Simplifying rational expressions: factorising, cancelling and combining

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Pure Mathematics P3 (WMA13/01) is the first of the two A Level pure papers, and algebra and functions is one of its two heaviest topics: across the three recent P3 papers studied for these notes it carried 59 of the 225 marks, about a quarter.

Rational expressions are where that algebra starts.

On P3 they rarely stand alone: a fraction is simplified so that it can be differentiated, integrated or solved in the next part.

P3 assumes everything in P1 and P2, and this section leans hardest on factorising quadratics (P1) and the factor theorem (P2).

The idea: fractions with letters obey the rules of fractions

A rational expression is a fraction whose top and bottom are polynomials, such as $\dfrac{x + 3}{x^2 - 9}$ or $\dfrac{2}{x + 1}$ .

Nothing new happens to fractions just because they contain $x$ .

You simplify $\frac{12}{18}$ by spotting the common factor 6 and dividing it out of top and bottom; you simplify $\dfrac{x + 3}{x^2 - 9}$ by spotting the common factor $(x + 3)$ :

\frac{x + 3}{x^2 - 9} = \frac{x + 3}{(x - 3)(x + 3)} = \frac{1}{x - 3}.

Adding, subtracting, multiplying and dividing work the same way too.

To add $\frac{1}{4} + \frac{1}{6}$ you use the lowest common denominator 12; to add $\dfrac{1}{x} + \dfrac{1}{x + 2}$ you use $x(x + 2)$ .

The only extra skill is factorising, because a common factor in algebra is hidden until the expression is written as a product.

Why bother?

A simpler form is easier to use.

The expression $\dfrac{1}{x - 3}$ is far quicker to differentiate, integrate, sketch or solve than the original.

On P3 the simplification is usually part (a) and the payoff is part (b).

Why cancelling works, and when it does not

Cancelling rests on one fact: $\dfrac{a \times c}{b \times c} = \dfrac{a}{b} \times \dfrac{c}{c} = \dfrac{a}{b}$ , because $\dfrac{c}{c} = 1$ .

That argument needs two things.

$c$ must multiply the whole top and the whole bottom. That is why you factorise first. In $\dfrac{x + 3}{x + 5}$ the $x$ is added, not multiplied, so nothing cancels.

$c$ must not be zero. $\dfrac{c}{c} = 1$ only when $c \neq 0$ .

The second point has a consequence that surprises students.

The two sides of $\dfrac{x + 3}{x^2 - 9} = \dfrac{1}{x - 3}$ are not quite the same function.

Try some numbers:

$x$

$\dfrac{x + 3}{x^2 - 9}$

$\dfrac{1}{x - 3}$

$\frac{7}{7} = 1$

$\frac{3}{-9} = -\frac{1}{3}$

$-\frac{1}{3}$

$-3$

$\frac{0}{0}$ , undefined

$-\frac{1}{6}$

They agree everywhere except at $x = -3$ , where the original has no value.

So a full answer says the simplified form holds for $x \neq -3$ (and $x \neq 3$ , where both fail).

P3 questions usually avoid the issue by giving a domain such as $x > 3$ in the stem; when they do, the cancelled factor is never zero and the cancelling is safe.

Combining fractions: the lowest common denominator

To add or subtract rational expressions:

1. Factorise every denominator.

2. Build the lowest common denominator (LCD): every factor that appears, each to the highest power it appears.

3. Rewrite each fraction over the LCD, combine the numerators, and simplify.

4. Factorise the new numerator and cancel any factor it shares with the denominator.

Step 4 is where marks are won or lost on P3.

A "show that" whose printed answer is simpler than the combined fraction is telling you that something cancels, and the mark scheme wants to see the factorised form before the cancelling.

Using the LCD and not just multiplying the denominators together keeps the numerator small: $(x + 1)(x + 2)$ and $(x + 2)(x + 5)$ have LCD $(x + 1)(x + 2)(x + 5)$ , not a product with $(x + 2)^2$ in it.

Multiplying and dividing

Multiplying is easier than adding: factorise everything, multiply tops and bottoms, cancel.

Dividing by a fraction is multiplying by its reciprocal, exactly as $\frac{2}{3} \div \frac{4}{9} = \frac{2}{3} \times \frac{9}{4}$ .

Cancel before you multiply out; expanding first hides the factors you need.

Two tempting wrong ideas

"Cancel anything that appears on both top and bottom." It is tempting because cancelling numbers like $\frac{6}{9}$ feels like crossing out matching symbols.

But cancelling divides by a factor, and a term inside a sum is not a factor.

Test it: $\dfrac{x + 3}{x + 5}$ at $x = 1$ is $\frac{4}{6}$ , while "cancel the $x$ " gives $\frac{3}{5}$ .

The right picture: only a bracket (or number, or letter) that multiplies everything on top and everything on the bottom can go.

"Simplified means expanded." Many students multiply out the denominator at the end.

That is not wrong, but it hides the factors, so a later part (finding where the expression is undefined, differentiating, or cancelling with another fraction) becomes harder.

Leave denominators factorised unless the question asks for a particular form.

Where this leads

The P2 work on dividing by a linear factor and the factor theorem is the toolkit for factorising cubic numerators here.

The next section turns an "improper" rational expression into a polynomial plus a proper fraction, the form that P3 integration and the quotient rule use.

In P4, partial fractions run this section in reverse: they split one fraction with a factorised denominator back into simpler fractions.

Rational functions also model real quantities, such as average cost per item, $\dfrac{\text{total cost}}{\text{number made}}$ , which falls towards the cost of one extra item as production grows.

How the marks work

M1: a correct method applied to this question, with your values in. A slip later does not lose it. dM1 is a method mark that needs the M mark before it, and ddM1 needs the two before it.

A1: a correct answer or line, which needs its M mark. A1* is the last mark of a "show that": the printed result, reached with no errors and with at least one line before it. A1cso: correct solution only.

B1: an independent mark for a correct value, fact or feature (a domain, an intercept, a constant). ft (follow through): credit for using your own earlier wrong value correctly.

awrt: any answer which rounds to the value given. isw: ignore subsequent working, so a correct answer followed by a harmless extra line keeps its marks.

Misread: copy a number wrongly and keep the character of the question, and you lose two of the A or B marks in that part, not everything.

Where a question says solutions relying on calculator technology are not acceptable, a bare answer scores nothing, and a quadratic must be solved visibly (factorising, the formula with values in, or completing the square).

Key terms

Rational expression: a fraction in which the numerator and the denominator are polynomials.

Simplest form: a rational expression in which the numerator and the denominator have no common factor.

Lowest common denominator: the simplest expression that every denominator divides into; it contains each factor of the denominators, to the highest power in which it appears.

Denominators on P3: denominators of rational expressions will be linear or quadratic, e.g. $\dfrac{1}{ax + b}$ , $\dfrac{ax + b}{px^2 + qx + r}$ , $\dfrac{x^3 + 1}{x^2 - 1}$ .

Worked example 1: factorise and cancel

Express $\dfrac{2x^2 + 5x - 3}{x^2 - 9}$ , $x \neq \pm 3$ , as a fraction in its simplest form.

(2)

1. $\dfrac{(2x - 1)(x + 3)}{(x - 3)(x + 3)}$ M1 (both top and bottom factorised)

2. $= \dfrac{2x - 1}{x - 3}$ A1

Answer: $\dfrac{2x - 1}{x - 3}$ .

Check with $x = 1$ : the original gives $\frac{4}{-8} = -\frac{1}{2}$ and the answer gives $\frac{1}{-2} = -\frac{1}{2}$ .

Worked example 2: dividing rational expressions

Express as a single fraction in its simplest form $\dfrac{x^2 - 4x}{x^2 - x - 6} \div \dfrac{x^2 - 16}{x + 2}$ , $x > 4$ .

(3)

1. $= \dfrac{x^2 - 4x}{x^2 - x - 6} \times \dfrac{x + 2}{x^2 - 16}$ M1 (turns the second fraction upside down and multiplies)

2. $= \dfrac{x(x - 4)}{(x - 3)(x + 2)} \times \dfrac{x + 2}{(x - 4)(x + 4)}$ M1 (every expression factorised)

3. $= \dfrac{x}{(x - 3)(x + 4)}$ A1

Answer: $\dfrac{x}{(x - 3)(x + 4)}$ , which may also be written $\dfrac{x}{x^2 + x - 12}$ .

The factors $(x + 2)$ and $(x - 4)$ cancel, which is safe because the domain $x > 4$ keeps both non-zero.

Worked example 3: combine, factorise, cancel ("show that", then "hence")

(a) Show that $\dfrac{2x}{x + 3} - \dfrac{30}{x^2 + x - 6} \equiv \dfrac{2(x - 5)}{x - 2}$ , $x > 2$ . (4)

(b) Hence solve $\dfrac{2x}{x + 3} - \dfrac{30}{x^2 + x - 6} = 1$ , $x > 2$ . (2)

1. (a) $x^2 + x - 6 = (x + 3)(x - 2)$ , so the LCD is $(x + 3)(x - 2)$ B1

2. $\dfrac{2x(x - 2) - 30}{(x + 3)(x - 2)}$ M1 (each fraction over the LCD, numerators combined)

3. $= \dfrac{2x^2 - 4x - 30}{(x + 3)(x - 2)} = \dfrac{2(x - 5)(x + 3)}{(x + 3)(x - 2)}$ M1 (numerator factorised, with $(x + 3)$ seen)

4. $= \dfrac{2(x - 5)}{x - 2}$ A1*

5. (b) $\dfrac{2(x - 5)}{x - 2} = 1 \Rightarrow 2x - 10 = x - 2$ M1 (uses the printed form and multiplies up)

6. $x = 8$ A1

Answer: (a) as printed (b) $x = 8$ , which satisfies $x > 2$ .

Part (b) uses part (a): the simplified fraction turns an equation with two fractions and a quadratic denominator into a linear one.

Check in the original: $\frac{16}{11} - \frac{30}{66} = \frac{16}{11} - \frac{5}{11} = 1$ .

Worked example 4: a three-term expression (exam length)

$\mathrm{f}(x) = 1 + \dfrac{4}{x - 3} - \dfrac{24}{x^2 - 9}$ , $x > 3$ .

(a) Show that $\mathrm{f}(x) = \dfrac{x + 7}{x + 3}$ . (4)

(b) Hence write $\mathrm{f}(x)$ in the form $A + \dfrac{B}{x + 3}$ , where $A$ and $B$ are integers to be found. (2)

1. (a) LCD $(x - 3)(x + 3)$ : $\mathrm{f}(x) = \dfrac{(x^2 - 9) + 4(x + 3) - 24}{(x - 3)(x + 3)}$ M1 (all three terms over the LCD)

2. $= \dfrac{x^2 + 4x - 21}{(x - 3)(x + 3)}$ A1

3. $= \dfrac{(x + 7)(x - 3)}{(x - 3)(x + 3)}$ M1 (numerator factorised, common factor shown)

4. $= \dfrac{x + 7}{x + 3}$ A1*

5. (b) $\dfrac{x + 7}{x + 3} = \dfrac{(x + 3) + 4}{x + 3} = 1 + \dfrac{4}{x + 3}$ , so $A = 1$ B1, $B = 4$ B1

6. (c) $\dfrac{x + 7}{x + 3} = \dfrac{5}{4} \Rightarrow 4x + 28 = 5x + 15$ M1

7. $x = 13$ A1

Answer: (b) $\mathrm{f}(x) = 1 + \dfrac{4}{x + 3}$ (c) $x = 13$ , which satisfies $x > 3$ .

The form in (b) shows at a glance that $\mathrm{f}(x)$ is always a little more than 1 when $x > 3$ , and gets closer to 1 as $x$ grows; that is the kind of fact the functions sections use to find a range.

Exam tip

When a "show that" ends in a simpler fraction than the one you have built, the printed answer is telling you a factor cancels.

Factorise the combined numerator and write the factorised form, with the common bracket visible on top and bottom, before the final line.

Jumping from an expanded numerator straight to the printed result can cost the A1*, even if the algebra in your head was right.

Common mistakes

Cancelling terms, not factors. $\dfrac{x + 3}{x + 5}$ does not simplify. Factorise first; only whole brackets that multiply the entire top and bottom cancel.

Losing a bracket when subtracting. $\dfrac{2x}{x + 3} - \dfrac{x - 4}{x + 3}$ has numerator $2x - (x - 4) = x + 4$ , not $x - 4$ . Put the second numerator in brackets before expanding.

Using the product of the denominators when they share a factor. It works but creates a cubic numerator that is hard to factorise. Build the LCD from the factorised denominators.

Forgetting where the original is undefined. A cancelled factor still makes the original expression undefined where it is zero; quote the domain where the question asks for one.

Check your understanding

1. (a) Express $\dfrac{3x^2 - 12}{3x^2 + 3x - 18}$ , $x > 2$ , as a fraction in its simplest form. (2)
(b) Hence solve $\dfrac{3x^2 - 12}{3x^2 + 3x - 18} = \dfrac{5}{6}$ , $x > 2$ . (2) Total [4]

2. Express as a single fraction in its simplest form $\dfrac{x + 5}{x^2 + 3x + 2} \div \dfrac{x^2 + 2x - 15}{x^2 - 9}$ , $x > 3$ . (3) Total [3]

3. (a) Show that $\dfrac{x}{x - 4} - \dfrac{12}{x^2 - 5x + 4} \equiv \dfrac{x + 3}{x - 1}$ , $x > 4$ . (3)
(b) Hence solve $\dfrac{x}{x - 4} - \dfrac{12}{x^2 - 5x + 4} = 2$ , $x > 4$ . (2) Total [5]

4. A student writes $\dfrac{x^2 + 5x}{x^2 + 5} = 1 + x$ , "cancelling the $x^2$ and the 5".
(a) By substituting $x = 1$ , show that the student is wrong. (1)
(b) State, with a reason, whether $\dfrac{x^2 + 5x}{x^2 + 5}$ can be simplified. (2) Total [3]

5. $\mathrm{f}(x) = \dfrac{2x^2 - x - 3}{x^2 - 1}$ , $x > 1$ .
(a) Show that $\mathrm{f}(x) = \dfrac{2x - 3}{x - 1}$ . (2)
(b) Hence show that $\mathrm{f}(x) = 2 - \dfrac{1}{x - 1}$ . (2)
(c) Show that $\mathrm{f}(x) < 2$ for every value of $x$ in the domain. (2) Total [6]

6. Show that $\dfrac{x^2 - 1}{x - 1}$ and $x + 1$ take the same value for every $x$ except one, and state that value of $x$ . (3) Total [3]

Answers

1. (a) $\dfrac{3(x - 2)(x + 2)}{3(x + 3)(x - 2)}$ M1; $= \dfrac{x + 2}{x + 3}$ A1
(b) $6(x + 2) = 5(x + 3)$ M1; $x = 3$ A1 (inside $x > 2$ ; check: $\frac{15}{18} = \frac{5}{6}$ )

2. $\dfrac{x + 5}{(x + 1)(x + 2)} \times \dfrac{(x - 3)(x + 3)}{(x + 5)(x - 3)}$ M1 (reciprocal) M1 (all factorised); $= \dfrac{x + 3}{(x + 1)(x + 2)}$ A1

3. (a) $x^2 - 5x + 4 = (x - 1)(x - 4)$ ; $\dfrac{x(x - 1) - 12}{(x - 1)(x - 4)}$ M1; $= \dfrac{x^2 - x - 12}{(x - 1)(x - 4)} = \dfrac{(x - 4)(x + 3)}{(x - 1)(x - 4)}$ M1; $= \dfrac{x + 3}{x - 1}$ A1*
(b) $x + 3 = 2(x - 1)$ M1; $x = 5$ A1

4. (a) Left side $\dfrac{1 + 5}{1 + 5} = 1$ , right side $1 + 1 = 2$ , which differ, so the student is wrong B1
(b) No B1: the numerator is $x(x + 5)$ and the denominator $x^2 + 5$ has no real factors, so top and bottom share no common factor B1

5. (a) $\dfrac{(2x - 3)(x + 1)}{(x - 1)(x + 1)}$ M1; $= \dfrac{2x - 3}{x - 1}$ A1*
(b) $\dfrac{2x - 3}{x - 1} = \dfrac{2(x - 1) - 1}{x - 1}$ M1; $= 2 - \dfrac{1}{x - 1}$ A1*
(c) For $x > 1$ , $x - 1 > 0$ , so $\dfrac{1}{x - 1} > 0$ B1; so $\mathrm{f}(x)$ is 2 minus a positive number, which is less than 2 B1*

6. $\dfrac{x^2 - 1}{x - 1} = \dfrac{(x - 1)(x + 1)}{x - 1}$ M1; $= x + 1$ whenever $x - 1 \neq 0$ A1; at $x = 1$ the fraction is $\frac{0}{0}$ and has no value, while $x + 1 = 2$ , so they differ only at $x = 1$ B1

Simplifying rational expressions: factorising, cancelling and combining

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