Factorising means taking a polynomial spread out as a sum and gathering it back into a product of factors, which is expansion run backwards.
Just as expanding (x + 3)² gives x² + 6x + 9, factorising is looking at x² + 6x + 9 and recognising (x + 3)².
In terms of area, the four pieces a², ab, ab and b² fit together without gaps into a square of side a + b, and that picture shows a² + 2ab + b² = (a + b)² directly.
Once you also learn to find two numbers whose product is the constant term and whose sum is the coefficient of x, most quadratic expressions at this level can be factorised.
Reading a product formula backwards
seegongsik.com
An area picture in which the four pieces of a² + 2ab + b² fit into a square of side a + b
3
2
🔄 Expanding and factorising
①Expanding spreads a product into a sum: (a + b)² = a² + 2ab + b²
②Factorising gathers a sum into a product: a² + 2ab + b² = (a + b)²
③With a = 3 and b = 2 in the picture, the piece areas 9 + 6 + 6 + 4 = 25 equal the area 5² of the square with side 5.
Factorising with product formulas
Perfect Square
a²+2ab+b² = (a+b)²
Check the middle term equals 2ab
Difference of two squares
a² − b² = (a + b)(a − b)
A difference of two squares factorises into a sum times a difference.
🔍 Choosing the right formula
①If there are three terms and both end terms are squares, suspect a perfect square.
②Then check that the middle term equals 2 × (square root of the first term) × (square root of the third term).
③If there are two terms, both squares, joined by a minus sign, factorise as a sum times a difference.
The form x² + (p + q)x + pq
seegongsik.com
Pairs of integers p, q whose product is the constant term, their sums, and the factorised form
6
Finding the two numbers
x² + (p + q)x + pq = (x + p)(x + q)
Find two numbers p, q whose product is the constant term and whose sum is the coefficient of x.
🧩 Steps for finding the two numbers
①List every pair of numbers whose product is the constant term, including negative pairs.
②Choose the pair whose sum equals the coefficient of x.
③If the product is positive and the sum negative, both numbers are negative; if the product is negative, the two numbers have opposite signs.
Common factors
Taking out a common factor
ma + mb = m(a + b)
First take out any factor that appears in every term.
📌 Order of factorising
①If there is a common factor, take it out first.
②Apply a product formula or the two-number method to what remains.
③Repeat until nothing more can be factorised.
④For example, 2x² + 4x + 2 = 2(x² + 2x + 1) = 2(x + 1)². The perfect square only appears after the common factor 2 is taken out.
Worked examples
Example 1
Factor x²+6x+9.
1
Check whether the end terms are squares: x² is the square of x and 9 is the square of 3.
x²+6x+9 = x²+2·x·3+3²
2
The middle term 6x equals 2 · x · 3, so this is a perfect square.
= (x+3)²
▸ (x+3)²
If the middle term equals 2 × (product of the square roots of the end terms), the expression is a perfect square.
Example 2
Factor x²−7x+12.
1
The two numbers with product 12 and sum −7 are −3 and −4.
(−3)×(−4)=12, (−3)+(−4)=−7
2
Put p = −3 and q = −4 into the form (x + p)(x + q).
= (x−3)(x−4)
▸ (x−3)(x−4)
If the product is positive and the sum negative, both numbers are negative.
Exam checklist
Factorisation formulas
a² ± 2ab + b² = (a ± b)², a² − b² = (a + b)(a − b)
The factorisation formulas are the product formulas read backwards.
G9 school-exam type
Which is the correct factorization of x²−25?
①(x+5)²
②(x−5)²
③(x+5)(x−5)
④(x+25)(x−1)
⑤5(x−5)
▸ ③ (x+5)(x−5)
1
Note x²−25 = x²−5² is a difference of squares.
x²−25 = x²−5²
2
Apply a²−b²=(a+b)(a−b).
= (x+5)(x−5)
🎯 Exam points
①Do not forget to take out a common factor first.
②A perfect square is recognised by checking that the middle term is 2 × (product of the square roots of the end terms).
③Two squares joined by a minus sign factorise straight into a sum times a difference.
④For x² + (p + q)x + pq, find the two numbers from their product and sum.
⑤Check the result by expanding it to see that the original expression comes back.