Mixed Algebraic Index Laws
Combine product, quotient, power-of-power and zero index rules in one expression. Plan the order. Track the coefficient. Land a single simplified term.
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Predict the simplified form of $\dfrac{2 x^3 \cdot 3 x^2}{6 x^4}$. Which law tackles the numerator? Which tackles the division? What happens to $x$ when the indices match?
Most algebra questions don't use just one law — they layer several. The trick is to follow an order: powers first, then products in the numerator and denominator, then the quotient, then check for any zero indices.
Plan: power $\to$ product $\to$ quotient $\to$ check $x^0 = 1$. Coefficients use normal arithmetic; indices follow the rule that matches the operation.
Know
- Product: $x^m \cdot x^n = x^{m+n}$
- Quotient: $\dfrac{x^m}{x^n} = x^{m-n}$
- Power: $(x^m)^n = x^{mn}$ · Zero: $x^0 = 1$
Understand
- Why you do powers before products and quotients
- Why coefficients use arithmetic, not index laws
- Why an index of $0$ collapses to $1$
Can Do
- Simplify $\dfrac{2x^3 \cdot 3x^2}{6x^4}$ to $x$
- Simplify $\dfrac{(2x^2)^3 \cdot 5x}{10 x^7}$ to $4$
- Handle multi-variable mixed expressions
Wrong: “$\dfrac{6 x^5}{6 x^4} = 0$” — forgetting that the $6$s cancel to $1$, not $0$.
Right: $\dfrac{6}{6} = 1$ and $x^{5-4} = x^1 = x$, so the answer is $x$.
Wrong: “Do the division before the bracket power.”
Right: Expand $(2x^2)^3 = 8x^6$ first, then multiply and divide.
Whenever an expression mixes laws, work the same order every time. It removes guesswork and stops you from skipping a step.
Step 1: expand all brackets using power-of-power. Step 2: combine factors in the numerator (and denominator) with the product rule. Step 3: divide with the quotient rule. Step 4: any index of $0$ becomes $1$.
If the indices match top and bottom, the variable index is $0$ — and $x^0 = 1$. The whole expression collapses to a coefficient.
$\dfrac{(2x^2)^3 \cdot 5 x}{10 x^7} = \dfrac{8 x^6 \cdot 5 x}{10 x^7} = \dfrac{40 x^7}{10 x^7} = 4 x^0 = 4$. The $x$ disappears because $x^{7-7} = x^0 = 1$.
Watch Me Solve It · 3 examples
- 1Combine the numerator (product rule)$2 \cdot 3 = 6$; $x^{3+2} = x^5$ → $6 x^5$
- 2Apply the quotient rule$\dfrac{6 x^5}{6 x^4} = \dfrac{6}{6} \cdot x^{5-4} = 1 \cdot x^1$
- 3Write the final term$x$
- 1Expand the bracket (power of a product)$(2 x^2)^3 = 2^3 \cdot x^{6} = 8 x^6$
- 2Multiply in the numerator$8 x^6 \cdot 5 x = 40 x^{7}$
- 3Divide and apply the zero index$\dfrac{40 x^7}{10 x^7} = 4 x^0 = 4$
- 1Expand the bracket$(3 a^2 b)^2 = 9 a^4 b^2$
- 2Multiply in the numerator$9 a^4 b^2 \cdot 4 a b^3 = 36 a^{5} b^{5}$
- 3Divide column by column$\dfrac{36}{6} \cdot a^{5-3} \cdot b^{5-4} = 6 a^2 b$
Common Pitfalls
Game plan
- Powers $\to$ products $\to$ quotient $\to$ zero check
Worked: product/quotient
- $\dfrac{2 x^3 \cdot 3 x^2}{6 x^4} = x$
Worked: all four laws
- $\dfrac{(2x^2)^3 \cdot 5x}{10 x^7} = 4$
- $x^0 = 1$
Multi-variable
- $\dfrac{(3 a^2 b)^2 \cdot 4 a b^3}{6 a^3 b^4} = 6 a^2 b$
How are you completing this lesson?
Brain Trainer · 4 problems
Apply every law you've learned. Watch the order.
1 Simplify $\dfrac{4 x^2 \cdot 3 x^5}{6 x^3}$.
Numerator $12 x^7$; quotient $12/6 = 2$; $x^{7-3} = x^4$.$2 x^4$2 Simplify $\dfrac{(2 a^3)^2}{a^4}$.
$(2 a^3)^2 = 4 a^6$; $a^{6-4} = a^2$.$4 a^2$3 Simplify $\dfrac{5 y^4 \cdot 2 y^3}{10 y^7}$.
Numerator $10 y^7$; divide by $10 y^7$ gives $y^0 = 1$.$1$4 Simplify $\dfrac{(2 m^2 n)^3 \cdot 3 m n^2}{12 m^5 n^4}$.
$(2 m^2 n)^3 = 8 m^6 n^3$; numerator $24 m^7 n^5$; quotient $\tfrac{24}{12} m^{7-5} n^{5-4} = 2 m^2 n$.$2 m^2 n$
Quick Check · 5 questions
Show Your Working · 3 questions
Q6. Simplify each: (a) $\dfrac{4 x^2 \cdot 3 x^5}{6 x^3}$, (b) $\dfrac{(2 a^3)^2}{a^4}$, (c) $\dfrac{5 y^4 \cdot 2 y^3}{10 y^7}$.
Q7. Simplify each, naming the law used at each step: (a) $\dfrac{(3 m^2)^2 \cdot 2 m}{m^3}$, (b) $\dfrac{(2 x y)^3 \cdot 5 x}{20 x^4 y^3}$, (c) $\dfrac{8 p^5 q^2 \cdot p}{(2 p^3 q)^2}$.
Q8. Simplify $\dfrac{(2 x^2 y)^3 \cdot 6 x y^2}{(3 x^3 y)^2}$ to a single term. Then evaluate when $x = 1$, $y = 2$.
Quick Check
1. B — $x$.
2. A — $4 a^2$.
3. D — $1$.
4. C — $4$.
5. A — $6 a^2 b$.
Show Your Working Model Answers
Q6 (3 marks): (a) Numerator $12 x^7$; $\tfrac{12}{6} = 2$, $x^{7-3} = x^4$; $2 x^4$ [1]. (b) $(2 a^3)^2 = 4 a^6$; $a^{6-4} = a^2$; $4 a^2$ [1]. (c) Numerator $10 y^7$; $\tfrac{10 y^7}{10 y^7} = y^0 = 1$ [1].
Q7 (3 marks): (a) Power: $(3 m^2)^2 = 9 m^4$; product: $9 m^4 \cdot 2 m = 18 m^5$; quotient: $\tfrac{18 m^5}{m^3} = 18 m^2$ [1]. (b) Power: $(2 x y)^3 = 8 x^3 y^3$; product: $40 x^4 y^3$; quotient: $\tfrac{40 x^4 y^3}{20 x^4 y^3} = 2$ [1]. (c) Numerator product: $8 p^6 q^2$; denominator power: $(2 p^3 q)^2 = 4 p^6 q^2$; quotient: $\tfrac{8 p^6 q^2}{4 p^6 q^2} = 2$ [1].
Q8 (3 marks): Numerator: $(2 x^2 y)^3 = 8 x^6 y^3$; $\times 6 x y^2 = 48 x^7 y^5$ [1]. Denominator: $(3 x^3 y)^2 = 9 x^6 y^2$ [1]. Quotient: $\tfrac{48}{9} x^{7-6} y^{5-2} = \tfrac{16}{3} x y^3$. At $x = 1$, $y = 2$: $\tfrac{16}{3} \cdot 1 \cdot 8 = \tfrac{128}{3}$ [1].
All Four Laws in One
Simplify $\dfrac{(2 x^2 y^3)^3 \cdot 5 x y^2}{20 x^7 y^{11}}$ as a single term. State which index law you used at each step, and explain why the final answer is a constant.
Reveal solution
Power: $(2 x^2 y^3)^3 = 8 x^6 y^9$. Product: $8 x^6 y^9 \cdot 5 x y^2 = 40 x^7 y^{11}$. Quotient: $\dfrac{40 x^7 y^{11}}{20 x^7 y^{11}} = 2 x^0 y^0 = 2$. Because every index in the numerator matches the denominator, both variables become $1$, leaving just the coefficient.
Order
Power $\to$ $\times$ $\to$ $\div$ $\to$ zero
Product
$x^m \cdot x^n = x^{m+n}$
Quotient
$\dfrac{x^m}{x^n} = x^{m-n}$
Power
$(x^m)^n = x^{mn}$
Zero
$x^0 = 1$ (for $x \ne 0$)
Coefficients
Use arithmetic, not index laws
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