2.2 - Algebraic fractions

You will simplify, multiply, divide, add and subtract fractions containing algebra. This Higher lesson uses numerical, linear and quadratic numerators and denominators. You will also keep track of the inputs for which the original calculation is defined.

Factors and excluded values (Higher)

If cancellation removes a denominator, does its forbidden input become allowed? For example, can simplifying $\frac{x^2-9}{x^2+x-6}$ make $x=-3$ safe to use? Keep that question beside the algebra: an answer includes both a simpler expression and its restrictions.

You already know

A fraction is division. Its denominator cannot be zero. Factorising a quadratic writes it as a product that you can check by expanding.

In $\frac{3x}{6x}$ , the whole numerator and denominator share a factor $x$ . For $x\ne0$ , dividing both by that factor gives $\frac12$ . In $\frac{x+3}{x}$ , the numerator is a sum with no common factor $x$ , so that cancellation is unavailable.

Algebraic fraction and excluded value

An algebraic fraction represents division involving algebraic expressions. An excluded value makes the original calculation undefined. Its domain is the set of allowed input values. Find denominator zeros before cancelling.

Cancellation divides the entire numerator and denominator by the same nonzero factor. It does not cross out matching terms in sums. Factor first so that you can see the whole products.

Cancelling a common factor

\frac{AC}{BC}=\frac AB\qquad\text{when }B\ne0\text{ and }C\ne0

Keep the original restrictions, including those hidden after cancellation.

Not in the Higher formulae sheet: learn it. Algebraic-fraction manipulation is not required at Foundation.

Now use quadratic factorisation to reveal the factors.

Worked example 1: cancelling a quadratic factor

Question. Simplify $\frac{x^2-9}{x^2+x-6}$ , stating the excluded values.

Plan. Factorise both polynomials, identify original denominator zeros, then cancel a nonzero common factor.

\begin{aligned} x^2+x-6&=(x+3)(x-2) &&\text{zeros at }-3,2\\ \frac{x^2-9}{x^2+x-6}&=\frac{(x-3)(x+3)}{(x+3)(x-2)} &&\text{factorise}\\ &=\frac{x-3}{x-2} &&\text{cancel }x+3 \end{aligned}

Why this step? Cancelling $x+3$ is division by it, so it requires $x\ne-3$ .

Answer. $\frac{x-3}{x-2}$ , with $x\ne-3,\ 2$ .

Check. At $x=0$ , the original is $\frac{-9}{-6}=\frac32$ , matching $\frac{-3}{-2}$ . At $x=-3$ , the original denominator is still zero.

Change only the numerator. The denominator and its excluded values stay the same.

Worked example 2: a common letter factor

Question. Simplify $\frac{x^2+3x}{x^2+x-6}$ , stating the excluded values.

Plan. Factorise the new numerator and cancel the same common bracket.

\begin{aligned} \frac{x^2+3x}{x^2+x-6}&=\frac{x(x+3)}{(x+3)(x-2)} &&\text{factorise both parts}\\ &=\frac{x}{x-2} &&\text{cancel }x+3 \end{aligned}

Why this step? The factor $x+3$ multiplies the whole numerator and denominator, so it may be cancelled on the original domain.

Answer. $\frac{x}{x-2}$ , with $x\ne-3,\ 2$ .

Check. At $x=1$ , the original gives $\frac4{-4}=-1$ , and the answer gives $\frac1{-1}=-1$ . Zero is allowed: both expressions give zero there.

Another useful cancellation is $\frac{2x^2+6x}{x^2-9}=\frac{2x(x+3)}{(x-3)(x+3)}=\frac{2x}{x-3}$ , with $x\ne-3,\ 3$ . A zero numerator is allowed when the denominator is nonzero.

Watch out

The line $\frac{x+3}{x}=3$ is wrong. The numerator is not a product containing a factor $x$ . At $x=2$ , the original is $\frac52$ , not 3. You may instead write $1+\frac3x$ , for $x\ne0$ .

List the restrictions from the original denominator, even if the final denominator is simpler.

Multiplying and dividing fractions (Higher)

Algebraic fractions multiply like numerical fractions. Keeping numerators and denominators factorised often lets you cancel before making large expansions.

Multiplication and division of fractions

\frac AB\times\frac CD=\frac{AC}{BD},\qquad \frac AB\div\frac CD=\frac AB\times\frac DC

Multiplication requires $B\ne0$ and $D\ne0$ . Division also requires $C\ne0$ , because the divisor must be nonzero.

Not in the Higher formulae sheet: learn them.

For a product, start with both original denominators. Do not exclude an input merely because it makes a numerator zero.

Worked example 3: multiplying with cancellation

Question. Simplify $\frac{x^2-4}{3x}\times\frac{6x}{x+2}$ , stating the excluded values.

Plan. Record the denominator restrictions, factorise, then cancel.

\begin{aligned} x&\ne0,\ -2 &&\text{original denominators}\\ \frac{x^2-4}{3x}\times\frac{6x}{x+2} &=\frac{(x-2)(x+2)6x}{3x(x+2)} &&\text{factorise and multiply}\\ &=2(x-2) &&\text{cancel common factors} \end{aligned}

Why this step? Cancelling before expansion shows both $x$ and $x+2$ as whole factors.

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

Check. At $x=3$ , $\frac59\times\frac{18}{5}=2$ , matching $2(1)$ . The input $x=2$ is allowed and gives zero.

Change only the numerical multiplier in the second numerator. The factors and restrictions are unchanged.

Worked example 4: changing a numerical factor

Question. Simplify $\frac{x^2-4}{3x}\times\frac{9x}{x+2}$ , stating the excluded values.

Plan. Use the same factorisation and simplify the coefficient $\frac93$ .

\begin{aligned} \frac{x^2-4}{3x}\times\frac{9x}{x+2} &=\frac{(x-2)(x+2)9x}{3x(x+2)} &&\text{factorise}\\ &=3(x-2) &&\text{cancel and divide }9\text{ by }3 \end{aligned}

Why this step? Cancelling letter factors must leave the correct numerical coefficient.

Answer. $3(x-2)$ , with $x\ne0,\ -2$ .

Check. At $x=3$ , $\frac59\times\frac{27}{5}=3$ , matching $3(1)$ .

Division has one extra condition. The second fraction must exist and must not equal zero. After recording those conditions, invert the whole second fraction, then multiply.

Worked example 5: dividing by a fraction

Question. Simplify $\frac{x^2-1}{x+3}\div\frac{x-1}{x+3}$ , stating the excluded values.

Plan. Exclude denominator zeros and a zero divisor, then multiply by the reciprocal.

\begin{aligned} x&\ne-3,\ 1 &&\text{defined, nonzero divisor}\\ \frac{x^2-1}{x+3}\div\frac{x-1}{x+3} &=\frac{(x-1)(x+1)}{x+3}\times\frac{x+3}{x-1} &&\text{invert the divisor}\\ &=x+1 &&\text{cancel factors} \end{aligned}

Why this step? The input 1 makes the divisor zero, so it is forbidden even though it is not a denominator zero initially.

Answer. $x+1$ , with $x\ne-3,\ 1$ .

Check. At $x=2$ , $\frac35\div\frac15=3$ , matching $2+1$ .

Change only the sign in the divisor's numerator. The zero-divisor restriction changes with it.

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Handling data

2.2 - Algebraic fractions

Factors and excluded values (Higher)

Multiplying and dividing fractions (Higher)

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