A4 - Advanced factorisation and algebraic fractions

Factorisation makes larger expressions easier to handle. Learn to expand several brackets in stages, factorise quadratics with a non-unit leading coefficient, and calculate with algebraic fractions while keeping track of their original restrictions.

Products of three or more binomials (Higher)

Can cancelling a bracket make a previously forbidden input allowed? For example, $\frac{x^2-9}{x^2+x-6}$ has a denominator of zero at $x=-3$ ; keep that input in mind as we develop the tools to simplify it.

Start with multiplication. More than two brackets need no new multiplication rule: expand a pair, simplify, then multiply the result by the next bracket.

You already know

Every term in one bracket multiplies every term in the other. Also, $(x-a)(x+a)=x^2-a^2$ ; a helpful pair can save work.

Multiplication is associative, so grouping the factors differently leaves their product unchanged. This multiplication law is not on the formula sheet: learn it. Choose an order that keeps the intermediate expression manageable.

Worked example 1: multiplying three binomials

Question. Expand and simplify $(x+1)(x+2)(x-3)$ .

Plan. Expand the first pair, then distribute all three resulting terms across the final bracket.

\begin{aligned} (x+1)(x+2)(x-3)&=(x^2+3x+2)(x-3) &&\text{expand a pair first}\\ &=x^3-3x^2+3x^2-9x+2x-6 &&\text{make all six products}\\ &=x^3-7x-6 &&\text{collect like terms} \end{aligned}

Why this step? The quadratic has three terms, so each must multiply both terms of $x-3$ .

Answer. $x^3-7x-6$ .

Check. At $x=1$ , $2\times3\times(-2)=-12$ , and $1-7-6=-12$ .

Change only the final constant from $-3$ to $-4$ . This time the squared terms will not cancel completely.

Worked example 2: retaining a squared term

Question. Expand and simplify $(x+1)(x+2)(x-4)$ .

Plan. Keep the same quadratic from the first pair and multiply it by the changed bracket.

\begin{aligned} (x+1)(x+2)(x-4)&=(x^2+3x+2)(x-4) &&\text{reuse the pair expansion}\\ &=x^3-4x^2+3x^2-12x+2x-8 &&\text{multiply every term}\\ &=x^3-x^2-10x-8 &&\text{collect each power separately} \end{aligned}

Why this step? A missing $x^2$ term in one example means its coefficient was zero, not that every cubic lacks that term.

Answer. $x^3-x^2-10x-8$ .

Check. At $x=1$ , the original gives $-18$ , and $1-1-10-8=-18$ .

A different order can help: $(x-2)(x+2)(x-1)=(x^2-4)(x-1)=x^3-x^2-4x+4$ . Four brackets can form two pairs:

\begin{aligned} (x-1)(x+1)(x-2)(x+2)&=(x^2-1)(x^2-4)\\ &=x^4-4x^2-x^2+4\\ &=x^4-5x^2+4. \end{aligned}

The same staged method continues for any finite number of brackets.

Watch out

Wrong: $(x+1)^3=x^3+1$ . Repeated multiplication gives $x^3+3x^2+3x+1$ . At $x=1$ , the original is $8$ , while the wrong expression is $2$ .

Change just the last constant once more.

Factorising a non-unit leading coefficient (Higher)

When the coefficient of $x^2$ is not $1$ , the letter terms in the brackets must multiply to that coefficient too.

For $6x^2+11x+3$ , trying $(3x+1)(2x+3)$ gets the end products right. Check the middle products as well: $9x+2x=11x$ , so this trial works.

Non-monic quadratic

A non-monic quadratic has a coefficient of $x^2$ different from $1$ . In $ax^2+bx+c$ , it is still quadratic only when $a\ne0$ .

A systematic route is to find two numbers with product $ac$ and sum $b$ , split the middle term, then factorise by grouping. The split creates a shared bracket. This factorisation method is not on the formula sheet: learn it.

Worked example 3: splitting the middle term

Question. Factorise $6x^2+11x+3$ .

Plan. Use product $6\times3=18$ and sum $11$ , then group the four terms.

\begin{aligned} 6x^2+11x+3&=6x^2+9x+2x+3 &&9\times2=18,\ 9+2=11\\ &=3x(2x+3)+1(2x+3) &&\text{factorise each pair}\\ &=(3x+1)(2x+3) &&\text{extract the shared bracket} \end{aligned}

Why this step? Splitting $11x$ as $9x+2x$ makes both groups contain $2x+3$ .

Answer. $(3x+1)(2x+3)$ .

Check. Expanding gives $6x^2+9x+2x+3=6x^2+11x+3$ .

Change only the middle coefficient from $11$ to $9$ . Now there is an overall common factor, so take it out first.

Worked example 4: a common factor before two brackets

Question. Factorise fully $6x^2+9x+3$ .

Plan. Extract $3$ , then split the middle term of the remaining quadratic.

\begin{aligned} 6x^2+9x+3&=3(2x^2+3x+1) &&\text{take out the common factor}\\ &=3(2x^2+2x+x+1) &&2\times1=2,\ 2+1=3\\ &=3\bigl[2x(x+1)+(x+1)\bigr] &&\text{group matching brackets}\\ &=3(2x+1)(x+1) &&\text{finish factorising} \end{aligned}

Why this step? Taking out the common factor first makes the smaller quadratic easier and prevents an incomplete final product.

Answer. $3(2x+1)(x+1)$ .

Check. $3(2x^2+3x+1)=6x^2+9x+3$ .

Negative constants use signed splits. For $2x^2-x-6$ , product $-12$ and sum $-1$ give $3,-4$ :

2x^2+3x-4x-6=x(2x+3)-2(2x+3)=(x-2)(2x+3).

Another common-factor example is $4x^2+10x+6=2(2x+3)(x+1)$ . A trial-bracket method and a grouping method are equally valid; expand to check all three coefficients. Some quadratics have no rational linear factors.

Watch out

Matching the first and last products is insufficient. The wrong trial $(3x+3)(2x+1)$ gives $6x^2+9x+3$ , not $6x^2+11x+3$ .

Change only the sign of the middle term from Worked example 4.

Cancelling factors and retaining restrictions (Higher)

An algebraic fraction is defined only where its original denominator is non-zero.

Cancellation removes a shared factor from the whole numerator and denominator. It cannot remove a term from a sum. For example, $\frac{x+3}{x}$ is $1+\frac3x$ for $x\ne0$ , not $3$ .

Cancelling a common factor

\frac{AB}{AC}=\frac BC

The original denominator requires $A\ne0$ and $C\ne0$ . These restrictions remain after cancellation.

Not on the formula sheet: learn it.

Factorising first reveals what can cancel. Record forbidden values before a factor disappears.

Worked example 5: simplifying quadratic factors

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

Plan. Factorise both quadratics, find the original denominator's zeros, then cancel their common factor.

\begin{aligned} \frac{x^2-9}{x^2+x-6}&=\frac{(x-3)(x+3)}{(x+3)(x-2)} &&\text{factorise top and bottom}\\ &=\frac{x-3}{x-2} &&\text{cancel the factor }x+3 \end{aligned}

Why this step? $x+3$ multiplies the whole remaining numerator and denominator, so it is a factor rather than a separate term.

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

Check. At $x=4$ , the original is $7/14=1/2$ , and the answer is $1/2$ . At $x=-3$ , the original is still undefined.

Change only the sign of the entire numerator, writing it as $9-x^2$ . Cancellation must preserve that negative factor.

Worked example 6: keeping a reversed sign

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

Plan. Take out the numerator's minus sign, then use the same factorisation.

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

Why this step? Reversing a subtraction changes its sign: $3-x=-(x-3)$ .

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

Check. At $x=4$ , the original is $-7/14=-1/2$ , matching $(3-4)/(4-2)$ .

A numerical and letter factor can cancel together:

\frac{6x^2+3x}{9x}=\frac{3x(2x+1)}{3x\times3}=\frac{2x+1}{3}\quad(x\ne0).

Also, $\frac{x-3}{3-x}=-1$ for $x\ne3$ . Neither cancellation permits the input that made the original denominator zero.

Watch out

Wrong: $\frac{x+3}{x}=3$ . The numerator is a sum, so $x$ is not a factor of the whole numerator. For $x\ne0$ , the correct form is $1+\frac3x$ . Cancelling whole factors also keeps every original excluded value.

Change only the coefficient of the numerator in Worked example 6.

Operations with algebraic fractions (Higher)

Algebraic fractions obey numerical fraction rules, with restrictions checked before simplifying. These fraction operation rules are not on the formula sheet: learn them.

For addition and subtraction, use a common denominator. If it is already shared, combine the numerators: $\frac{x+4}{x-2}-\frac3{x-2}=\frac{x+1}{x-2}$ , for $x\ne2$ . With different denominators, multiply top and bottom by the missing factor.

Worked example 7: adding with different denominators

Question. Write $\frac2{x+1}+\frac3{x-1}$ as one fraction.

Plan. Use denominator $(x+1)(x-1)$ , keeping each original denominator's restrictions.

\begin{aligned} \frac2{x+1}+\frac3{x-1}&=\frac{2(x-1)+3(x+1)}{(x+1)(x-1)} &&\text{multiply by missing factors}\\ &=\frac{2x-2+3x+3}{x^2-1} &&\text{expand the numerator}\\ &=\frac{5x+1}{x^2-1} &&\text{collect like terms} \end{aligned}

Why this step? Multiplying both top and bottom by the same non-zero factor preserves each fraction's value.

Answer. $\frac{5x+1}{x^2-1}$ , for $x\ne-1,1$ .

Check. At $x=2$ , $2/3+3=11/3$ , matching $(10+1)/(4-1)$ .

Change only the operation from addition to subtraction. Group the entire second numerator after the minus sign.

Worked example 8: subtracting a whole numerator

Question. Write $\frac2{x+1}-\frac3{x-1}$ as one fraction.

Plan. Use the same denominator and subtract the complete second numerator.

\begin{aligned} \frac2{x+1}-\frac3{x-1}&=\frac{2(x-1)-3(x+1)}{(x+1)(x-1)} &&\text{form a common denominator}\\ &=\frac{2x-2-3x-3}{x^2-1} &&\text{distribute the subtraction}\\ &=\frac{-x-5}{x^2-1} &&\text{collect terms} \end{aligned}

Why this step? The minus applies to $3x+3$ , so both terms are subtracted.

Answer. $\frac{-x-5}{x^2-1}$ , for $x\ne-1,1$ .

Check. At $x=2$ , $2/3-3=-7/3$ , matching the answer.

Factor denominators before choosing a common multiple. For $\frac1x+\frac2{x(x+1)}$ , the smaller common denominator is already $x(x+1)$ , giving $\frac{x+3}{x(x+1)}$ , with $x\ne0,-1$ .

Multiplication needs no common denominator. Multiply the numerators and denominators, cancelling whole factors where possible.

Worked example 9: multiplying factored fractions

Question. Simplify $\frac{x^2-9}{x^2+5x+6}\times\frac{x+2}{x-3}$ .

Plan. Record the original exclusions and factorise before multiplying.

\begin{aligned} \frac{x^2-9}{x^2+5x+6}\times\frac{x+2}{x-3}&=\frac{(x-3)(x+3)}{(x+2)(x+3)}\times\frac{x+2}{x-3} &&\text{factorise the quadratics}\\ &=1 &&\text{cancel all shared factors} \end{aligned}

Why this step? Every remaining factor appears once above and once below; all are non-zero on the permitted domain.

Answer. $1$ , for $x\ne-3,-2,3$ .

Check. At $x=0$ , $(-9/6)(2/-3)=1$ . A fraction can simplify to $1$ yet retain three forbidden inputs.

Change only multiplication to division. The second fraction must now be inverted, and the divisor must be non-zero.

Worked example 10: dividing by a fraction

Question. Simplify $\frac{x^2-9}{x^2+5x+6}\div\frac{x+2}{x-3}$ .

Plan. Multiply by the reciprocal of the second fraction, then cancel factors.

\begin{aligned} \frac{x^2-9}{x^2+5x+6}\div\frac{x+2}{x-3}&=\frac{(x-3)(x+3)}{(x+2)(x+3)}\times\frac{x-3}{x+2} &&\text{invert the divisor}\\ &=\frac{(x-3)^2}{(x+2)^2} &&\text{cancel only }x+3 \end{aligned}

Why this step? Dividing by a fraction means multiplying by its reciprocal; leaving it unchanged would calculate a product instead.

Answer. $\frac{(x-3)^2}{(x+2)^2}$ , for $x\ne-3,-2,3$ . The divisor is zero at $-2$ , which is already excluded.

Check. At $x=0$ , $(-9/6)\div(2/-3)=9/4$ , matching $(-3)^2/2^2$ .

Division can add an exclusion that no original denominator reveals. For example,

\frac{x^2-4}{x+1}\div\frac{x-2}{x+1} =\frac{(x-2)(x+2)}{x+1}\times\frac{x+1}{x-2}=x+2.

Here $x\ne-1$ because of the original denominator, and $x\ne2$ because the divisor must not be zero. However, $x=-2$ is allowed: zero divided by a non-zero number is zero.

Watch out

You cannot add denominators: $\frac1a+\frac1b\ne\frac2{a+b}$ in general. For division, check both the original denominators and the divisor's numerator before cancellation hides them.

Change one numerator coefficient in Worked example 8 and one in Worked example 10. Give simplified answers on the stated domains.

Quick Recap

Expand several brackets in manageable stages, choosing useful pairs first.
For non-monic quadratics, trial brackets or split the middle term using product $ac$ and sum $b$ .
Cancel whole factors after factorising; retain the original restrictions.
Add and subtract with a common denominator. Multiply factors directly; divide by multiplying by the reciprocal of a non-zero divisor.

Cancelling $x+3$ in the opening fraction never permits $x=-3$ : the original denominator is still zero there.

Last Accessed

Papers 1, 2 and 3

A4 - Advanced factorisation and algebraic fractions

Products of three or more binomials (Higher)

Factorising a non-unit leading coefficient (Higher)

Cancelling factors and retaining restrictions (Higher)

Operations with algebraic fractions (Higher)

Quick Recap

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