Mathematics • Year 9 • Unit 1 • Lesson 12
Product Rule — Mixed Challenge
Bring it all together: multi-variable products, sign work, one-step expansions and three-term distribution. Spot the mistake in someone else's algebra and tackle an open-ended challenge.
1. Mixed problems — choose your tool
Each question uses some mix of the product rule, sign work and expansion. Decide which steps to use BEFORE you start. Show working. 3 marks each
1.1 Simplify $2 a^2 b \times 7 a^3 b^4$.
1.2 Simplify $-3 m^4 n \times 2 m n^2$.
1.3 Expand and simplify $4y(y^2 - 2y + 1)$.
1.4 A rectangle has area $A = (2x^2 y)(5x y^3)$. Write $A$ as a single simplified term, then evaluate $A$ when $x = 2$ and $y = 1$.
1.5 Expand and simplify $-3xy(2x^2 - 4xy + 5 y^2)$. Watch every sign carefully — the answer has three terms.
1.6 Find the missing term: $4 a^3 \times \square = 12 a^7$. Show the working that gets you to the missing factor.
2. Find the mistake
A student has tried to expand and simplify $2x(3x^2 - 4x)$. Their working is shown below. Exactly one line contains a mistake. Spot it, explain it, then redo the working correctly. 3 marks
Student's working — expand $2x(3x^2 - 4x)$:
Line 1: $2x(3x^2 - 4x) = 2x \cdot 3x^2 - 2x \cdot 4x$
Line 2: $2x \cdot 3x^2 = 6 x^3$
Line 3: $2x \cdot 4x = 8 x$
Line 4: So $2x(3x^2 - 4x) = 6 x^3 - 8 x$.
(a) Which line contains the mistake?
(b) Explain in one or two sentences why that line is wrong.
(c) Write out the corrected working in full, including the corrected final answer.
Stuck? On Line 3, count the indices on the $x$s. There are two $x$s being multiplied — what should the resulting index be?3. Open-ended challenge — build your own product
This question has many valid answers — your job is to find two that fit the brief. 4 marks
3.1 Find two different pairs of algebraic terms (each pair of the form $\text{coefficient} \times a^? b^?$) that multiply to give exactly $\mathbf{24 a^5 b^3}$. The two pairs must NOT be the same, and neither term in any pair can be a plain number with no variables.
For each pair you choose:
(i) Write the two terms.
(ii) Multiply them out, showing the coefficient and each variable's working.
(iii) Confirm the result equals $24 a^5 b^3$.
Bonus: One of your pairs must include a negative coefficient on at least one term.
How did this worksheet feel?
What I'll revisit before next class:
1.1 — $2 a^2 b \times 7 a^3 b^4$
$(2 \times 7) \cdot a^{2+3} \cdot b^{1+4} = \mathbf{14 a^5 b^5}$.
1.2 — $-3 m^4 n \times 2 m n^2$
Sign $(-)(+) = (-)$. $(3 \times 2) \cdot m^{4+1} \cdot n^{1+2} = \mathbf{-6 m^5 n^3}$.
1.3 — $4y(y^2 - 2y + 1)$
$4y \cdot y^2 - 4y \cdot 2y + 4y \cdot 1 = 4 y^3 - 8 y^2 + 4 y = \mathbf{4 y^3 - 8 y^2 + 4 y}$.
1.4 — Rectangle $A = (2x^2 y)(5 x y^3)$
$A = (2 \times 5) \cdot x^{2+1} \cdot y^{1+3} = \mathbf{10 x^3 y^4}$.
At $x=2, y=1$: $A = 10(2)^3(1)^4 = 10 \times 8 \times 1 = \mathbf{80}$.
1.5 — $-3xy(2x^2 - 4xy + 5 y^2)$
Term 1: $-3xy \cdot 2x^2 = -6 x^{1+2} y = -6 x^3 y$.
Term 2: $-3xy \cdot (-4xy) = +12 x^{1+1} y^{1+1} = +12 x^2 y^2$.
Term 3: $-3xy \cdot 5 y^2 = -15 x y^{1+2} = -15 x y^3$.
Final: $\mathbf{-6 x^3 y + 12 x^2 y^2 - 15 x y^3}$.
1.6 — Missing term: $4a^3 \times \square = 12 a^7$
Coefficient: $\square$ has coefficient $12 \div 4 = 3$. Index: $a^{3+?} = a^7$, so $? = 4$. Missing term $= \mathbf{3 a^4}$.
Check: $4 a^3 \times 3 a^4 = 12 a^7$. ✓
2 — Find the mistake
(a) The mistake is on Line 3.
(b) The student forgot to add the indices on the $x$s: $2x \cdot 4x = 8 x^{1+1} = 8 x^2$, not $8x$. They handled the coefficient correctly ($2 \times 4 = 8$) but treated $x \times x$ as if it were just $x$.
(c) Corrected working:
$2x(3x^2 - 4x) = 2x \cdot 3x^2 - 2x \cdot 4x$
$= 6 x^3 - 8 x^2$
$= \mathbf{6 x^3 - 8 x^2}$. (Two terms; they are NOT like terms since the indices differ.)
3 — Open-ended challenge (sample solution)
We need two terms whose coefficients multiply to $24$, whose $a$-indices add to $5$, and whose $b$-indices add to $3$.
Pair 1 (positive): $4 a^2 b$ and $6 a^3 b^2$.
Working: $(4 \times 6) \cdot a^{2+3} \cdot b^{1+2} = 24 a^5 b^3$ ✓.
Pair 2 (with a negative — bonus): $-3 a b$ and $-8 a^4 b^2$.
Working: sign $(-)(-) = (+)$; $(3 \times 8) \cdot a^{1+4} \cdot b^{1+2} = 24 a^5 b^3$ ✓.
Other valid pairs: $2a \times 12 a^4 b^3$, $8 a^3 b \times 3 a^2 b^2$, $-4 a^2 b \times -6 a^3 b^2$, etc. — any factorisation of $24$ paired with any split of the indices $5$ and $3$.
Marking: 2 marks per valid pair (1 for the pair, 1 for the multiplication shown). One of the pairs MUST include a negative coefficient for full marks.