2.9 - Simplifying, multiplying and dividing algebraic fractions

An algebraic fraction obeys the same rules as a numerical fraction, but its factors sometimes hide inside brackets. You will simplify, multiply and divide fractions with numeric, linear and quadratic denominators, and keep track of values that make the original calculation undefined.

Cancelling factors, not terms

Can $\dfrac{x^2-1}{x-1}$ really become $x+1$ , and if so, what happens when $x=1$ ? The answer depends on exactly what cancellation means.

For ordinary numbers, $\frac{6\times5}{9\times5}=\frac69=\frac23$ . You divide the entire numerator and denominator by the same non-zero factor. The letters change nothing about that rule.

You already know

Factorising rewrites a sum as a product. For example, $6x+12=6(x+2)$ , whereas $6x+12\ne6x\times12$ . Products have factors; sums have terms.

Writing a numerator as a product makes the permitted cancellation visible.

Algebraic fraction

An algebraic fraction is a quotient of algebraic expressions, with a denominator that must not be zero. For example, $\frac{x+2}{x-3}$ is defined when $x\ne3$ . The expression $\frac{x+2}{0}$ is undefined for every $x$ .

The rule below applies equally to a number, a letter or a whole bracket used as a factor.

Cancelling a common factor

\frac{AC}{BC}=\frac AB \qquad B\ne0,\ C\ne0

The common factor must multiply the whole numerator and the whole denominator. Keep the original exclusions after cancellation.
Not on the formulae sheet: learn it.

First use a numeric denominator, so that the only common factor is a number.

Worked example 1: a numeric common factor

Question. Simplify fully $\dfrac{6x+12}{9}$ .

Plan. Factorise the numerator, then reduce the numeric factors.

\begin{aligned} \frac{6x+12}{9}&=\frac{6(x+2)}9 &&\text{factorise every term}\\ &=\frac{2(x+2)}3 &&\text{divide top and bottom by }3 \end{aligned}

Why this step? The factor $3$ divides the entire numerator, including the constant term.

Answer. $\dfrac{2(x+2)}3$ , or $\dfrac{2x+4}3$ .

Check. At $x=1$ , both forms give $18\div9=2$ .

The next example changes the common factor from a number to a product containing $x$ . This introduces an excluded value.

Worked example 2: a common factor containing a letter

Question. Simplify fully $\dfrac{6x^2+12x}{9x}$ .

Plan. Record the denominator restriction, then factorise and cancel $3x$ .

\begin{aligned} 9x\ne0&\Longrightarrow x\ne0 &&\text{original denominator}\\ \frac{6x^2+12x}{9x}&=\frac{6x(x+2)}{9x} &&\text{factorise the numerator}\\ &=\frac{2(x+2)}3 &&\text{divide by }3x \end{aligned}

Why this step? Cancelling $x$ is division by $x$ , so it is permitted only when $x\ne0$ .

Answer. $\dfrac{2(x+2)}3$ , for $x\ne0$ .

Check. At $x=2$ , the original is $48/18=8/3$ , matching the answer. At $x=0$ , the original has denominator zero even though the simplified expression does not.

A sum must first become a product; crossing out matching-looking terms is not a substitute.

Watch out

Wrong: $\frac{x+6}{x}=6$ . The $x$ is a term upstairs, not a factor of the whole numerator. Right: $\frac{x+6}{x}=1+\frac6x$ , for $x\ne0$ .

In this question the common factor contains a letter again. Record its restriction before it disappears.

Revealing bracket factors

A quadratic denominator is a reason to factorise before calculating. Its brackets may also occur in the numerator, saving several lines of expansion.

Worked example 3: quadratics above and below

Question. Simplify fully $\dfrac{x^2+7x+12}{x^2+5x+6}$ .

Plan. Factorise both quadratics, then cancel the common bracket.

\begin{aligned} \frac{x^2+7x+12}{x^2+5x+6}&=\frac{(x+3)(x+4)}{(x+3)(x+2)} &&\text{factorise}\\ &=\frac{x+4}{x+2} &&\text{cancel }x+3 \end{aligned}

Why this step? The whole bracket $x+3$ is one factor, just as $5$ was a factor in the numerical example.

Answer. $\dfrac{x+4}{x+2}$ , for $x\ne-3,-2$ .

Check. At $x=0$ , $12/6=2$ and $4/2=2$ . Expanding the factorisations recovers the original quadratics.

The next example changes only the numerator, multiplying it by $x$ . It is now cubic, but taking out $x$ leaves the same quadratic to factorise.

Worked example 4: a cubic numerator

Question. Simplify fully $\dfrac{x^3+7x^2+12x}{x^2+5x+6}$ .

Plan. Take out $x$ first, then factorise the remaining quadratic.

\begin{aligned} \frac{x^3+7x^2+12x}{x^2+5x+6}&=\frac{x(x^2+7x+12)}{x^2+5x+6} &&\text{take out }x\\ &=\frac{x(x+3)(x+4)}{(x+3)(x+2)} &&\text{factorise the quadratics}\\ &=\frac{x(x+4)}{x+2} &&\text{cancel }x+3 \end{aligned}

Why this step? The extra $x$ remains in the numerator; there is no factor of $x$ in the denominator to cancel it.

Answer. $\dfrac{x(x+4)}{x+2}$ , for $x\ne-3,-2$ .

Check. At $x=2$ , the original is $60/20=3$ , matching $2(6)/4=3$ . Zero is allowed: both forms then give $0$ .

Factorisation can turn a calculation with enormous integers into a small one. For example,

\frac{10^{12}-1}{10^6-1} =\frac{(10^6-1)(10^6+1)}{10^6-1} =1\,000\,001

The difference of two squares does all the work; no long division is needed.

Watch out

Opposite brackets leave a minus sign: $2-x=-(x-2)$ , so $\frac{2-x}{x-2}=-1$ for $x\ne2$ . Cancelling them to give $1$ loses that sign.

Now the cubic numerator contains a different repeated bracket.

Multiplying algebraic fractions

Multiplication needs no common denominator. Factorise first, then cancel common factors anywhere between the combined numerator and denominator.

Multiplying fractions

\frac AB\times\frac CD=\frac{AC}{BD}\qquad B\ne0,\ D\ne0

Cancel common non-zero factors before multiplying out.
Not on the formulae sheet: learn it.

Leaving products unexpanded makes those factors easier to see.

Worked example 5: cancelling across a product

Question. Simplify fully $\dfrac{x+2}{3x}\times\dfrac{6x^2}{x+2}$ .

Plan. Combine the products, then cancel numeric, letter and bracket factors.

\begin{aligned} \frac{x+2}{3x}\times\frac{6x^2}{x+2}&=\frac{6x^2(x+2)}{3x(x+2)} &&\text{multiply tops and bottoms}\\ &=2x &&\text{cancel }3x(x+2) \end{aligned}

Why this step? Both denominators become factors of one denominator, so cancellation can cross the original fraction boundary.

Answer. $2x$ , for $x\ne0,-2$ .

Check. At $x=1$ , the original is $(3/3)(6/3)=2$ ; the answer gives $2(1)=2$ .

The next example changes only the first numerator, replacing $x+2$ by $x^2-4$ . Its factorisation reveals the same cancellable bracket and leaves an additional factor.

Worked example 6: a bracket cancels too

Question. Simplify fully $\dfrac{x^2-4}{3x}\times\dfrac{6x^2}{x+2}$ .

Plan. Factorise $x^2-4$ , then cancel numeric, letter and bracket factors.

\begin{aligned} \frac{x^2-4}{3x}\times\frac{6x^2}{x+2}&=\frac{(x-2)(x+2)6x^2}{3x(x+2)} &&\text{difference of squares}\\ &=2x(x-2) &&\text{cancel }3x(x+2) \end{aligned}

Why this step? Factoring the quadratic exposes $x+2$ ; expanding the whole product would hide it again.

Answer. $2x(x-2)$ , for $x\ne0,-2$ .

Check. At $x=3$ , the original is $(5/9)(54/5)=6$ , and $2(3)(3-2)=6$ .

The simplified answer may be a polynomial, but its original excluded values still matter.

Exam tip

Show the factorised fraction before cancelling. Correct factorisations can earn method marks even when the final simplification goes wrong.

Use the same structure with a new difference of squares.

Dividing by an algebraic fraction

To divide by a fraction, multiply by its reciprocal. The fraction you divide by must itself be defined and non-zero.

Dividing fractions

\frac AB\div\frac CD=\frac AB\times\frac DC

Here $B\ne0$ , $D\ne0$ and $C\ne0$ . The last condition prevents division by zero.
Not on the formulae sheet: learn it.

Turn the division into multiplication before deciding what can cancel.

Worked example 7: reversing the divisor

Question. Simplify fully $\dfrac{3x^2-8x-3}{2x-3}\div\dfrac{x-3}{2x}$ .

Plan. Reverse the second fraction, factorise, then cancel.

\begin{aligned} \frac{3x^2-8x-3}{2x-3}\div\frac{x-3}{2x}&=\frac{(3x+1)(x-3)}{2x-3}\times\frac{2x}{x-3} &&\text{factorise and reverse divisor}\\ &=\frac{2x(3x+1)}{2x-3} &&\text{cancel }x-3 \end{aligned}

Why this step? Reversing the second fraction alone changes division into the equivalent multiplication.

Answer. $\dfrac{2x(3x+1)}{2x-3}$ , for $x\ne0,\frac32,3$ . At $x=3$ , the divisor is zero.

Check. At $x=1$ , the original is $(-8/-1)\div(-2/2)=-8$ , matching $2(1)(4)/(-1)=-8$ .

The next example multiplies both original denominators by the same extra factor, $2x+3$ , making them quadratic. Watch how that factor cancels in the quotient while adding an excluded value.

Worked example 8: quadratic denominators

Question. Simplify fully $\dfrac{3x^2-8x-3}{4x^2-9}\div\dfrac{x-3}{4x^2+6x}$ .

Plan. Factorise each expression, reverse the divisor and cancel complete factors.

\begin{aligned} 3x^2-8x-3&=(3x+1)(x-3) &&\text{factorise the numerator}\\ 4x^2-9&=(2x-3)(2x+3) &&\text{difference of squares}\\ 4x^2+6x&=2x(2x+3) &&\text{common factor}\\ \frac{3x^2-8x-3}{4x^2-9}\div\frac{x-3}{4x^2+6x} &=\frac{(3x+1)(x-3)}{(2x-3)(2x+3)}\times\frac{2x(2x+3)}{x-3}\\ &=\frac{2x(3x+1)}{2x-3} &&\text{cancel shared brackets} \end{aligned}

Why this step? Keeping every factor in brackets shows that $x-3$ and $2x+3$ each occur once upstairs and once downstairs.

Answer. $\dfrac{2x(3x+1)}{2x-3}$ , for $x\ne-\frac32,0,\frac32,3$ .

Check. At $x=1$ , the original is $(-8/-5)\div(-2/10)=-8$ , and the answer gives $2(4)/(-1)=-8$ .

Return to the opening question: $x^2-1=(x-1)(x+1)$ , so $\frac{x^2-1}{x-1}=x+1$ for $x\ne1$ . At $x=1$ , the original remains undefined. Cancellation changes the appearance, not which original inputs were allowed.

Watch out

Wrong: reverse both fractions, or cancel before changing $\div$ to $\times$ . Right: keep the first fraction, reverse only the divisor, then factorise and cancel.

In the final question the denominator quadratics change, but the plan stays the same.

Quick Recap

Cancel common factors of the whole numerator and denominator; factorise sums first.
Keep exclusions from the original denominators, even after their factors disappear.
Multiply numerators and denominators, cancelling factors before expanding.
Divide by multiplying by the reciprocal of the second fraction; the divisor must be non-zero.

The opening fraction becomes $x+1$ for $x\ne1$ ; cancellation does not make its missing input valid.

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2.9 - Simplifying, multiplying and dividing algebraic fractions

Cancelling factors, not terms

Revealing bracket factors

Multiplying algebraic fractions

Dividing by an algebraic fraction

Quick Recap

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