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Mathematics · Year 10 (+10A) · Chapter 1

Squaring a sum

Practice

(x+2)(x+3): x² + 5x + 6
(x+1)(x+4): x² + 5x + 4
(x+3)²: x² + 6x + 9 — NOT x² + 9
(x−4)²: x² − 8x + 16
(x+5)(x−5): x² − 25 — difference of squares
(x+a)(x+b): x² + (a+b)x + ab
x² − 9 factorised: (x+3)(x−3)
x² + 5x + 6 factorised: (x+2)(x+3)
(2x+1)(x+3): 2x² + 7x + 3
2(x+3): 2x + 6

Easy questions

(x+2)(x+3): x² + 5x + 6
(x+1)(x+4): x² + 5x + 4
(x+5)(x+2): x² + 7x + 10
(x−2)(x+2): x² − 4
2(x+3): 2x + 6

Medium questions

(x+3)²: x² + 6x + 9
(x−4)²: x² − 8x + 16
(x+6)(x−2): x² + 4x − 12
(2x+1)(x+2): 2x² + 5x + 2
x² − 16 factorised: (x+4)(x−4)

Hard questions

x² + 7x + 12 factorised: (x+3)(x+4) — factors of 12 summing to 7
(3x−2)²: 9x² − 12x + 4
x² − x − 6 factorised: (x−3)(x+2)
(x+a)² expanded: x² + 2ax + a²
Sam claims (x+4)² = x² + 16. Bust it with x = 1.: (1+4)² = 25 but 1 + 16 = 17 — the missing 8x

Lesson

Every term meets every term

(x + 3)² means (x + 3)(x + 3) — and when you multiply brackets, EVERY term in the first must meet EVERY term in the second. Four meetings: x·x = x², x·3 = 3x, 3·x = 3x, 3·3 = 9. Collect: x² + 6x + 9. Sam's x² + 9 kept only the first and last meetings — the two middle ones, the 6x, went missing. Squaring does not distribute over addition; (a+b)² is a² + 2ab + b², always with the middle term. The five-second lie detector: substitute a small number. x = 1 gives 16; x² + 9 gives 10 — busted. Meanwhile (x+5)(x−5) is the one case where the middles genuinely cancel: +5x − 5x = 0, leaving x² − 25.

(x+3)² = (x+3)(x+3) = x² + 3x + 3x + 9 = x² + 6x + 9.
(x+2)(x+3) = x² + 3x + 2x + 6 = x² + 5x + 6.
(x+5)(x−5) = x² − 5x + 5x − 25 = x² − 25 — the middles cancel.
Check (x+3)² at x = 1: true value 16; x² + 9 gives 10 — the missing 6x, caught.

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