1.02k - Rational expressions and algebraic division | OCR A-Level Mathematics A

Simplify algebraic fractions while keeping track of their allowed inputs. Combine and divide rational expressions, separate a quotient from its remainder and use factorisation to solve polynomial equations through degree four.

Cancelling factors and retaining restrictions

Can $(x^2-1)/(x-1)=2$ have no solution even though the fraction simplifies to $x+1$ ? Try $x=1$ : the simplified expression gives $2$ , but the fraction gives $0/0$ , which is undefined. One missing input can change an equation's answer completely.

You already know

A denominator cannot be zero. Factorisation rewrites a sum as a product, and cancelling means dividing the whole numerator and denominator by the same non-zero factor.

An expression includes its restrictions, not just its visible formula.

Rational expression

A rational expression is a quotient of polynomials, defined only where its denominator is non-zero. For example, $(x+1)/(x-2)$ requires $x\ne2$ . The expression $(x+1)/(x+3)$ cannot simplify to $1/3$ : the $x$ terms are not factors of the whole numerator and denominator.

List restrictions before cancelling; afterwards a vanished factor may hide one.

Worked example 1: a cancelled factor

Question. Simplify $(x^2-9)/(x^2+x-6)$ , retaining all restrictions.

Plan. Factorise both polynomials, identify denominator zeros and cancel the shared factor.

\begin{aligned} \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? Cancellation is valid only when $x+3\ne0$ ; its zero stays excluded.

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

Check. At $x=0$ , both give $3/2$ . At $x=-3$ , the first is undefined, even though the simplified formula alone would give $6/5$ .

Change one feature: the shared factor appears with the opposite sign.

Worked example 2: opposite factors

Question. Simplify $(9-x^2)/(x^2+x-6)$ , retaining all restrictions.

Plan. Extract a minus sign before cancelling the same factor.

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

Why this step? Opposite factors differ by $-1$ , so cancelling them does not remove that sign.

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

Check. At $x=0$ , both forms give $-3/2$ .

These factorisation and cancellation rules are not in the formula sheet: learn them. Two formulae obtained by cancellation agree on the earlier expression's domain; they do not automatically describe functions with the same domain.

Watch out

Wrong: $(x+3)/(x+5)=3/5$ . Right: there is no common factor to cancel. At $x=1$ , the fraction is $4/6=2/3$ , not $3/5$ .

Keep the opposite-sign issue, with different factors.

Multiplying and dividing rational expressions

Products let you cancel across factors. Division adds one more condition: the expression you divide by must itself be non-zero.

Multiplication and division of fractions

For defined fractions,

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

Multiplication requires $B,D\ne0$ ; division also requires $C\ne0$ .

Not in the formula sheet: learn it.

Determine the restrictions before turning a divisor upside down. Otherwise an input at which it was undefined may disappear from sight.

Worked example 3: multiply factored expressions

Question. Simplify $(x^2-9)/(x-1)\times(x+3)/(x-2)$ , with restrictions.

Plan. Record denominator zeros, factorise and multiply.

\begin{aligned} \frac{x^2-9}{x-1}\times\frac{x+3}{x-2} &=\frac{(x-3)(x+3)^2}{(x-1)(x-2)} && \text{factorise and multiply} \end{aligned}

Why this step? Both copies of $x+3$ are in the numerator; neither cancels.

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

Check. At $x=0$ , the product is $9\times(-3/2)=-27/2$ , matching the result.

Change only the operation from multiplication to division. The second fraction must be inverted, and its zero becomes forbidden.

Worked example 4: divide by the whole fraction

Question. Simplify $(x^2-9)/(x-1)\div(x+3)/(x-2)$ , with restrictions.

Plan. Exclude original denominator zeros and the divisor's zero, then multiply by its reciprocal.

\begin{aligned} \frac{x^2-9}{x-1}\div\frac{x+3}{x-2} &=\frac{(x-3)(x+3)}{x-1}\times\frac{x-2}{x+3} && \text{invert whole divisor}\\ &=\frac{(x-3)(x-2)}{x-1} && \text{cancel common factor} \end{aligned}

Why this step? At $x=-3$ , the divisor is zero; at $x=2$ , it is undefined. Both remain excluded after inversion.

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

Check. At $x=0$ , $9\div(-3/2)=-6$ , matching the result.

For a product with common factors, factorise first: $(x^2-1)/(x^2-4)\times(x+2)/(x-1)=(x+1)/(x-2)$ , still with $x\ne-2,2,1$ .

Watch out

Wrong: keep only zeros of the final denominator. Right: retain the original denominator zeros and, for division, zeros of the divisor too.

Try the division decision with a new pair of brackets.

Adding fractions and checking equations

Addition needs a common denominator, because it combines whole fractions rather than factors. For example,

\frac{x}{x-2}-\frac2{x+2}=\frac{x(x+2)-2(x-2)}{(x-2)(x+2)}=\frac{x^2+4}{x^2-4},\qquad x\ne\pm2.

The brackets protect the subtracted numerator. This rule is not in the formula sheet: learn it.

Worked example 5: add fractions in an equation

Question. Solve $2/(x-1)+3/(x+1)=0$ .

Plan. Exclude $\pm1$ , combine the fractions and set the numerator equal to zero.

\begin{aligned} \frac2{x-1}+\frac3{x+1}&=\frac{2(x+1)+3(x-1)}{(x-1)(x+1)} && \text{common denominator}\\ &=\frac{5x-1}{x^2-1} && \text{collect numerator}\\ 5x-1&=0\quad\Rightarrow\quad x=\tfrac15 && \text{non-zero denominator} \end{aligned}

Why this step? A defined fraction is zero exactly when its numerator is zero.

Answer. $x=1/5$ .

Check. $2/(-4/5)+3/(6/5)=-5/2+5/2=0$ ; the input is allowed.

Change only the plus sign to a minus sign. It changes every term in the second numerator.

Worked example 6: subtract fractions in an equation

Question. Solve $2/(x-1)-3/(x+1)=0$ .

Plan. Use the same denominator, with brackets around the subtracted numerator.

\begin{aligned} \frac2{x-1}-\frac3{x+1}&=\frac{2(x+1)-3(x-1)}{(x-1)(x+1)} && \text{common denominator}\\ &=\frac{5-x}{x^2-1} && \text{distribute the minus}\\ 5-x&=0\quad\Rightarrow\quad x=5 && \text{solve numerator equation} \end{aligned}

Why this step? The numerator is $2x+2-3x+3$ , not $2x+2-3x-3$ .

Answer. $x=5$ , with $\pm1$ excluded from the outset.

Check. $2/4-3/6=0$ .

Multiplying an equation by its denominator is reversible only where that denominator is non-zero. For $(x^2-4)/(x-2)=5$ , cross-multiplication gives $(x-2)(x-3)=0$ . Reject $2$ , where the fraction was undefined; the sole solution is $3$ . Cancellation within $x\ne2$ gives $x+2=5$ directly.

Watch out

Wrong: $1/(x-1)+1/(x+1)=2/(2x)$ . Right: use $(x-1)(x+1)$ as the common denominator and adjust both numerators. Denominators do not add.

Keep subtraction, with new numerators and denominator zeros.

Quotient and remainder form

An improper rational expression has numerator degree at least as large as denominator degree. Division separates its polynomial part from a proper fraction.

Rational quotient and remainder form

If $a\ne0$ and $\mathrm f(x)\equiv(ax-b)q(x)+r$ , then

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

The polynomial identity holds everywhere; the fractional equation excludes the divisor's zero.

Not in the formula sheet: learn it.

Use the divisor's actual leading coefficient when choosing quotient terms. A non-monic divisor is still a linear divisor.

Worked example 7: division with no remainder

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

Plan. Divide successively using $x^2,x,-1$ .

\begin{aligned} (2x^3+3x^2-x-1)-(2x+1)x^2&=2x^2-x-1 && \text{first subtraction}\\ (2x^2-x-1)-(2x+1)x&=-2x-1 && \text{second subtraction}\\ (-2x-1)-(2x+1)(-1)&=0 && \text{last subtraction} \end{aligned}

Why this step? Division by $2x$ determines the quotient terms; division by $x$ alone would double them incorrectly.

Answer. $x^2+x-1$ , with $x\ne-1/2$ ; quotient $x^2+x-1$ , remainder $0$ .

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

Change only the constant in the numerator, from $-1$ to $4$ . The quotient survives; the remainder changes.

Worked example 8: retain the remainder

Question. Write $(2x^3+3x^2-x+4)/(2x+1)$ in quotient-and-remainder form.

Plan. Repeat the same quotient terms while retaining the new constant.

\begin{aligned} \mathrm f(x)-(2x+1)x^2&=2x^2-x+4 && \text{remove cubic term}\\ (2x^2-x+4)-(2x+1)x&=-2x+4 && \text{remove quadratic term}\\ (-2x+4)-(2x+1)(-1)&=5 && \text{constant remainder} \end{aligned}

Why this step? A remainder does not disappear just because the quotient looks simpler.

Answer. $x^2+x-1+5/(2x+1)$ , with $x\ne-1/2$ .

Check. $(2x+1)(x^2+x-1)+5=2x^3+3x^2-x+4$ ; also $\mathrm f(-1/2)=5$ .

The quotient is $x^2+x-1$ , and the remainder is $5$ , rather than $5/(2x+1)$ . Coefficient comparison or any correct linear-division method can produce this identity.

Watch out

Wrong: drop the remainder. Right: keep it divided by the divisor. Multiplication back must recover every term in the numerator.

Change the constant once more.

Solving equations through degree four

A quartic has degree $4$ , but its factors may be familiar. Look first for a quadratic in $x^2$ ; otherwise try useful linear factors and divide.

Worked example 9: a quadratic in a square

Question. Solve $x^4-5x^2+4=0$ .

Plan. Set $u=x^2$ , solve the quadratic and recover all real square roots.

\begin{aligned} u^2-5u+4&=(u-1)(u-4)=0 && \text{set }u=x^2\\ x^2&=1\text{ or }4 && \text{return to }x\\ x&=\pm1\text{ or }\pm2 && \text{both square roots} \end{aligned}

Why this step? Solving for $u$ is not the last step; each positive value gives two possible values of $x$ .

Answer. $x=-2,-1,1,2$ .

Check. The two squared values give $1-5+4=0$ and $16-20+4=0$ .

Change one feature: odd powers are now present, so $u=x^2$ cannot capture the whole equation. Use successive linear divisions instead.

Worked example 10: a general quartic

Question. Solve $x^4+x^3-7x^2-x+6=0$ .

Plan. Test $1,-1$ , divide out their factors and solve the remaining quadratic.

\begin{aligned} \mathrm f(1)&=1+1-7-1+6=0 && \text{first factor}\\ \mathrm f(-1)&=1-1-7+1+6=0 && \text{second factor}\\ \mathrm f(x)&=(x-1)(x^3+2x^2-5x-6) && \text{divide by }x-1\\ &=(x-1)(x+1)(x^2+x-6) && \text{divide by }x+1\\ &=(x-1)(x+1)(x+3)(x-2) && \text{factorise quadratic} \end{aligned}

Why this step? A full factorisation proves that the list is complete. Two successful substitutions alone do not.

Answer. $x=1,-1,-3,2$ .

Check. Multiplying $(x^2-1)(x^2+x-6)$ restores the quartic.

All these divisions are by linear expressions. If a substituted square is negative, it gives no real roots: $x^4+3x^2-4=0$ becomes $(x^2+4)(x^2-1)=0$ , so only $x=\pm1$ is real. Quartic-solving methods are not in the formula sheet: learn them.

Watch out

Wrong: $x^2=4\Rightarrow x=4$ . Right: $x=\pm2$ . Also, a repeated factor can give fewer distinct roots than the degree.

Keep the successive-factor method with new coefficients and given starting roots.

Quick Recap

Cancel whole factors and retain every original excluded input.
Division by a fraction also excludes its zeros; invert the whole divisor.
Add over a common denominator; solve within its non-zero domain and reject forbidden candidates.
Polynomial division gives a quotient plus a remainder fraction; quartics may reduce to quadratics or successive linear factors.

The opening equation has no solution: cancellation gives $x+1=2$ , hence $x=1$ , precisely the input at which the fraction is undefined.

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1.02k - Rational expressions and algebraic division

Cancelling factors and retaining restrictions

Multiplying and dividing rational expressions

Adding fractions and checking equations

Quotient and remainder form

Solving equations through degree four

Quick Recap

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