Factorising Special Binomial Products
Two special patterns make factorising quick: the difference of two squares, \(a^2 - b^2 = (a - b)(a + b)\), and perfect squares such as \(x^2 + 8x + 16 = (x + 4)^2\). This NSW Year 9 Advanced maths subtopic practises both (NESA outcome MA5‑ALG‑P‑02).
You will factorise expressions such as \(36 - 49x^2\) and \(9x^2 - 12x + 4\), take out a common factor first, and factorise \(p^4 - 16\) in two steps. You will treat a bracket as one term in \((x + 3)^2 - 16\), evaluate \(73^2 - 27^2\) without a calculator, find the values of \(k\) that make \(x^2 + kx + 36\) a perfect square, and prove a fact about consecutive squares. It is part of Year 9 Advanced Algebra: Expanding and Factorising.
Theory
Two patterns factorise instantly: 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 Year 9 Advanced maths page covers content that matters if you are heading for Advanced or Extension maths in Years 11 and 12: taking out the HCF first, squares of brackets and using the patterns to calculate.
A perfect square is a term that is something squared: \(49\), \(16x^2 = (4x)^2\), \(y^6 = (y^3)^2\).
A difference of two squares is one perfect square minus another, such as \(x^2 - 81\) or \(4 - 9y^2\). It factorises into a sum times a difference: \(x^2 - 81 = (x - 9)(x + 9)\).
A perfect square trinomial is the expansion of a squared binomial: \(x^2 - 14x + 49 = (x - 7)^2\). Its first and last terms are squares and its middle term is twice their product.
A sum of two squares, such as \(x^2 + 16\), does not factorise using these patterns.
How to factorise a special product
- HCF first: take out any common factor, including \(-1\) when the first term is negative.
- Two terms with a minus between them? Check both are squares, then use \((a - b)(a + b)\).
- Three terms? Check the first and last are squares and the middle is \(\pm 2ab\). Then write \((a \pm b)^2\) with the sign of the middle term.
- Look again: a bracket may factorise further. \(x^4 - 81\) \(= (x^2 - 9)(x^2 + 9)\) \(= (x - 3)(x + 3)(x^2 + 9)\).
- Brackets as the squares: \((x - 2)^2 - 25\) \(= (x - 2 - 5)(x - 2 + 5)\) \(= (x - 7)(x + 3)\). Simplify inside each bracket.
- Proofs: \((n + 2)^2 - n^2\) \(= (2)(2n + 2)\) \(= 4(n + 1)\), so the difference of the squares of two numbers \(2\) apart is \(4\) times the number between them.
Take out the HCF \(3m\) first. What is left is a difference of two squares.
| \(75m^3 - 12m\) | \(=\) | \(3m(25m^2 - 4)\) |
| \(\) | \(=\) | \(3m\big((5m)^2 - 2^2\big)\) |
| \(\) | \(=\) | \(\boldsymbol{3m(5m - 2)(5m + 2)}\) |
\(75\) and \(12\) are not squares, but \(25\) and \(4\) are, once the \(3\) is out.
Take out the HCF \(5\), then test the trinomial against \((p - q)^2 = p^2 - 2pq + q^2\).
| \(20a^2 - 60ab + 45b^2\) | \(=\) | \(5(4a^2 - 12ab + 9b^2)\) |
| \(4a^2\) | \(=\) | \((2a)^2,\quad 9b^2 = (3b)^2\) |
| \(2 \times 2a \times 3b\) | \(=\) | \(12ab\ \checkmark\) |
| \(\therefore\ 20a^2 - 60ab + 45b^2\) | \(=\) | \(\boldsymbol{5(2a - 3b)^2}\) |
The middle term is negative, so the bracket has a minus sign.
This is \(A^2 - B^2\) with \(A = 3x - 2\) and \(B = x + 4\). Use \((A - B)(A + B)\), keeping \(B\) in brackets.
| \(\) | \(\) | \(\big[(3x - 2) - (x + 4)\big]\big[(3x - 2) + (x + 4)\big]\) |
| \(\) | \(=\) | \((2x - 6)(4x + 2)\) |
| \(\) | \(=\) | \(2(x - 3) \times 2(2x + 1)\) |
| \(\) | \(=\) | \(\boldsymbol{4(x - 3)(2x + 1)}\) |
Both new brackets had a common factor of \(2\), so take each out to factorise fully.
(i) Evaluate \(86^2 - 14^2\) without a calculator.
(ii) Find the values of \(k\) that make \(9x^2 + kx + 25\) a perfect square.
(i) Use \(a^2 - b^2 = (a - b)(a + b)\).
| \(86^2 - 14^2\) | \(=\) | \((86 - 14)(86 + 14)\) |
| \(\) | \(=\) | \(72 \times 100 = \mathbf{7200}\) |
(ii) The square roots of the end terms are \(3x\) and \(5\). The middle term must be \(\pm 2 \times 3x \times 5\).
| \(kx\) | \(=\) | \(\pm 30x\) |
| \(k\) | \(=\) | \(\mathbf{30}\ \text{or}\ \mathbf{-30}\) |
\(k = 30\) gives \((3x + 5)^2\) and \(k = -30\) gives \((3x - 5)^2\).
Common pitfalls
Frequently asked questions
What is the difference of two squares?
It is one squared term minus another, such as x squared minus 64. It always factorises as (x minus 8)(x plus 8): the difference of the square roots times their sum.
How do you know if a quadratic is a perfect square?
Check that the first and last terms are perfect squares and the last term is positive, then check that the middle term is twice the product of their square roots. For x squared plus 14x plus 49, the roots are x and 7, and 2 times x times 7 is 14x, so it is (x + 7) squared.
Can you factorise x squared plus 9?
Not with whole numbers or these patterns. A sum of two squares does not factorise into real brackets. Only a difference of two squares does.
How do you factorise a difference of two squares with a number in front?
Take out the highest common factor first. For example 2x squared minus 50 is 2 times (x squared minus 25), which is 2(x minus 5)(x plus 5).
How can the difference of two squares help with mental arithmetic?
Write a difference of squares as a product. For example 61 squared minus 39 squared is (61 minus 39) times (61 plus 39), which is 22 times 100, or 2200.