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Year 9 Maths Algebraic Techniques

Factorising the Difference of Perfect squares

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Theory

A difference of two squares such as \(x^2 - 81\) factorises in one line: \(a^2 - b^2 = (a - b)(a + b)\). This Year 9 maths page shows how to spot the two squares, take out a common factor first, factorise fully, treat a bracket as one of the squares, and use the pattern to calculate and prove.

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: \(x^2 - 81 = (x - 9)(x + 9)\).

The pattern comes from expanding: \((a - b)(a + b)\) \(= a^2 + ab - ab - b^2\) \(= a^2 - b^2\). The middle terms cancel, so factorising runs this backwards.

A sum of two squares, such as \(x^2 + 16\), does not factorise.

Difference of two squares as an areaLeft: a square of side a with a small square of side b removed from one corner, leaving area a squared minus b squared, cut into a blue and a red rectangle. Right: the two pieces rearranged into one rectangle with sides a plus b and a minus b. a b a a² − b² a + b a − b
Cut \(b^2\) from \(a^2\) and rearrange the rest: a rectangle \((a + b)\) by \((a - b)\). So \(a^2 - b^2 = (a + b)(a - b)\).
Spotting a difference of two squares25 y squared minus 36 is written as 5y all squared minus 6 squared. It then factorises as (5y minus 6)(5y plus 6), the pattern (a minus b)(a plus b). 25y2 − 36 a square a square (5y)2 − 62 (a − b)(a + b) (5y − 6)(5y + 6)
Write each term as a square, \((5y)^2\) and \(6^2\), then use \((a - b)(a + b)\) with \(a = 5y\) and \(b = 6\).
\[a^2 - b^2 = (a - b)(a + b)\]
a2−b2=(a−b)(a+b)
ExpressionWrite asFactorised
\(x^2 - 100\)\(x^2 - 10^2\)\((x - 10)(x + 10)\)
\(9a^2 - 1\)\((3a)^2 - 1^2\)\((3a - 1)(3a + 1)\)
\(4p^2 - 49q^2\)\((2p)^2 - (7q)^2\)\((2p - 7q)(2p + 7q)\)
\(y^6 - 16\)\((y^3)^2 - 4^2\)\((y^3 - 4)(y^3 + 4)\)
Order does not matter in the product. \(16 - x^2 = (4 - x)(4 + x)\). Write the square that comes first in the question first in each bracket.
Negative first term. Take out a negative common factor: \(-2k^2 + 72\) \(= -2(k^2 - 36)\) \(= -2(k - 6)(k + 6)\). Writing it as \(72 - 2k^2 = 2(6 - k)(6 + k)\) gives the same expression.

How to factorise a difference of two squares

  1. HCF first: take out any common factor, including \(-1\) when the first term is negative.
  2. Two terms with a minus between them? Check both are squares, then use \((a - b)(a + b)\).
  3. Check the brackets: the same two terms in each, one with a minus and one with a plus. Expanding should give back the question, with the middle terms cancelling.
  4. Look again: a bracket may factorise further. \(x^4 - 81\) \(= (x^2 - 9)(x^2 + 9)\) \(= (x - 3)(x + 3)(x^2 + 9)\).
  5. Brackets as the squares: \((x - 2)^2 - 25\) \(= (x - 2 - 5)(x - 2 + 5)\) \(= (x - 7)(x + 3)\). Simplify inside each bracket.
  6. 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.
Example 1 — HCF first, then two squares
Factorise fully \(75m^3 - 12m\).
Solution

Take out the HCF \(3m\) first. What is left is a difference of two squares.

\(75m^3 - 12m\)\(=\)\(3m(25m^2 - 4)\)
\(\)\(=\)\(3m\big((5m)^2 - 2^2\big)\)
\(\)\(=\)\(\boldsymbol{3m(5m - 2)(5m + 2)}\)
3m(5m−2)(5m+2)

\(75\) and \(12\) are not squares, but \(25\) and \(4\) are, once the \(3\) is out.

Example 2 — Two pronumerals, and a number first
Factorise fully
(i) \(81a^2 - 25b^2\)
(ii) \(18 - 2k^2\).
Solution

(i) Both terms are squares: \(81a^2 = (9a)^2\) and \(25b^2 = (5b)^2\).

\(81a^2 - 25b^2\)\(=\)\((9a)^2 - (5b)^2\)
\(\)\(=\)\(\boldsymbol{(9a - 5b)(9a + 5b)}\)
(9a−5b)(9a+5b)

(ii) \(18\) is not a square, but take out the HCF \(2\) and a difference of squares appears. The number square comes first, so it goes first in each bracket.

\(18 - 2k^2\)\(=\)\(2(9 - k^2)\)
\(\)\(=\)\(2(3^2 - k^2)\)
\(\)\(=\)\(\boldsymbol{2(3 - k)(3 + k)}\)
2(3−k)(3+k)

Check by expanding: \(2(9 + 3k - 3k - k^2) = 18 - 2k^2\).

Example 3 — Squares of brackets
Factorise fully \((3x - 2)^2 - (x + 4)^2\).
Solution

This is \(A^2 - B^2\) with \(A = 3x - 2\) and \(B = x + 4\). Use \((A - B)(A + B)\), keeping \(B\) in brackets.

\(\)\(\)\(\big[(3x - 2) - (x + 4)\big]\)
\(\times\, \big[(3x - 2) + (x + 4)\big]\)
\(\)\(=\)\((2x - 6)(4x + 2)\)
\(\)\(=\)\(2(x - 3) \times 2(2x + 1)\)
\(\)\(=\)\(\boldsymbol{4(x - 3)(2x + 1)}\)
4(x−3)(2x+1)

Both new brackets had a common factor of \(2\), so take each out to factorise fully.

Example 4 — Using the patterns

(i) Evaluate \(86^2 - 14^2\) without a calculator.
(ii) Show that the difference between the squares of the odd numbers \(2n + 1\) and \(2n - 1\) is always \(8n\).
Solution

(i) Use \(a^2 - b^2 = (a - b)(a + b)\).

\(86^2 - 14^2\)\(=\)\((86 - 14)(86 + 14)\)
\(\)\(=\)\(72 \times 100 = \mathbf{7200}\)
862−142=7200

(ii) Treat each bracket as one term. \(A = 2n + 1\) and \(B = 2n - 1\).

\(A^2 - B^2\)\(=\)\(\big[(2n + 1) - (2n - 1)\big]\)
\(\times\, \big[(2n + 1) + (2n - 1)\big]\)
\(\)\(=\)\((2)(4n)\)
\(\)\(=\)\(\boldsymbol{8n}\)
(2n+1)2−(2n−1)2=8n

Check with \(n = 3\): the odd numbers are \(7\) and \(5\), and \(49 - 25 = 24 = 8 \times 3\).

Common pitfalls

A difference, not a sum. \(x^2 - 36 = (x - 6)(x + 6)\), but \(x^2 + 36\) cannot be factorised this way. Writing \((x + 6)^2\) for it is wrong: that expands to \(x^2 + 12x + 36\).
Square the whole term. \(16y^2 = (4y)^2\), so \(16y^2 - 1 = (4y - 1)(4y + 1)\), not \((16y - 1)(16y + 1)\).
Both terms must be squares. \(x^2 - 7\) does not factorise this way with whole numbers, because \(7\) is not a square. But \(7x^2 - 7 = 7(x^2 - 1)\) \(= 7(x - 1)(x + 1)\) does, once the \(7\) is out.
Factorise fully. \(10x^2 - 360\) \(= 10(x^2 - 36)\) \(= 10(x - 6)(x + 6)\). Stopping at \(10(x^2 - 36)\) is not fully factorised.

Frequently asked questions

What is the difference of two squares?

It is one squared term minus another, such as \(x^2 - 64\). It factorises as \((x - 8)(x + 8)\): the difference of the square roots times their sum.

How can you check a difference of two squares answer?

Expand the brackets. The two middle terms cancel: \((x - 8)(x + 8)\) \(= x^2 + 8x - 8x - 64\) \(= x^2 - 64\).

Can you factorise x squared plus 9?

Not with whole numbers or this pattern. A sum of two squares such as \(x^2 + 9\) 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: \(2x^2 - 50 = 2(x^2 - 25) = 2(x - 5)(x + 5)\).

How can the difference of two squares help with mental arithmetic?

Write a difference of squares as a product: \(61^2 - 39^2 = (61 - 39)(61 + 39)\) \(= 22 \times 100 = 2200\).

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