Test at p = 1, q = 1: –3(–5 + 2) = 9, and 15 – 6 = 9 ✓ (the printed answer p – 2q gives –1 ✗)
Why it happens: distributivity says a(b + c) = ab + ac — every term inside gets multiplied by the whole outside factor, letters and all, not just touched by its sign.
Q2.
2(x – 1) + 3 (x + 4) = 2x – 1 + 3x + 4 = 5x + 3
Answer
Mistake. The 2 and the 3 were multiplied into the first term of each bracket but not the second.
2(x – 1) + 3(x + 4) = 2x – 2 + 3x + 12 = 5x + 10
Test at x = 1: 2(0) + 3(5) = 15, and 5 + 10 = 15 ✓ (the printed answer gives 8 ✗)
Tip: when you distribute, count the terms — a bracket with two terms must produce two products.
Q3.
y + 2 (y + 2) = (y + 2)² = y² + 4y + 4
Answer
Mistake. y + 2(y + 2) is a sum, not a product; “y” and “2” were wrongly gathered into a factor (y + 2).
y + 2(y + 2) = y + 2y + 4 = 3y + 4
Test at y = 1: 1 + 2(3) = 7, and 3 + 4 = 7 ✓ (the printed answer gives 9 ✗)
Why it happens: in y + 2(y + 2) the multiplication by 2 happens first; the lone y is added afterwards. Reading it as (y + 2) × (y + 2) changes the order of operations.
Q4.
(5m + 6n)² = 25m² + 36n²
Answer
Mistake. The cross term is missing — squaring a sum is not squaring each part.
Test at m = 1, n = 1: 11² = 121, and 25 + 60 + 36 = 121 ✓ (the printed answer gives 61 ✗)
Why it happens: the area picture on page 145 shows two rectangles of 5m × 6n sitting between the two squares. Dropping them is the same as pretending a square of side 11 is just a 5-square plus a 6-square.
Q5.
(– q + 2)² = q² – 4q + 4
Answer
No mistake — this is correct.
(–q + 2)² = (–q)² + 2(–q)(2) + 2² = q² – 4q + 4 ✓
Test at q = 3: (–3 + 2)² = 1, and 9 – 12 + 4 = 1 ✓
Tip: (–q + 2) and (q – 2) differ only in sign, so their squares agree — you may expand whichever form you find easier.
Q6.
3a (2b × 3c) = 6ab × 9ac = 54a²bc
Answer
Mistake. The distributive property is multiplication over addition. Inside the bracket here there is a product, not a sum, so nothing may be distributed.
3a (2b × 3c) = 3a × 6bc = 18abc
Test at a = b = c = 1: 3 × (2 × 3) = 18, and 18abc = 18 ✓ (the printed answer gives 54 ✗)
Why it happens: writing 3a into both factors multiplies by 3a twice over, which triples-and-triples instead of tripling once. Compare: 2 × (3 × 4) = 24, not (2 × 3) × (2 × 4) = 48.
Q7.
1⁄2 (10s – 6) + 3 = 5s – 3 + 3 = 5s
Answer
No mistake — this is correct.
1⁄2 (10s – 6) + 3 = 5s – 3 + 3 = 5s ✓
Test at s = 2: 1⁄2 (20 – 6) + 3 = 7 + 3 = 10, and 5 × 2 = 10 ✓
Tip: both terms inside the bracket were halved, and the “+3” outside was left alone — exactly right.
Q8.
5w² + 6w = 11w²
Answer
Mistake. w² and w are unlike terms, so their coefficients cannot be added.
5w² + 6w is already in simplest form. (If you wish, factorise: 5w² + 6w = w(5w + 6).)
Test at w = 2: 20 + 12 = 32, while 11w² = 44 ✗
Why it happens: collecting terms is distributivity backwards: 5w² + 6w² = (5 + 6)w². That step needs the same letter part in both terms. Here one term has w × w and the other only w.
Q9.
2a³ + 3a³ + 6a²b + 6ab² = 5a³ + 12a²b²
Answer
Mistake. 2a³ and 3a³ were combined correctly, but 6a²b and 6ab² are unlike terms and were wrongly merged into 12a²b².
Tip: this is Identity 1 with a = x, m = 2, b = x, n = 5: ab + mb + an + mn = x² + 2x + 5x + 10.
Q11.
(a + 2) (b + 4) = ab + 8
Answer
Mistake. Only the first-with-first and last-with-last products were taken; the two cross products are missing.
(a + 2)(b + 4) = ab + 4a + 2b + 8 Correct: ab + 4a + 2b + 8
Test at a = 1, b = 1: 3 × 5 = 15, and 1 + 4 + 2 + 8 = 15 ✓ (the printed answer gives 9 ✗)
Why it happens: Identity 1 has four terms because a rectangle cut both ways has four pieces. Two of them, 4a and 2b, are the strips — the same ones that go missing in question 4.
Q12.
ab² + a²b + a²b² = ab (a + b + ab)
Answer
No mistake — this is correct. Check by expanding again:
ab(a + b + ab) = ab × a + ab × b + ab × ab = a²b + ab² + a²b² ✓
Test at a = 2, b = 3: LHS = 18 + 12 + 36 = 66; RHS = 6(2 + 3 + 6) = 6 × 11 = 66 ✓
Why it happens: every one of the three terms contains at least one a and at least one b, so ab can be taken out of each. Taking a common factor out is distributivity read from right to left.