More factorising · Quadratics & their graphs

Beyond the basic $x^2 + bx + c$ , JAMB often tests four more factorising skills: spotting a difference of two squares, grouping four terms, seeing a quadratic in disguise, and simplifying fractions.

Difference of two squares

$a^2 - b^2 = (a - b)(a + b)$ . Look for it whenever two perfect squares are subtracted. JAMB asks for this often.

$x^2 - 16 = (x - 4)(x + 4)$
$25x^2 - 4y^2 = (5x - 2y)(5x + 2y)$
The squares can be whole brackets: $(p + 1)^2 - q^2 = (p + 1 - q)(p + 1 + q)$ .

It makes some arithmetic quick too: $57^2 - 43^2 = (57 - 43)(57 + 43)$ , which is $14 \times 100 = 1400$ .

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

Why it works: cut a small square from a big one, and the piece that’s left rearranges into a rectangle.

The difference of two squaresSet a and b, then rearrange

49 − 9 = 40a² − b²10 × 4 = 40(a + b)(a − b)

a = 7b = 3

A square of side a = 7, with a corner b × b = 3² cut away. What is left has area a² − b² = 40. Press Rearrange to see it as a rectangle.

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Factorising by grouping

Four terms with no common factor can often be factorised in pairs. Take a common factor out of each pair, and the same bracket appears twice. For $ax - ay + 3x - 3y$ :

take a common factor from each pair: ${a(x - y) + 3(x - y)}$ ;
take out the shared bracket: ${(a + 3)(x - y)}$ .

If the brackets don’t match, try pairing the terms differently.

ax + ay + bx + by

= a(x + y) + b(x + y)

= (a + b)(x + y)

Pair the terms, take a common factor out of each pair; the two brackets must match

Factorising by groupingTwo pairs, one common bracket

More: grouping and common factors

Quadratics in disguise

Some equations are quadratics in disguise. Look for one piece that repeats, such as $\sqrt y$ or $3^x$ , with its square also there. Call that piece $u$ . Solve the quadratic in $u$ , then turn each $u$ back.

y − 7√y + 12 = 0

let u = √y: u² − 7u + 12 = 0

u = 3 or 4, so y = 9 or 16

Swap in u, solve the quadratic, then turn u back into the original

Quadratics in disguiseLet u stand for the repeated piece, solve for u, then go back

Worked example · JAMB 1990

JAMB 1990 · UME · Q21

Find the solution of the equation $x - 8\sqrt x + 15 = 0$ .

Spot the repeated piece
$x = (\sqrt x)^2$ , so let $u = \sqrt x$ . The equation becomes $u^2 - 8u + 15 = 0$ .
Think first. x = (√x)². What should u be?
Solve for u
$(u - 3)(u - 5) = 0$ , so $u = 3$ or $u = 5$ .
Think first. Two numbers that multiply to 15 and add to −8.
Turn u back
$x = 9$ or $x = 25$ . Both check in the original: $9 - 24 + 15 = 0$ ✓ and $25 - 40 + 15 = 0$ ✓. Option C.
Think first. √x = 3 and √x = 5.

When you square both sides to remove a square root, always check your answers in the original equation. Squaring can bring in an extra answer that doesn’t work.

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Using factorising: algebraic fractions

To simplify a fraction, factorise the top and the bottom, then cancel any bracket they share (see algebraic fractions):

\begin{aligned} &\frac{x^2 - 4}{x^2 + 5x + 6} \\[4pt] &= \frac{(x - 2)(x + 2)}{(x + 2)(x + 3)} \\[4pt] &= \frac{x - 2}{x + 3} \end{aligned}

The second line factorises the top and the bottom; the third cancels $(x + 2)$ .

A fraction is undefined when its bottom is 0, because you can’t divide by 0. To find those values, solve “bottom = 0”.

More: fractions and factorising

Your turn

WAEC 2025 · Paper 2 · Q1 (a)✱✱

(a)
Factorize $px - 2qx - 4qy + 2py$ .

Worked solution (try it first)

(a)

Group the terms in $x$ and the terms in $y$ : $(px - 2qx) + (2py - 4qy) = x(p - 2q) + 2y(p - 2q)$ .
$(p - 2q)$ is a common factor: $px - 2qx - 4qy + 2py = (p - 2q)(x + 2y)$ .

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