Algebra · Ordinary Level
Factorising
Ordinary Level · Class 3 of 8 · Tap NEXT to begin
Section 1 of 3
Finding Factors & HCF
Factorising is the reverse of multiplying out. Section 2 turned $x(x+3)$ into $x^2+3x$; now we go back the other way. First, spot what’s common across the terms.
Worked example — spot the factors
Find the factors: of $12$, of $x^2$, and factorise $x^2 + 7x$ and $2x + 4y$.
$12 = 1\times12 = 2\times6 = 3\times4$
$x^2 = x \cdot x$, so factors are $1, x, x^2$
$x^2 + 7x = x(x + 7)$; $2x + 4y = 2(x + 2y)$
$x(x+7)$ and $2(x+2y)$
Highest common factor
Take out the biggest thing every term shares — a number, a letter, or both — and write what’s left in a bracket. Multiply back out to check.
Worked example — number outside
Factorise $10a + 15b$ and $20x + 30y$.
$10a + 15b = 5(2a + 3b)$
$20x + 30y = 10(2x + 3y)$
$5(2a+3b)$ and $10(2x+3y)$
Worked example — letter outside
Factorise $xy + xz$ and $2x^2 + 5x$.
$xy + xz = x(y + z)$
$2x^2 + 5x = x(2x + 5)$
$x(y+z)$ and $x(2x+5)$
You try
Factorise $3x^2 + 6x$, $12ab + 16ac$ and $3x^2 + 12x$.
Take out the biggest number and letter every term shares.
$3x^2 + 6x = 3x(x + 2)$
$12ab + 16ac = 4a(3b + 4c)$
$3x^2 + 12x = 3x(x + 4)$
$3x(x+2);\ 4a(3b+4c);\ 3x(x+4)$
Section 2 of 3
Factorising by Grouping
Four terms with nothing common to all? Split into two pairs and factorise each. If you group well, an identical bracket falls out of both — lift it.
Grouping
Split into pairs. Factorise each pair. Lift the matching bracket.
Worked example — all positive
Factorise $x^2 + 4x + 3x + 12$.
$= x(x + 4) + 3(x + 4)$
common bracket $(x + 4)$
$(x + 4)(x + 3)$
Worked example — negative third term
Factorise $x^2 + 7x - 2x - 14$.
$= x(x + 7) - 2(x + 7)$
$-2$ makes the inside $(x+7)$, matching the first
$(x + 7)(x - 2)$
You try
Factorise $x^2 - 3x - 2x + 6$.
Pair them: $x(x-3)$ from the first two. Take out $-2$ from the last two so the bracket matches.
$= x(x - 3) - 2(x - 3)$
common bracket $(x - 3)$
$(x - 3)(x - 2)$
Section 3 of 3
Quadratic Trinomials
Three terms like $x^2 + 7x + 12$? Make a fourth term by splitting the middle: find two numbers that multiply to the constant and add to the middle coefficient, then group.
Split the middle
Find two numbers: their product = the constant, their sum = the middle coefficient. Split the middle term with them, then factorise by grouping.
Worked example — split and group
Factorise $x^2 + 7x + 12$.
Multiply to $12$, add to $7$: pairs of $12$ are $1\times12, 2\times6, 3\times4$; take $3$ and $4$
$x^2 + 3x + 4x + 12 = x(x + 3) + 4(x + 3)$
$(x + 3)(x + 4)$
You try
Factorise $x^2 + 9x + 20$.
Two numbers that multiply to $20$ and add to $9$ — then split the middle and group.
Multiply to $20$, add to $9$: $4$ and $5$
$x^2 + 4x + 5x + 20 = x(x + 4) + 5(x + 4)$
$(x + 4)(x + 5)$
That’s Class 3.
HCF, grouping, and splitting the middle for trinomials — always multiply back to check. Class 4: algebraic fractions.