Ma16 - Rational expressions and algebraic division | OCR A-Level Mathematics B (MEI)

Make a complicated algebraic fraction simpler without changing the inputs it allows. You will factorise and cancel, combine fractions and use linear algebraic division to reveal a polynomial part and a remainder.

Cancelling factors and retaining restrictions

Does cancelling a bracket make a forbidden input safe? Consider $(x^2-4x)/(x^2-16)$ . At $x=0$ it gives $0$ , and at $x=8$ it gives $2/3$ . Its simpler form will give both answers too, but it will also appear to allow an input the fraction forbids.

You already know

A fraction is undefined when its denominator is zero. Factorisation turns sums into products; cancellation divides the whole numerator and denominator by the same non-zero factor.

That last word matters. Cancelling a zero factor would be division by zero.

Rational expression

A rational expression is a quotient of polynomials with a non-zero denominator. For example, $x/(x+4)$ requires $x\ne-4$ . In $(x+2)/(x+5)$ , the $x$ terms cannot cancel because neither is a factor of its whole bracket.

Write denominator restrictions before simplifying, while all the factors are still visible.

Worked example 1: a cancelled bracket

Question. Simplify $(x^2-4x)/(x^2-16)$ and state its restrictions.

Plan. Factorise, exclude the denominator zeros and cancel only the common factor.

\begin{aligned} \frac{x^2-4x}{x^2-16}&=\frac{x(x-4)}{(x-4)(x+4)} && \text{factorise}\\ &=\frac{x}{x+4} && \text{cancel }x-4 \end{aligned}

Why this step? The cancellation requires $x\ne4$ , even though $4$ is absent from the final denominator.

Answer. $x/(x+4)$ , with $x\ne4,-4$ .

Check. At $x=8$ , the fraction is $32/48=2/3$ , matching $8/12$ . At $x=4$ , the fraction is undefined, although $4/(4+4)=1/2$ .

Change one feature: reverse the sign of the numerator. A minus sign survives the cancellation.

Worked example 2: opposite brackets

Question. Simplify $(4x-x^2)/(x^2-16)$ , stating its restrictions.

Plan. Extract a minus sign and use the same denominator factors.

\begin{aligned} \frac{4x-x^2}{x^2-16}&=\frac{-x(x-4)}{(x-4)(x+4)} && \text{factorise}\\ &=-\frac{x}{x+4} && \text{cancel }x-4 \end{aligned}

Why this step? $4-x=-(x-4)$ ; opposite brackets cancel to $-1$ , not $1$ .

Answer. $-x/(x+4)$ , with $x\ne4,-4$ .

Check. At $x=8$ , both forms give $-2/3$ .

Here is the surprise: two formulae can agree at every allowed input and still have different domains. The input $4$ stays forbidden in the opening expression. A common factor cancels; a restriction does not. These algebraic rules are not in the formula sheet: learn them.

Watch out

Wrong: $(x+2)/(x+5)=2/5$ . At $x=1$ , the actual value is $3/6=1/2$ , so that claimed cancellation fails. Factorise first; if no common factor exists, stop.

Try the opposite-sign decision with a different common bracket.

Multiplying and dividing rational expressions

Factorise before multiplying. Keeping the brackets intact often lets most of the work disappear.

Products and quotients

For defined fractions,

\frac AB\times\frac CD=\frac{AC}{BD},\qquad \frac AB\div\frac CD=\frac{AD}{BC}.

Require $B,D\ne0$ , and for division also $C\ne0$ .

Not in the formula sheet: learn it.

In division, record where the divisor is undefined or zero before inverting it.

Worked example 3: a factored product

Question. Simplify $(x^2-4)/(x-1)\times(x+2)/(x-3)$ , with restrictions.

Plan. Factorise the quadratic and multiply the numerators and denominators.

\frac{x^2-4}{x-1}\times\frac{x+2}{x-3} =\frac{(x-2)(x+2)^2}{(x-1)(x-3)}\qquad\text{factorise and multiply}.

Why this step? The two copies of $x+2$ are both on top, so they multiply rather than cancel.

Answer. $(x-2)(x+2)^2/((x-1)(x-3))$ , for $x\ne1,3$ .

Check. At $x=0$ , the product is $4\times(-2/3)=-8/3$ , as is the result.

Change only the operation to division. The divisor's zero now has a different role.

Worked example 4: a rational quotient

Question. Simplify $(x^2-4)/(x-1)\div(x+2)/(x-3)$ , with restrictions.

Plan. Exclude $1,3$ and the divisor's zero $-2$ , then invert the whole divisor.

\begin{aligned} \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{take reciprocal}\\ &=\frac{(x-2)(x-3)}{x-1} && \text{cancel }x+2 \end{aligned}

Why this step? Turning a fraction over does not make its earlier undefined input $3$ allowable. Nor may you divide by its zero at $-2$ .

Answer. $(x-2)(x-3)/(x-1)$ , for $x\ne1,3,-2$ .

Check. At $x=0$ , $4\div(-2/3)=-6$ , matching the result.

For cancellation in a product, factors can cross between the two fractions. For example,

\frac{x^2-1}{x^2-9}\times\frac{x+3}{x-1} =\frac{x+1}{x-3},\qquad x\ne-3,3,1.

Watch out

Wrong: invert only the numerator of a divisor. Right: invert the complete fraction. Keep all its earlier restrictions, as well as its zero exclusion.

Keep division but change the brackets.

Adding fractions and checking equations

To add fractions, give them the same denominator. Each numerator must gain exactly the factor its own denominator was missing.

Worked example 5: a common denominator

Question. Express $3/(x-2)+2/(x+1)$ as a single fraction.

Plan. Use the common denominator $(x-2)(x+1)$ , keeping $x\ne2,-1$ .

\begin{aligned} \frac3{x-2}+\frac2{x+1}&=\frac{3(x+1)+2(x-2)}{(x-2)(x+1)} && \text{match denominators}\\ &=\frac{5x-1}{(x-2)(x+1)} && \text{collect numerator} \end{aligned}

Why this step? Multiplying numerator and denominator by the same factor preserves each fraction's value.

Answer. $(5x-1)/((x-2)(x+1))$ , for $x\ne2,-1$ .

Check. At $x=0$ , $-3/2+2=1/2$ , matching $(-1)/(-2)$ .

Change only addition to subtraction. Brackets around the second numerator now prevent a sign error.

Worked example 6: subtract a whole numerator

Question. Express $3/(x-2)-2/(x+1)$ as a single fraction.

Plan. Use the same denominator and subtract the whole adjusted numerator.

\begin{aligned} \frac3{x-2}-\frac2{x+1}&=\frac{3(x+1)-2(x-2)}{(x-2)(x+1)} && \text{common denominator}\\ &=\frac{x+7}{(x-2)(x+1)} && \text{distribute the minus} \end{aligned}

Why this step? Subtracting $2x-4$ gives $-2x+4$ , so the constant is $7$ , not $-1$ .

Answer. $(x+7)/((x-2)(x+1))$ , with $x\ne2,-1$ .

Check. At $x=0$ , $-3/2-2=-7/2$ , matching $7/(-2)$ .

The common-denominator rule is not in the formula sheet: learn it. If a fraction equation is zero, set its numerator to zero only within the allowed domain. Here the difference is zero at $x=-7$ , which is allowed.

Restrictions can reject an apparent solution. For $(x^2-1)/(x-1)=3$ , multiplying through gives $(x-1)(x-2)=0$ . Reject $1$ , where the starting fraction is undefined; only $2$ solves the equation. Cancellation within $x\ne1$ gives $x+1=3$ directly. Both routes are valid.

Watch out

Wrong: add the denominators. Right: use a common denominator and adjust both numerators. In an equation, check every candidate against the initial restrictions.

Keep the subtraction and change a numerator.

Quotient and remainder form

An improper fraction has numerator degree at least its denominator degree. Linear algebraic division reveals its polynomial part, and the remainder stays as a fraction.

Quotient and remainder form

For $a\ne0$ , if $\mathrm f(x)\equiv(ax+b)q(x)+r$ , then

\frac{\mathrm f(x)}{ax+b}=q(x)+\frac r{ax+b},\qquad x\ne-\frac ba.

Here $q(x)$ is the quotient and $r$ is the constant remainder. Division by non-linear expressions is outside Ma16.

Not in the formula sheet: learn it.

MEI permits any correct division method. Coefficient comparison is a useful alternative to long division: propose a quotient of the right degree and match coefficients after multiplying back.

Worked example 7: coefficient comparison

Question. Write $(3x^2+2x-1)/(3x-1)$ in quotient-and-remainder form.

Plan. Set the numerator equal to $(3x-1)(Ax+B)+r$ and compare powers.

\begin{aligned} (3x-1)(Ax+B)+r&=3Ax^2+(3B-A)x+(r-B) && \text{expand}\\ 3A&=3\quad\Rightarrow\quad A=1 && \text{match }x^2\\ 3B-1&=2\quad\Rightarrow\quad B=1 && \text{match }x\\ r-1&=-1\quad\Rightarrow\quad r=0 && \text{match constant} \end{aligned}

Why this step? A quadratic divided by a linear expression has a linear quotient, so $Ax+B$ includes every possible term.

Answer. $x+1$ , for $x\ne1/3$ ; quotient $x+1$ , remainder $0$ .

Check. $(3x-1)(x+1)=3x^2+2x-1$ .

Change only the numerator's constant from $-1$ to $5$ . The comparison of the higher powers stays unchanged.

Worked example 8: a remainder changes the form

Question. Write $(3x^2+2x+5)/(3x-1)$ in quotient-and-remainder form.

Plan. Keep $A=B=1$ , then match the changed constant.

\begin{aligned} 3x^2+2x+5&=(3x-1)(x+1)+r && \text{higher powers matched}\\ 5&=-1+r\quad\Rightarrow\quad r=6 && \text{constant comparison}\\ \frac{3x^2+2x+5}{3x-1}&=x+1+\frac6{3x-1} && \text{divide the identity} \end{aligned}

Why this step? A non-zero remainder contributes an actual fraction; dropping it changes the expression.

Answer. $x+1+6/(3x-1)$ , with $x\ne1/3$ .

Check. The numerator at $1/3$ is $1/3+2/3+5=6$ , agreeing with the remainder. Multiplying back checks the complete quotient too.

For larger numerators the quotient needs more terms. For example,

2x^3+3x^2-x+4=(2x+1)(x^2+x-1)+5.

Each method still uses only a linear divisor. Compare every coefficient, including zero coefficients for missing powers, to avoid losing a quotient term.

Watch out

Wrong: remainder $6/(3x-1)$ . Right: remainder $6$ , with remainder term $6/(3x-1)$ . Also retain $x\ne1/3$ even when the remainder is zero.

Change the constant again and keep the same coefficient-comparison route.

Quick Recap

Cancel factors, and retain all inputs excluded before cancellation.
Products multiply numerator by numerator; division inverts the complete divisor and excludes its zeros.
Addition and subtraction need a common denominator; check equation roots in the initial domain.
A linear divisor gives a polynomial quotient and constant remainder, checked by multiplying back.

Cancellation does not make the opening expression safe at $x=4$ : $x/(x+4)$ matches it only with both $4$ and $-4$ still excluded.

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Ma16 - Rational expressions and algebraic division

Cancelling factors and retaining restrictions

Multiplying and dividing rational expressions

Adding fractions and checking equations

Quotient and remainder form

Quick Recap

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