Factorising Special Products
Theory
Two patterns make factorising quick: 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 10A maths page (Victorian Curriculum 10A) shows how to spot them, how to use them with powers, surds and brackets, how to combine them with grouping, and how to use them with numbers and proofs. Curriculum: VC2M10A04 / AC9M10A01.
A difference of two squares is one square minus another, such as \(x^2 - 49\) or \(16a^2 - 9b^2\). It factorises as the difference of the square roots times their sum.
A perfect square trinomial is the expansion of a squared bracket. Its first and last terms are squares, and its middle term is twice the product of their square roots, such as \(x^2 + 14x + 49 = (x + 7)^2\).
A sum of two squares, such as \(x^2 + 9\), does not factorise with real numbers. It stays as it is.
The patterns work for any terms: numbers, pronumerals, powers such as \(x^4 = (x^2)^2\), surds such as \(5 = (\sqrt{5})^2\), and whole brackets such as \((x + 1)^2\).
How to factorise with the special products
- Take out any common factor (including \(-1\) or a pronumeral) first: \(-x^2 + 8x - 16 = -(x - 4)^2\).
- Two terms with a minus sign: check both are squares, then write \((a - b)(a + b)\).
- Three terms: check the ends are squares and the middle term is \(2ab\); then write \((a \pm b)^2\) with the sign of the middle term. If the middle term does not match, use the ordinary trinomial method.
- Keep going until no bracket factorises further, then check by expanding.
Harder forms
- Brackets as terms: treat \((x + y)\) or \((2x + 1)\) as a single letter, factorise, then simplify inside each bracket (Example 3).
- Four terms: group in pairs to find a common bracket, then look for a difference of two squares.
- Areas: the area left when one shape is cut from another is a difference. Take out any common factor, such as \(\pi\) or \(4\), then look for a difference of two squares: \(4X^2 - 4Y^2 = 4(X - Y)(X + Y)\).
- Numbers: \(a^2 - b^2\) with large numbers is easier as \((a - b)(a + b)\). Integer solutions of \(x^2 - y^2 = k\) come from factor pairs of \(k\) (Example 4).
- Proofs about divisibility: factorise fully, then reason about each factor, e.g. for odd \(n\), \(n - 1\) and \(n + 1\) are consecutive even numbers.
(i) \(81 - 4y^2\)
(ii) \(5x^3 - 20x\)
(iii) \(x^2 - 7\), using surds
(i) Both terms are squares: \(81 = 9^2\) and \(4y^2 = (2y)^2\).
\(81 - 4y^2\)
| \(\) | \(=\) | \(9^2 - (2y)^2 = \boldsymbol{(9 - 2y)(9 + 2y)}\) |
(ii) Take out the common factor \(5x\) first.
| \(5x^3 - 20x\) | \(=\) | \(5x(x^2 - 4)\) |
| \(\) | \(=\) | \(\boldsymbol{5x(x - 2)(x + 2)}\) |
(iii) \(7\) is not a perfect square, but \(7 = (\sqrt{7})^2\).
\(x^2 - 7\)
| \(\) | \(=\) | \(x^2 - (\sqrt{7})^2 = \boldsymbol{(x - \sqrt{7})(x + \sqrt{7})}\) |
(i) \(9x^2 - 42x + 49\)
(ii) \(x^2 - 3x + \dfrac{9}{4}\)
(iii) \(x^2 + 10x + 16\)
(i) \(9x^2 = (3x)^2\), \(49 = 7^2\) and \(2 \times 3x \times 7 = 42x\). The middle term is negative.
| \(9x^2 - 42x + 49\) | \(=\) | \(\boldsymbol{(3x - 7)^2}\) |
(ii) \(\dfrac{9}{4} = \left(\dfrac{3}{2}\right)^2\) and \(2 \times x \times \dfrac{3}{2} = 3x\).
| \(x^2 - 3x + \dfrac{9}{4}\) | \(=\) | \(\boldsymbol{\left(x - \dfrac{3}{2}\right)^2}\) |
(iii) \(16 = 4^2\), but \(2 \times x \times 4 = 8x\), not \(10x\). It is not a perfect square. Factorise as an ordinary trinomial instead: product \(16\), sum \(10\).
| \(x^2 + 10x + 16\) | \(=\) | \(\boldsymbol{(x + 2)(x + 8)}\) |
(i) \((2x + 3)^2 - (x - 1)^2\)
(ii) \(x^3 + 2x^2 - 25x - 50\)
(i) A difference of two squares with \(A = 2x + 3\) and \(B = x - 1\). Keep \(B\) in a bracket when you subtract it.
| \(A^2 - B^2\) | \(=\) | \((A - B)(A + B)\) |
| \(\) | \(=\) | \(\big[(2x + 3) - (x - 1)\big]\) |
| \(\) | \(\) | \({}\times \big[(2x + 3) + (x - 1)\big]\) |
| \(\) | \(=\) | \(\boldsymbol{(x + 4)(3x + 2)}\) |
(ii) Group in pairs, then use the difference of two squares.
\(x^3 + 2x^2 - 25x - 50\)
| \(\) | \(=\) | \(x^2(x + 2) - 25(x + 2)\) |
| \(\) | \(=\) | \((x + 2)(x^2 - 25)\) |
| \(\) | \(=\) | \(\boldsymbol{(x + 2)(x - 5)(x + 5)}\) |
(i) Find \(53^2 - 47^2\) without a calculator.
(ii) Find all pairs of positive integers \(x\) and \(y\) with \(x^2 - y^2 = 21\).
(i) Factorise instead of squaring.
| \(53^2 - 47^2\) | \(=\) | \((53 - 47)(53 + 47)\) |
| \(\) | \(=\) | \(6 \times 100 = \boldsymbol{600}\) |
(ii) \((x - y)(x + y) = 21\), where \(x - y\) and \(x + y\) are positive integers and \(x - y\) is the smaller. The factor pairs of \(21\) are \(1 \times 21\) and \(3 \times 7\).
\(x - y = 1\) and \(x + y = 21\): adding gives \(2x = 22\), so \(x = 11\) and \(y = 10\).
\(x - y = 3\) and \(x + y = 7\): adding gives \(2x = 10\), so \(x = 5\) and \(y = 2\).
So \((x, y) = \boldsymbol{(11, 10)}\) or \(\boldsymbol{(5, 2)}\). Check: \(121 - 100 = 21\) and \(25 - 4 = 21\).
Common pitfalls
Frequently asked questions
How do you factorise a difference of two squares?
Write each term as a square, \(a^2\) and \(b^2\), then the factorised form is \((a - b)(a + b)\). For example, \(16x^2 - 25 = (4x - 5)(4x + 5)\).
How do you know if a trinomial is a perfect square?
The first and last terms must be perfect squares, and the middle term must be twice the product of their square roots. The sign of the middle term goes inside the bracket.
Can x squared + 4 be factorised?
Not with real numbers. A sum of two squares, such as \(x^2 + 4\), has no real factors. Only a difference of two squares factorises this way.
How do you factorise x squared minus a number that is not a square?
Write the number as the square of a surd: \(x^2 - 3\) \(= x^2 - (\sqrt{3})^2\) \(= (x - \sqrt{3})(x + \sqrt{3})\).
What does factorise fully mean?
Keep factorising until no bracket can be broken down further. Take out common factors first, and use the difference of two squares again if a bracket is itself a difference of squares.