x+2)}x\cdot\frac{3x}{x+2}=3(x-2)$, with $x\ne0,-2$ from the original expressions.
- $x\ne5$ The divisor cannot equal zero, so its numerator cannot vanish.
- No It is not a factor of the entire denominator sum.
Synthesis and transfer
In a ratio-of-rates model, factor before inverting the divisor; a value that makes the divisor zero remains forbidden even if its factor later cancels.
Suppose one rate is $(x^2-1)/(x-2)$ and it is divided by $(x+1)/(x-2)$. Original denominators exclude $x=2$, and division also excludes $x=-1$ because that value makes the divisor zero. Factoring and multiplying by the reciprocal leaves $x-1$ only on the restricted domain. The missing inputs cannot be restored by the simple final formula. At a legal value such as $x=3$, multiplying the quotient by the original divisor returns the original dividend, which checks the operation. This example shows why the divisor's numerator is part of the domain audit before any cancellation begins.
Related topics
Teaching and accessibility note
Try it yourself
Hints are part of learning. Open one whenever it makes the next step feel possible.
Simplify $\frac{x^2-4}{x}\div\frac{x+2}{3x}$.
- $\frac{(x-2)(x+2)}x\cdot\frac{3x}{x+2}=3(x-2)$, with $x\ne0,-2$ from the original expressions.
End of lesson
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