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Upper secondary · selected topics

Squaring identities and factorisation – with examples

Recognising the pattern in an expression makes it easier to expand or factorise it. Here you can see the rules in use.

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The squaring identities

First squaring identity

(a + b)2 = a2 + 2ab + b2

Example: (x + 3)2 = x2 + 2 · x · 3 + 32 = x2 + 6x + 9

Example: (2x + 5)2 = (2x)2 + 2 · 2x · 5 + 52 = 4x2 + 20x + 25

Second squaring identity

(a − b)2 = a2 − 2ab + b2

Example: (x − 4)2 = x2 − 8x + 16

Example: (3x − 2)2 = 9x2 − 12x + 4

Difference of two squares

(a + b)(a − b) = a2 − b2

Example: (x + 7)(x − 7) = x2 − 49

Example: (3x + 4)(3x − 4) = 9x2 − 16

Identities

The squaring rules and difference of two squares are identities — they hold for all values of a and b. An identity is not an equation to solve, but a rewriting that is always true.

First squaring identity
(a+b)2 = a2+2ab+b2
Second squaring identity
(a−b)2 = a2−2ab+b2
Difference of two squares
(a+b)(a−b) = a2−b2
Common mistake: (a + b)2 ≠ a2 + b2. People often forget the double product 2ab!

Completing the square

To write x² + bx as a square, take half of b, square that number, and both add and subtract the same value. This keeps the value of the expression unchanged.

x² + bx = (x + b/2)² − (b/2)²

For x² + 8x half of eight is four: x² + 8x = (x + 4)² − 16. Check by expanding the brackets: x² + 8x + 16 − 16.

For x² − 6x + 5 half of −6 is −3: (x − 3)² − 9 + 5 = (x − 3)² − 4. The 5 from the original expression must be kept.

Factorising and simplifying

Factorising quadratic expressions

When we know the roots x1 and x2, we can write:

ax2 + bx + c = a(x − x1)(x − x2)

Example 1: Factorise x2 − 5x + 6

Find the roots: Sum 5, product 6 → the numbers 2 and 3. x² − 5x + 6 = (x − 2)(x − 3)

Example 2: Factorise x2 − 4

Difference of two squares: x² − 4 = x² − 2² = (x + 2)(x − 2)

Example 3: Factorise 2x2 + 6x

Common factor: 2x² + 6x = 2x(x + 3)

Simplifying algebraic fractions

Factorise the numerator and denominator, and cancel common factors:

Example: Simplify x2 − 9x + 3

Factorise the numerator using the difference of two squares: x² − 9 = (x + 3)(x − 3)
Simplify: (x + 3)(x − 3) / (x + 3) = x − 3   (for x ≠ −3)

Example: Simplify x2 − x − 6x2 − 4

Factorise the numerator: x² − x − 6 = (x − 3)(x + 2)
Factorise the denominator: x² − 4 = (x + 2)(x − 2)
Cancel (x + 2): (x − 3)(x + 2) / ((x + 2)(x − 2)) = (x − 3) / (x − 2)   (for x ≠ −2)

Equations with the variable in the denominator

When x is in the denominator, we must state restrictions on x (the denominator cannot be zero) and check for extraneous solutions.

Example 1 (valid solution): Solve 1x − 2 + 3 = xx − 2

Restriction: x ≠ 2 (the denominator cannot be zero).
Multiply both sides by (x − 2): 1 + 3(x − 2) = x
Expand and solve: 1 + 3x − 6 = x   ⇒   3x − 5 = x   ⇒   2x = 5   ⇒   x = 2.5
Check the restriction: x = 2.5 ≠ 2 ✓ The solution is valid.

Example 2 (extraneous solution): Solve xx − 3 = 3x − 3

Restriction: x ≠ 3 (the denominator cannot be zero).
Multiply both sides by (x − 3): x = 3
Check the restriction: x = 3 violates x ≠ 3!   Extraneous solution.
Conclusion: The equation has no solution.
Why is the solution extraneous? When we multiplied by (x − 3), we assumed it was allowed — meaning x ≠ 3. If the answer is x = 3, we have performed an invalid operation. Always check the restriction at the end.

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Selected explanations for practice and revision. Levels may vary between examples.