Factorising Special Binomial Products
Two patterns make factorising fast: the difference of two squares, \(a^2 - b^2 = (a - b)(a + b)\), and the perfect square, \(a^2 \pm 2ab + b^2 = (a \pm b)^2\). This NSW Year 9 Advanced maths subtopic shows how to spot and use them (NESA outcome MA5‑ALG‑P‑02).
You will factorise special products in one and two pronumerals, with fractions and with brackets as terms, take out a common factor first, and keep going until higher powers are fully factorised. You will multiply numbers mentally as a difference of squares, find the constant that makes a perfect square, find the size of a photo frame from its area, and prove a result about odd numbers. It is part of Year 9 Advanced Quadratics.
Theory
Two patterns save a lot of work: \(a^2 - b^2 = (a - b)(a + b)\) and \(a^2 \pm 2ab + b^2 = (a \pm b)^2\). Once you can spot them, expressions that look hard factorise in one line. This Year 9 Advanced maths page shows how to recognise each pattern, including when the terms are brackets or higher powers. It leads on to Advanced and Extension maths.
A perfect square number or term is something squared: \(64 = 8^2\), \(9y^2 = (3y)^2\), \(x^6 = (x^3)^2\).
A difference of two squares is one square minus another, such as \(x^2 - 25\) or \(16a^2 - b^2\). It factorises into two brackets with the same terms and opposite signs.
A perfect square trinomial has three terms: its first and last terms are perfect squares and its middle term is twice their product, such as \(x^2 + 8x + 16 = (x + 4)^2\).
A sum of two squares, such as \(x^2 + 9\), does not factorise with real numbers.
The terms \(a\) and \(b\) can be numbers, pronumerals, fractions such as \(\dfrac{x}{3}\), or whole brackets such as \((x - 2)\).
| Expression | Write as | Factorised |
|---|---|---|
| \(x^2 - 36\) | \(x^2 - 6^2\) | \((x - 6)(x + 6)\) |
| \(\dfrac{x^2}{9} - 1\) | \(\left(\dfrac{x}{3}\right)^2 - 1^2\) | \(\left(\dfrac{x}{3} - 1\right)\left(\dfrac{x}{3} + 1\right)\) |
| \(9p^2 - 12pq + 4q^2\) | \((3p)^2 - 2(3p)(2q) + (2q)^2\) | \((3p - 2q)^2\) |
| \(36 - (x + 1)^2\) | \(6^2 - (x + 1)^2\) | \((5 - x)(7 + x)\) |
How to factorise a special product
- HCF first. Numbers like \(45\) or \(72\) may not be squares until a common factor is taken out.
- Two terms with a minus sign? Write each as a square and use \((a - b)(a + b)\).
- Three terms? Check the ends are squares and the middle is \(2 \times\) their product. The sign of the middle term goes in the bracket.
- Brackets as terms: write \([\,a - b\,][\,a + b\,]\) with the brackets kept, then expand inside and simplify.
- Four terms? Look for three that form a perfect square, then a difference of squares.
- Factorise fully: check every bracket again, and take out common factors from each bracket at the end.
Using the patterns
- Arithmetic: \(98 \times 102 = 100^2 - 2^2\); \(a^2 - b^2\) with big numbers is \((a - b)(a + b)\).
- Areas: a square border is outer square \(-\) inner square, a difference of two squares.
- Missing constant: for \((px)^2 + bx + c\) to be a perfect square, \(c = \left(\dfrac{b}{2p}\right)^2\).
(i) \(49 - 4y^2\)
(ii) \(5x^2 - 45\)
(iii) \(81a^4 - 16\)
(i) Both terms are squares: \(49 = 7^2\) and \(4y^2 = (2y)^2\).
| \(49 - 4y^2\) | \(=\) | \(7^2 - (2y)^2 = \boldsymbol{(7 - 2y)(7 + 2y)}\) |
(ii) \(45\) is not a square, but take out the HCF \(5\) and a difference of squares appears.
| \(5x^2 - 45\) | \(=\) | \(5(x^2 - 9) = \boldsymbol{5(x - 3)(x + 3)}\) |
(iii) \(81a^4 = (9a^2)^2\) and \(16 = 4^2\).
| \(81a^4 - 16\) | \(=\) | \((9a^2 - 4)(9a^2 + 4)\) |
| \(\) | \(=\) | \(\boldsymbol{(3a - 2)(3a + 2)(9a^2 + 4)}\) |
\(9a^2 - 4\) is another difference of squares, so it splits again. \(9a^2 + 4\) is a sum of squares and does not factorise.
(i) Factorise \(4x^2 + 28x + 49\).
(ii) Factorise fully \(2x^2 - 24x + 72\).
(iii) Find \(c\) so that \(25x^2 - 30x + c\) is a perfect square, and write it as one.
(i) Test the pattern: are the ends squares, and is the middle twice their product?
| \(4x^2\) | \(=\) | \((2x)^2,\quad 49 = 7^2\) |
| \(2 \times 2x \times 7\) | \(=\) | \(28x\ \checkmark\) |
| \(4x^2 + 28x + 49\) | \(=\) | \(\boldsymbol{(2x + 7)^2}\) |
(ii) HCF \(2\) first. Then \(36 = 6^2\) and \(2 \times x \times 6 = 12x\); the middle term is negative.
| \(2x^2 - 24x + 72\) | \(=\) | \(2(x^2 - 12x + 36) = \boldsymbol{2(x - 6)^2}\) |
(iii) \(25x^2 = (5x)^2\), and the middle term \(-30x\) must be \(-2 \times 5x \times b\).
| \(10b\) | \(=\) | \(30 \ \Rightarrow\ b = 3\) |
| \(c\) | \(=\) | \(b^2 = \boldsymbol{9}\) |
| \(25x^2 - 30x + 9\) | \(=\) | \(\boldsymbol{(5x - 3)^2}\) |
(i) \((2x + 1)^2 - 9\)
(ii) \((x + 5)^2 - (x - 1)^2\)
Treat each bracket as a single term \(a\) or \(b\) in \(a^2 - b^2 = (a - b)(a + b)\). Keep the brackets until you simplify.
| \((2x + 1)^2 - 3^2\) | \(=\) | \(\big[(2x + 1) - 3\big]\big[(2x + 1) + 3\big]\) |
| \(\) | \(=\) | \((2x - 2)(2x + 4)\) |
| \(\) | \(=\) | \(2(x - 1) \times 2(x + 2)\) |
| \(\) | \(=\) | \(\boldsymbol{4(x - 1)(x + 2)}\) |
(ii) Here \(b = x - 1\). The minus sign in front of \(b\) changes both of its signs.
| \(\) | \(\) | \(\big[(x + 5) - (x - 1)\big]\big[(x + 5) + (x - 1)\big]\) |
| \(\) | \(=\) | \((x + 5 - x + 1)(2x + 4)\) |
| \(\) | \(=\) | \(6 \times 2(x + 2) = \boldsymbol{12(x + 2)}\) |
Take out the common factor of each bracket at the end: \(2x - 2 = 2(x - 1)\).
(i) Factorise \(x^2 + 10x + 25 - 4y^2\).
(ii) Without a calculator, evaluate \(47 \times 53\) and \(64^2 - 36^2\).
(i) The first three terms form a perfect square. Group them, then use the difference of two squares.
| \(x^2 + 10x + 25 - 4y^2\) | \(=\) | \((x + 5)^2 - (2y)^2\) |
| \(\) | \(=\) | \(\boldsymbol{(x + 5 - 2y)(x + 5 + 2y)}\) |
(ii) Both numbers are \(3\) from \(50\), so the product is a difference of squares.
| \(47 \times 53\) | \(=\) | \((50 - 3)(50 + 3) = 2500 - 9 = \boldsymbol{2491}\) |
| \(64^2 - 36^2\) | \(=\) | \((64 - 36)(64 + 36)\) |
| \(\) | \(=\) | \(28 \times 100 = \boldsymbol{2800}\) |
Common pitfalls
Frequently asked questions
What is the difference of two squares?
An expression of the form \(a^2 - b^2\). It factorises as \((a - b)(a + b)\).
How do you recognise a perfect square trinomial?
The first and last terms are perfect squares and the middle term is twice the product of their square roots. Then it factorises as \((a \pm b)^2\).
Can a sum of two squares be factorised?
No, not with real numbers. For example \(x^2 + 9\) has no real factors.
Why should you take out a common factor first?
Often the expression only becomes a difference of squares or a perfect square after the common factor is removed, for example \(2x^2 - 50 = 2(x^2 - 25)\).
How do you factorise a difference of squares with brackets?
Treat each bracket as one term. Write \([a - b][a + b]\) with the brackets kept, then simplify inside each bracket.