Mathematics • Year 9 • Unit 1 • Lesson 14
Mixed Algebraic Index Laws
Build fluency with the four-step game plan: power $\to$ product $\to$ quotient $\to$ zero. Watch one fully worked example, fill in a guided one, then complete eight independent practice problems.
1. I do — fully worked example
Powers come FIRST. Then product. Then quotient. Then check for any $x^0 = 1$.
Problem. Simplify $\dfrac{(2 x^2)^3 \cdot 5 x}{10 x^7}$.
Step 1 — POWER first: expand the bracket.
$(2 x^2)^3 = 2^3 \cdot x^{2 \times 3} = 8 x^6$
Reason: every factor inside (the $2$ and the $x^2$) gets the outer index.
Step 2 — PRODUCT: multiply in the numerator.
$8 x^6 \cdot 5 x = 40 x^{6+1} = 40 x^7$
Reason: coefficients multiply; same-base indices add.
Step 3 — QUOTIENT: divide by the denominator.
$\dfrac{40 x^7}{10 x^7} = \dfrac{40}{10} \cdot x^{7-7} = 4 x^0$
Reason: coefficients divide; same-base indices subtract.
Step 4 — ZERO: collapse $x^0$.
$4 x^0 = 4 \times 1 = 4$
Reason: any non-zero base to the power $0$ equals $1$.
Answer: $\mathbf{4}$.
2. We do — fill in the missing steps
Same four-step plan with the working faded. Fill in each blank. 4 marks
Problem. Simplify $\dfrac{2 x^3 \cdot 3 x^2}{6 x^4}$.
Step 1 — POWER: no brackets to expand, so __________________ .
Step 2 — PRODUCT (numerator):
$2 \times 3 = \_\_\_\_$ and $x^{3+2} = x^{\_\_\_}$, giving $\_\_\_\_ x^{\_\_\_}$
Step 3 — QUOTIENT:
$\dfrac{6 x^5}{6 x^4} = \dfrac{6}{6} \cdot x^{\,5 - \_\_\,} = \_\_\_\_ \cdot x^{\_\_\_}$
Step 4 — ZERO (if needed):
No $x^0$ here. Final answer: $\dfrac{2 x^3 \cdot 3 x^2}{6 x^4} = \_\_\_\_\_\_$
3. You do — independent practice
Show working under each problem. Foundation = two rules. Standard = three rules. Extension = all four rules including a zero-index collapse.
Foundation — two rules
3.1 Simplify $\dfrac{4 x^2 \cdot 3 x^5}{6 x^3}$. 1 mark
3.2 Simplify $\dfrac{(2 a^3)^2}{a^4}$. 1 mark
3.3 Simplify $\dfrac{5 y^4 \cdot 2 y^3}{10 y^7}$ (watch for $y^0$). 1 mark
3.4 Simplify $\dfrac{40 x^7}{10 x^7}$. 1 mark
Standard — three rules
3.5 Simplify $\dfrac{(3 m^2)^2 \cdot 2 m}{m^3}$. 2 marks
3.6 Simplify $\dfrac{8 p^5 q^2 \cdot p}{(2 p^3 q)^2}$. 2 marks
Extension — all four including $x^0$
3.7 Simplify $\dfrac{(2 m^2 n)^3 \cdot 3 m n^2}{12 m^5 n^4}$. 3 marks
3.8 Simplify $\dfrac{(3 a^2 b)^2 \cdot 4 a b^3}{6 a^3 b^4}$. 2 marks
How did this worksheet feel?
What I'll revisit before next class:
Section 2 — We do (faded $\dfrac{2 x^3 \cdot 3 x^2}{6 x^4}$)
Step 1: no brackets, so skip.
Step 2: $2 \times 3 = \mathbf{6}$ and $x^{3+2} = \mathbf{x^5}$, giving $\mathbf{6 x^5}$.
Step 3: $\dfrac{6}{6} \cdot x^{5-4} = \mathbf{1} \cdot x^{1} = \mathbf{x}$.
Step 4: no $x^0$. Final answer: $\mathbf{x}$.
3.1 — $\dfrac{4 x^2 \cdot 3 x^5}{6 x^3}$
Numerator $12 x^7$; quotient $\dfrac{12}{6} \cdot x^{7-3} = \mathbf{2 x^4}$.
3.2 — $\dfrac{(2 a^3)^2}{a^4}$
$(2 a^3)^2 = 4 a^6$; quotient $\dfrac{4 a^6}{a^4} = \mathbf{4 a^2}$.
3.3 — $\dfrac{5 y^4 \cdot 2 y^3}{10 y^7}$
Numerator $10 y^7$; quotient $\dfrac{10}{10} \cdot y^{7-7} = 1 \cdot y^0 = \mathbf{1}$.
3.4 — $\dfrac{40 x^7}{10 x^7}$
$\dfrac{40}{10} \cdot x^{7-7} = 4 x^0 = \mathbf{4}$.
3.5 — $\dfrac{(3 m^2)^2 \cdot 2 m}{m^3}$
$(3 m^2)^2 = 9 m^4$; numerator $9 m^4 \cdot 2 m = 18 m^5$; quotient $\dfrac{18 m^5}{m^3} = \mathbf{18 m^2}$.
3.6 — $\dfrac{8 p^5 q^2 \cdot p}{(2 p^3 q)^2}$
Denominator $(2 p^3 q)^2 = 4 p^6 q^2$. Numerator $8 p^5 q^2 \cdot p = 8 p^6 q^2$.
Quotient $\dfrac{8 p^6 q^2}{4 p^6 q^2} = 2 \cdot p^0 \cdot q^0 = \mathbf{2}$.
3.7 — $\dfrac{(2 m^2 n)^3 \cdot 3 m n^2}{12 m^5 n^4}$
Power: $(2 m^2 n)^3 = 8 m^6 n^3$.
Product: $8 m^6 n^3 \cdot 3 m n^2 = 24 m^7 n^5$.
Quotient: $\dfrac{24 m^7 n^5}{12 m^5 n^4} = 2 \cdot m^{7-5} \cdot n^{5-4} = \mathbf{2 m^2 n}$.
3.8 — $\dfrac{(3 a^2 b)^2 \cdot 4 a b^3}{6 a^3 b^4}$
Power: $(3 a^2 b)^2 = 9 a^4 b^2$.
Product: $9 a^4 b^2 \cdot 4 a b^3 = 36 a^5 b^5$.
Quotient: $\dfrac{36 a^5 b^5}{6 a^3 b^4} = 6 \cdot a^{5-3} \cdot b^{5-4} = \mathbf{6 a^2 b}$.