Mathematics • Year 9 • Unit 1 • Lesson 13
Quotient and Power Rules — Mixed Challenge
Combine quotient, power-of-a-product and product rules in one go. Pick the right tool for each problem, spot a real Year 9 mistake, and build your own expression that lands a target answer.
1. Mixed problems — pick the right rule
Each one uses some mix of quotient ($\div$, subtract indices), power-of-a-product (everything inside the bracket gets the outer index) and product ($\times$, add indices). Show working. 3 marks each
1.1 Simplify $\dfrac{-14 x^5}{2 x^2}$. (Watch the sign on the coefficient.)
1.2 Expand $(2 m n^3)^4$ fully.
1.3 Expand $(5 x^3 y)^2$ fully.
1.4 Simplify $\dfrac{(3 x^2)^3 \times 2 x}{6 x^4}$ as a single term. Then evaluate it when $x = 2$.
1.5 Find $n$ if $(2 x^n)^3 = 8 x^{15}$. Show how you got there.
1.6 Simplify $\dfrac{(2 x y)^3 \cdot 5 x}{20 x^4 y^3}$ fully.
2. Find the mistake
A student has tried to simplify $\dfrac{(2 a^3)^2}{a^4}$. Exactly one line contains a mistake. Spot it, explain it, then redo the working correctly. 3 marks
Student's working — simplify $\dfrac{(2 a^3)^2}{a^4}$:
Line 1: $(2 a^3)^2 = 2 \cdot a^{3 \times 2}$
Line 2: $= 2 a^6$
Line 3: $\dfrac{2 a^6}{a^4} = 2 a^{6-4}$
Line 4: $= 2 a^2$.
(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 1, what happens to the $2$ when there's a power outside the bracket?3. Open-ended challenge — hit the target
This question has many valid answers. 4 marks
3.1 Build a quotient of the form $\dfrac{\text{numerator}}{\text{denominator}}$ that simplifies to exactly $\mathbf{4 x^3}$. Both the numerator and the denominator must contain at least one bracketed power (like $(2x)^2$ or $(3x^2)^3$) so that you use the power-of-a-product rule somewhere.
Find two different valid quotients. For each:
(i) Write the quotient.
(ii) Show the working step by step (power $\to$ multiply $\to$ divide).
(iii) Confirm the answer is $4 x^3$.
Bonus: Try to make your two quotients use different powers (e.g. one uses $(\ldots)^2$ and the other uses $(\ldots)^3$).
How did this worksheet feel?
What I'll revisit before next class:
1.1 — $\dfrac{-14 x^5}{2 x^2}$
$\dfrac{-14}{2} \cdot x^{5-2} = \mathbf{-7 x^3}$. Coefficient division keeps the negative sign.
1.2 — $(2 m n^3)^4$
$2^4 \cdot m^{1 \times 4} \cdot n^{3 \times 4} = \mathbf{16 m^4 n^{12}}$.
1.3 — $(5 x^3 y)^2$
$5^2 \cdot x^{3 \times 2} \cdot y^{1 \times 2} = \mathbf{25 x^6 y^2}$.
1.4 — $\dfrac{(3 x^2)^3 \times 2 x}{6 x^4}$
Step 1 (power): $(3 x^2)^3 = 27 x^6$.
Step 2 (product): $27 x^6 \times 2x = 54 x^7$.
Step 3 (quotient): $\dfrac{54 x^7}{6 x^4} = 9 x^{7-4} = \mathbf{9 x^3}$.
At $x = 2$: $9 \times 2^3 = 9 \times 8 = \mathbf{72}$.
1.5 — Solve $(2 x^n)^3 = 8 x^{15}$
$(2 x^n)^3 = 2^3 \cdot x^{3n} = 8 x^{3n}$. Match indices: $3n = 15$, so $\mathbf{n = 5}$.
Check: $(2 x^5)^3 = 8 x^{15}$. ✓
1.6 — $\dfrac{(2xy)^3 \cdot 5 x}{20 x^4 y^3}$
$(2xy)^3 = 8 x^3 y^3$. Numerator: $8 x^3 y^3 \cdot 5 x = 40 x^4 y^3$. Quotient: $\dfrac{40 x^4 y^3}{20 x^4 y^3} = 2 \cdot x^0 \cdot y^0 = \mathbf{2}$.
2 — Find the mistake
(a) The mistake is on Line 1.
(b) The student forgot that the coefficient $2$ is INSIDE the bracket and so also gets the outer power. The line should be $(2 a^3)^2 = 2^2 \cdot a^{3 \times 2} = 4 a^6$, not $2 \cdot a^{3 \times 2}$.
(c) Corrected working:
$(2 a^3)^2 = 2^2 \cdot a^{3 \times 2} = 4 a^6$
$\dfrac{4 a^6}{a^4} = 4 a^{6-4} = \mathbf{4 a^2}$.
Even though their final index $a^2$ was correct by luck, the coefficient should be $4$, not $2$.
3 — Open-ended challenge (sample solution)
We want the simplified result to be $4 x^3$. Strategy: pick any "extra" factor, then divide by it.
Quotient 1 (uses $(\ldots)^2$): $\dfrac{(2x)^2 \cdot 4 x^3}{(2x)^2}$.
Working: $(2x)^2 = 4 x^2$. Numerator $= 4 x^2 \cdot 4 x^3 = 16 x^5$. Quotient $= \dfrac{16 x^5}{4 x^2} = \mathbf{4 x^3}$ ✓.
Quotient 2 (uses $(\ldots)^3$): $\dfrac{(2 x^2)^3 \cdot 2 x}{4 x^4}$.
Working: $(2x^2)^3 = 8 x^6$. Numerator $= 8 x^6 \cdot 2 x = 16 x^7$. Quotient $= \dfrac{16 x^7}{4 x^4} = \mathbf{4 x^3}$ ✓.
Other valid quotients: $\dfrac{(2x)^3 \cdot x^2}{2 x^2}$, $\dfrac{(4 x^2)^2}{4 x}$, and many more. Any quotient where powers $\to$ products $\to$ division simplifies to $4 x^3$.
Marking: 2 marks per valid quotient (1 for the quotient, 1 for clear working showing it equals $4 x^3$). Up to 4 in total. Bonus only awarded if the two quotients use different powers.