Factorising and algebraic fractions · Expressions, formulae & change of subject

Factorising is expanding in reverse: you write an expression as things multiplied together. It’s how algebraic fractions get simplified, and many objective questions are nothing else.

Three ways to factorise

1. Take out a common factor. Find what divides every term, and write it outside a bracket:

6x^2 - 9x = 3x(2x - 3)

Check by expanding: $3x \times 2x = 6x^2$ and $3x \times (-3) = -9x$ ✓.

2. Group in pairs. With four terms, take a common factor from each pair; the same bracket should appear twice, and then it’s a common factor itself:

\begin{aligned} ax + ay + bx + by &= a(x + y) + b(x + y) \\ &= (a + b)(x + y) \end{aligned}

3. Difference of two squares. Anything of the form (square) − (square):

a^2 - b^2 = (a - b)(a + b)

So $x^2 - 25 = (x - 5)(x + 5)$ and $12x^2 - 75y^2 = 3(4x^2 - 25y^2) = 3(2x - 5y)(2x + 5y)$ .

Difference of two squaresCut b² from a², and the rest rearranges into (a + b)(a − b)

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Simplifying algebraic fractions

You may cancel a factor: something that multiplies the whole top and the whole bottom. You may never cancel a single term out of a sum. So: factorise first, then cancel whole brackets.

Test it with numbers. Pick an $x$ , compare the original with each “simplified” version, then try another $x$ :

Test it with numbersChange x

(2x + 6) ÷ 2 when x = 3 = 6

Simplified to	Value	Same?
x + 3	6	✓
2x + 3	9	✗
x + 6	9	✗

Simplified to

Value

Same?

x + 3

✓

2x + 3

✗

x + 6

✗

x = 3

A simplification is only right if it gives the same value as the original for every x. Wrong ones can agree by luck at one value, so try a few. Where the original is undefined (dividing by 0), skip that x.

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When is a fraction undefined?

A fraction is undefined when its denominator is 0, because you can’t divide by 0. Set the bottom equal to 0 and solve. The top doesn’t matter. If the bottom is a product or a quadratic, factorise it: $\frac{1}{x(x - 4)}$ is undefined at $x = 0$ and at $x = 4$ .

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Adding and subtracting algebraic fractions

It’s the same as with numbers: use a common denominator, then combine the tops.

\begin{aligned} \frac{1}{x - 3} - \frac{1}{x + 2} &= \frac{(x + 2) - (x - 3)}{(x - 3)(x + 2)} \\ &= \frac{5}{(x - 3)(x + 2)} \end{aligned}

Factorise the denominators first: the common denominator only needs each bracket once. For $\frac{1}{2x + 2} + \frac{1}{3x + 3} = \frac{1}{2(x + 1)} + \frac{1}{3(x + 1)}$ , it is $6(x + 1)$ , not $(2x + 2)(3x + 3)$ . And for a fraction built out of fractions, multiply its top and bottom by the common denominator of all the small ones.

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Partial fractions

Adding fractions can be run backwards: a fraction whose denominator factorises splits into simpler fractions, one for each bracket. These are called partial fractions. To split $\frac{5x - 1}{(x - 1)(x + 1)}$ , write

\frac{5x - 1}{(x - 1)(x + 1)} = \frac{A}{x - 1} + \frac{B}{x + 1}

and multiply both sides by $(x - 1)(x + 1)$ to get $5x - 1 = A(x + 1) + B(x - 1)$ . This is true for every $x$ , so choose values that make one bracket zero:

$x = 1$ : $4 = 2A$ , so $A = 2$ ;
$x = -1$ : $-6 = -2B$ , so $B = 3$ .

Partial fractionsMake each bracket zero in turn to find A and B

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Your turn

WAEC 2019 · Paper 1 · Q19

Simplify $\dfrac{x^2 - 5x - 14}{x^2 - 9x + 14}$ .

Worked solution (try it first)

Top: two numbers multiplying to $-14$ and adding to $-5$ are $-7$ and $2$ , so $x^2 - 5x - 14 = (x - 7)(x + 2)$ .
Bottom: $-7$ and $-2$ multiply to 14 and add to $-9$ , so $x^2 - 9x + 14 = (x - 7)(x - 2)$ .
Cancel $x - 7$ : $\dfrac{x + 2}{x - 2}$ , option D.

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