Factorising is expanding in reverse: it regroups a single polynomial as a product of several polynomials.
Think of finding the length and width of a rectangle whose area is x² + 5x + 6: you look for two numbers that add to 5 and multiply to 6, namely 2 and 3, and split it as (x + 2)(x + 3).
An expression with x⁴ that looks daunting becomes a familiar quadratic once you set x² = t, and for cubics and beyond the way in is to find one root with the factor theorem.
In the first picture the four pieces of the area model change with the numbers a and b; in the second, the steps of factorising by substitution change with the coefficient k.
Factorising is multiplication in reverse
Factorising is expanding in reverse. It is like recovering the length and width of a rectangle when you know its area as an expression.
2
3
seegongsik.com
Area model: a rectangle of area x² + (a + b)x + ab split into four pieces to find its sides x + b and x + a
💡 Find two numbers from their sum and product
①For x² + px + q, look for two numbers that add to p and multiply to q
②If those numbers are a and b, then x² + px + q = (x + a)(x + b)
③In the area model, the top-right rectangle bx and the bottom-left rectangle ax add up to px, and the bottom-right rectangle of area ab is q
Factorising formulas
Formulas
Key factorising formulas
Factorising by substitution
Even complicated expressions turn into quadratics with a well-chosen substitution. If x⁴ appears, set x² = t; if the same block keeps repeating, replace that block with one letter.
3
seegongsik.com
Substituting x² = t in x⁴ + kx² + (k − 1), factorising the quadratic in t, then putting x² back
🔑 Steps for factorising by substitution
①If you spot a repeated block, replace it with a single letter
②An expression of the form x⁴ + ax² + b becomes a quadratic in t when x² = t
③Factorise the quadratic in t, then put x² back in place of t
④If a factor such as x² − 4 can be split further, keep going
Where to start
Take out common factors first
common factor → formulas → substitution or factor theorem
Checking in this order handles most expressions
Difference of two squares
a² − b² = (a + b)(a − b)
The factorising formula you will use most often
Worked examples and unit test
Example 1
Factorise x³ + 8.
1
Since 8 = 2³, use the sum of cubes a³ + b³ = (a + b)(a² − ab + b²).
a³ + b³ = (a + b)(a² − ab + b²)
2
Substitute a = x and b = 2.
x³ + 2³ = (x + 2)(x² − 2x + 4)
▸ (x + 2)(x² − 2x + 4)
The discriminant of x² − 2x + 4 is 4 − 16 < 0, so it does not split any further over the real numbers.
Example 2
Factorise x⁴ − 13x² + 36.
1
Setting x² = t gives a quadratic in t.
t² − 13t + 36 = (t − 4)(t − 9)
2
Put x² back in place of t and split each factor again with the difference of two squares.
(x² − 4)(x² − 9) = (x − 2)(x + 2)(x − 3)(x + 3)
▸ (x − 2)(x + 2)(x − 3)(x + 3)
Stopping at (x² − 4)(x² − 9) leaves the job half done. After putting x² back, always check whether the factors split further.
Summary
Factorising a quadratic
x² + (a + b)x + ab = (x + a)(x + b)
Find the two factors from a sum and a product
Sum and difference of cubes
a³ ± b³ = (a ± b)(a² ∓ ab + b²)
The middle term of the second bracket takes the opposite sign
Grade-10 school exam type
Factor the polynomial x³ − 2x² − 5x + 6.
①(x − 1)(x − 2)(x + 3)
②(x − 1)(x + 2)(x − 3)
③(x + 1)(x − 2)(x − 3)
④(x − 1)(x − 2)(x − 3)
⑤(x + 1)(x + 2)(x − 3)
▸ ② (x − 1)(x + 2)(x − 3)
1
Since substituting x = 1 gives 0, by the factor theorem x − 1 is a factor.
f(1) = 1 − 2 − 5 + 6 = 0
2
Divide to get the quotient, then factor again.
(x − 1)(x² − x − 6) = (x − 1)(x + 2)(x − 3)
🎯 Points that often come up
①First check for a common factor
②If both terms are squares, use the difference of two squares
③Check whether three terms fit a² ± 2ab + b² and can be written as a perfect square
④Try x² = t for expressions with x⁴
⑤For cubics and higher, substitute divisors of the constant term to find a value that gives 0, then use the factor theorem to get a linear factor