B6 - Rational expressions and algebraic division | AQA A-Level Mathematics

Learn to simplify and combine algebraic fractions while retaining every excluded input. You will also divide by a linear expression to separate an improper fraction into a polynomial and a proper fractional remainder.

Cancelling factors and retaining exclusions

Can cancelling a factor make an undefined fraction become defined? At $x=-3$ , the fraction $\dfrac{x^2-9}{x^2+x-6}$ has numerator and denominator both zero, so it is undefined. Yet cancelling a common factor will produce a shorter formula whose value there is $\dfrac65$ . The formulas agree on the original domain, but cancellation cannot add a missing input.

You already know

Factorisation rewrites a polynomial as a product. The difference of two squares gives $x^2-9=(x-3)(x+3)$ , and a quadratic can be checked by expanding its proposed factors.

A denominator is a divisor, so it must be nonzero before you simplify anything. Find its zeros from the original expression and keep them as excluded values.

Rational expression and domain

A rational expression is a quotient of two polynomials whose denominator is not the zero polynomial. Its domain consists of inputs where the denominator is nonzero. For example, $\dfrac{x+2}{x-1}$ excludes $x=1$ ; a zero numerator is allowed when the denominator is nonzero.

Cancellation uses $\dfrac{ab}{ac}=\dfrac bc$ only where $a\ne0$ and $c\ne0$ . It removes a common factor from the written formula, not a restriction from the original domain. This cancellation rule and its conditions are not supplied in the formula booklet.

Worked example 1: a disappearing denominator factor

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

Plan. Factorise the original denominator to find its zeros, then cancel a complete common factor.

\begin{aligned} x^2+x-6&=(x+3)(x-2) && \text{denominator factorisation} \\ (x+3)(x-2)\ne0&\ \Rightarrow\ x\ne-3,\ 2 && \text{record exclusions} \\ \frac{x^2-9}{x^2+x-6} &=\frac{(x-3)(x+3)}{(x+3)(x-2)} && \text{factorise numerator} \\ &=\frac{x-3}{x-2} && \text{cancel nonzero }x+3 \end{aligned}

Why this step? The cancelled factor is zero at $-3$ . Recording exclusions first prevents that input being mistakenly restored.

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

Check. At $x=0$ , both forms equal $\dfrac32$ . At $x=-3$ , the shorter formula would give $\dfrac65$ , but the original is undefined, so this input remains excluded.

Worked example 2 changes only the numerator. The denominator, and therefore its original exclusions, stay fixed.

Worked example 2: a different numerator, the same domain

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

Plan. Extract $x$ from the numerator and cancel the shared bracket while retaining the same restrictions.

\begin{aligned} x^2+3x&=x(x+3) && \text{common factor }x \\ \frac{x^2+3x}{x^2+x-6} &=\frac{x(x+3)}{(x+3)(x-2)} && \text{factorise both polynomials} \\ &=\frac{x}{x-2},\qquad x\ne-3,\ 2 && \text{cancel on the original domain} \end{aligned}

Why this step? Changing the numerator does not remove a zero of the original denominator. The cancelled factor still imposes $x\ne-3$ .

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

Check. $x=0$ is allowed: the original expression is $0/(-6)=0$ , as is the simplified form. A zero numerator is not an exclusion.

Only complete factors cancel. In $\dfrac{x+2}{x}$ , the numerator is a sum and does not have a factor $x$ . Instead, splitting the numerator gives $\dfrac{x+2}{x}=1+\dfrac2x$ , with $x\ne0$ . This rewriting rule is not supplied in the formula booklet.

Watch out

Wrong: $\dfrac{x+2}{x}=2$ by crossing out the two written $x$ terms. They are not common factors of the whole numerator and denominator. Right: $\dfrac{x+2}{x}=1+\dfrac2x$ , with $x\ne0$ .

The next question changes the numerator by a factor of $2$ . Check whether that factor can cancel and keep both original exclusions.

Multiplying and dividing rational expressions

Products combine numerator factors and denominator factors; division first takes a reciprocal and adds the requirement that the divisor itself is nonzero.

For ordinary numbers, $\dfrac23\div\dfrac45=\dfrac23\times\dfrac54=\dfrac56$ . The same reciprocal rule works for expressions, but the expression you are dividing by must both exist and have a nonzero value.

Products and quotients of fractions

Where all the displayed original fractions are defined,

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

For division, also require $C\ne0$ :

\frac AB\div\frac CD=\frac AB\times\frac DC=\frac{AD}{BC}

Here $A,B,C,D$ may be polynomial expressions. The conditions are checked before cancellation.

Not in the formula booklet: learn it.

List original denominator zeros first. For a division, then exclude any additional input making the divisor's numerator zero while its denominator is nonzero.

Worked example 3: multiplying factorised fractions

Question. Simplify $\dfrac{x^2-4}{x-1}\times\dfrac{x+2}{x-3}$ , stating all excluded values.

Plan. Record the original denominator restrictions, factorise and multiply.

\begin{aligned} x-1\ne0,\quad x-3\ne0&\ \Rightarrow\ x\ne1,\ 3 && \text{original restrictions} \\ \frac{x^2-4}{x-1}\times\frac{x+2}{x-3} &=\frac{(x-2)(x+2)(x+2)}{(x-1)(x-3)} && \text{factorise and multiply} \\ &=\frac{(x-2)(x+2)^2}{(x-1)(x-3)} && \text{collect repeated factors} \end{aligned}

Why this step? There is no common numerator and denominator factor to cancel. A repeated numerator factor stays in the product.

Answer. $\dfrac{(x-2)(x+2)^2}{(x-1)(x-3)}$ , with $x\ne1,3$ .

Check. At $x=0$ , the original product is $4\times\left(-\dfrac23\right)=-\dfrac83$ , matching the simplified form. The input $-2$ is allowed and makes the product zero.

Worked example 4 changes only multiplication to division. Taking the reciprocal now exposes a common factor and a further exclusion.

Worked example 4: the zero divisor restriction

Question. Simplify $\dfrac{x^2-4}{x-1}\div\dfrac{x+2}{x-3}$ , stating all excluded values.

Plan. Keep both original denominator restrictions, exclude a zero divisor and multiply by the reciprocal.

\begin{aligned} x&\ne1,\ 3 && \text{original denominators nonzero} \\ x&\ne-2 && \text{divisor must not equal zero} \\ \frac{x^2-4}{x-1}\div\frac{x+2}{x-3} &=\frac{(x-2)(x+2)}{x-1}\times\frac{x-3}{x+2} && \text{reciprocate the divisor} \\ &=\frac{(x-2)(x-3)}{x-1} && \text{cancel nonzero }x+2 \end{aligned}

Why this step? The original divisor is undefined at $3$ and zero at $-2$ . Both make the division invalid, even though cancellation moves or removes the corresponding factors.

Answer. $\dfrac{(x-2)(x-3)}{x-1}$ , with $x\ne1,3,-2$ .

Check. At $x=0$ , $4\div\left(-\dfrac23\right)=-6$ , matching the simplified form. At $x=2$ , the dividend is zero and the divisor is $-4$ , so this input is allowed and gives zero.

Keep the distinction between a zero dividend and a zero divisor. A numerator becoming zero is harmless in a product; in a divisor, it can make the operation impossible.

Watch out

Wrong: after reciprocating, allow $x=3$ because $x-3$ has moved to the numerator. Right: $x=3$ made the original divisor undefined, so it remains excluded.

The next division changes the linear factors. Make the restriction list from the original expression before simplifying it.

Adding and subtracting algebraic fractions

Addition and subtraction need a common denominator, so first rewrite both fractions as multiples of the same fractional unit.

For example, $\dfrac12+\dfrac13=\dfrac36+\dfrac26=\dfrac56$ , rather than $\dfrac25$ . With variable denominators, multiply each numerator by the factor missing from its own denominator, just as with numbers.

A common denominator

For $B\ne0$ and $D\ne0$ ,

\frac AB\pm\frac CD=\frac{AD\pm CB}{BD}

The product $BD$ is a valid common denominator, though a smaller common multiple may be available. Retain all original exclusions.

Not in the formula booklet: learn it.

Keep brackets around a subtracted numerator until the minus sign has reached every term. Inspect the completed fraction for cancellable factors only after the numerator has been combined.

Worked example 5: adding two algebraic fractions

Question. Express $\dfrac3{x-2}+\dfrac1{x+2}$ as one simplified fraction, stating excluded values.

Plan. Use $(x-2)(x+2)$ as the common denominator and add the transformed numerators.

\begin{aligned} \frac3{x-2}+\frac1{x+2} &=\frac{3(x+2)+(x-2)}{(x-2)(x+2)} && \text{supply missing factors} \\ &=\frac{3x+6+x-2}{x^2-4} && \text{expand} \\ &=\frac{4(x+1)}{x^2-4},\qquad x\ne2,\ -2 && \text{collect and factorise} \end{aligned}

Why this step? Multiplying the first numerator by $x+2$ compensates for multiplying its denominator by that same factor.

Answer. $\dfrac{4(x+1)}{x^2-4}$ , with $x\ne2,-2$ .

Check. At $x=0$ , the original sum is $-\dfrac32+\dfrac12=-1$ , and the single fraction is $\dfrac4{-4}=-1$ .

Worked example 6 changes only the addition sign to subtraction. The common denominator stays the same; the entire second numerator now changes sign.

Worked example 6: subtracting a complete numerator

Question. Express $\dfrac3{x-2}-\dfrac1{x+2}$ as one simplified fraction, stating excluded values.

Plan. Use the same common denominator, keeping brackets around the second numerator.

\begin{aligned} \frac3{x-2}-\frac1{x+2} &=\frac{3(x+2)-(x-2)}{(x-2)(x+2)} && \text{transform both numerators} \\ &=\frac{3x+6-x+2}{x^2-4} && \text{distribute the minus sign} \\ &=\frac{2(x+4)}{x^2-4},\qquad x\ne2,\ -2 && \text{collect and factorise} \end{aligned}

Why this step? Subtracting $x-2$ contributes $-x+2$ . Changing only the first sign would subtract a different expression.

Answer. $\dfrac{2(x+4)}{x^2-4}$ , with $x\ne2,-2$ .

Check. At $x=0$ , the original difference is $-\dfrac32-\dfrac12=-2$ , matching $\dfrac8{-4}=-2$ .

If a denominator already contains the other as a factor, avoid unnecessary duplication. For $x\ne1$ ,

\frac1{x-1}+\frac2{(x-1)^2} =\frac{x-1+2}{(x-1)^2} =\frac{x+1}{(x-1)^2}

The denominator $(x-1)^2$ is already a common multiple. This is the same fraction-addition rule, not a separate booklet formula.

Watch out

Wrong: $3(x+2)-(x-2)=3x+6-x-2$ . The outside minus sign reverses both signs in the bracket. Right: $3x+6-x+2=2x+8$ .

The next difference changes the factors and one numerator. Keep the subtraction visible until the brackets are expanded.

Improper expressions and linear division

When a fraction's numerator has degree at least as large as its denominator, polynomial division separates a polynomial part from a smaller fractional remainder.

For ordinary numbers, $\dfrac{17}{5}=3+\dfrac25$ , because $17=5(3)+2$ . The polynomial division identity provides exactly the same rearrangement for expressions.

Proper and improper rational expressions

A rational expression is proper when its nonzero numerator has lower degree than its denominator. It is improper when its numerator's degree is at least the denominator's degree. For example, $\dfrac2{x-1}$ is proper, while $\dfrac{x^2+1}{x-1}$ is improper. A zero remainder simply leaves the polynomial part.

Division by a linear expression leaves a constant remainder. Divide the identity by that expression, retaining the restriction that it is nonzero.

A polynomial plus a proper fraction

If $D(x)=mx+n$ , where $m\ne0$ , and

P(x)=D(x)Q(x)+r

then, where $D(x)\ne0$ ,

\frac{P(x)}{D(x)}=Q(x)+\frac{r}{D(x)}

The remainder $r$ is constant. If $r\ne0$ , the last fraction is proper.

Not in the formula booklet: learn it.

A quotient alone is insufficient when the remainder is nonzero. The remainder must stay divided by the original linear divisor.

Worked example 7: separating a monic linear divisor

Question. Express $\dfrac{2x^3+x^2-7x+5}{x-1}$ as a polynomial plus a proper fraction, and state the excluded value.

Plan. Divide the numerator by $x-1$ , then put the constant remainder over $x-1$ .

\begin{aligned} (2x^3+x^2-7x+5)-2x^2(x-1)&=3x^2-7x+5 && \text{first quotient term }2x^2 \\ (3x^2-7x+5)-3x(x-1)&=-4x+5 && \text{next term }3x \\ (-4x+5)-(-4)(x-1)&=1 && \text{constant remainder} \end{aligned}

Why this step? Dividing the identity $P(x)=(x-1)(2x^2+3x-4)+1$ by $x-1$ makes the remainder contribute $\dfrac1{x-1}$ .

Answer.

\frac{2x^3+x^2-7x+5}{x-1} =2x^2+3x-4+\frac1{x-1},\qquad x\ne1

Check. Multiplying back gives $(x-1)(2x^2+3x-4)+1=2x^3+x^2-7x+5$ . At $x=0$ , both fractional forms equal $-5$ .

Worked example 8 changes only the coefficient of $x$ in the divisor, from $1$ to $2$ . Each leading-term division must now use $2x$ .

Worked example 8: a non-monic linear divisor

Question. Express $\dfrac{2x^3+x^2-7x+5}{2x-1}$ as a polynomial plus a proper fraction, and state the excluded value.

Plan. Match the leading terms using the divisor $2x-1$ , retaining the final remainder.

\begin{aligned} (2x^3+x^2-7x+5)-x^2(2x-1)&=2x^2-7x+5 && \text{first quotient term }x^2 \\ (2x^2-7x+5)-x(2x-1)&=-6x+5 && \text{next term }x \\ (-6x+5)-(-3)(2x-1)&=2 && \text{remainder }2 \end{aligned}

Why this step? The quotient term comes from dividing leading terms: $2x^3\div2x=x^2$ , rather than $2x^2$ . The divisor's coefficient must be included each time.

Answer.

\frac{2x^3+x^2-7x+5}{2x-1} =x^2+x-3+\frac2{2x-1},\qquad x\ne\frac12

Check. $(2x-1)(x^2+x-3)+2=2x^3+x^2-7x+5$ . At $x=2$ , both forms equal $\dfrac{11}{3}$ .

Coefficient matching or inspection can also produce these identities. Check the completed expression by multiplying by the divisor and recovering every coefficient. The original exclusion remains because the division identity was divided by a quantity that must be nonzero.

Watch out

Wrong: replace the answer in Worked example 8 by $x^2+x-3+2$ . The remainder is part of the numerator, so it must still be divided by $2x-1$ . Right: $x^2+x-3+\dfrac2{2x-1}$ , with $x\ne\dfrac12$ .

For the next question, the linear divisor's leading coefficient changes to $3$ , and the numerator gives a new quotient and remainder.

Quick Recap

Find excluded values before cancelling any factors.
Cancel complete common factors, not individual terms of a sum.
For division, require the original divisor to be defined and nonzero.
Use a common denominator for addition or subtraction, keeping brackets around a subtracted numerator.
Linear division gives a polynomial plus the constant remainder over the original divisor.

Cancellation does not repair the opening fraction at $x=-3$ : its shorter formula has a value there, but equality with the original expression holds only for $x\ne-3,2$ .

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B6 - Rational expressions and algebraic division

Cancelling factors and retaining exclusions

Multiplying and dividing rational expressions

Adding and subtracting algebraic fractions

Improper expressions and linear division

Quick Recap

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