Lesson 13 ~25 min Unit 1 · Index Laws +85 XP

Quotient and Power Rules (Algebra)

Divide algebraic terms, apply power-of-a-power to coefficients and bases, and combine product, quotient and power rules in one expression.

Today's hook: $(2a^3)^4$ — does the $2$ get raised to the power $4$ as well? Yes — and many students forget that!

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Think First

warm-up

Predict the simplified form of $\dfrac{15 x^5}{3 x^2}$. What do you do with the $15$ and the $3$? What do you do with the $5$ and the $2$? They use different operations — can you say why?

Record in your workbook.

The Big Idea

+5 XP

For quotients, divide the coefficients and subtract the indices. For powers, raise everything inside the bracket — including the coefficient — to that power.

$\dfrac{a x^m}{b x^n} = \dfrac{a}{b}\, x^{m-n}$ — divide coefficients, subtract indices. $(a x^m)^n = a^n \, x^{m n}$ — raise the coefficient to the power too, then multiply the index inside by the outside.

$(2 a^3)^4 = 2^4 a^{12} = 16 a^{12}$

Quotient: divide $\div$

Coefficients divide, indices subtract.

Power: applies to all

Coefficient gets the power too.

Combine carefully

Powers $\to$ then product/quotient.

What You'll Master

objectives

Know

$\dfrac{a x^m}{b x^n} = \dfrac{a}{b} x^{m-n}$
$(a x^m)^n = a^n x^{mn}$
$(x y)^n = x^n y^n$

Understand

Why coefficients divide while indices subtract
Why the outside power lands on the coefficient too
Why all factors inside the bracket get raised

Can Do

Simplify $\dfrac{15x^5}{3x^2}$ to $5x^3$
Expand $(2x^3)^4$ to $16 x^{12}$
Simplify $\dfrac{(2x^2)^3 \cdot x^4}{4 x^3}$ to $2x^7$

Words You Need

vocabulary

QuotientResult of dividing.

Coefficient divided$\dfrac{15}{3} = 5$ — uses normal division.

Power of a product$(xy)^n = x^n y^n$ — each factor gets the power.

Bracketed coefficient$(2a^3)^4$ — the $2$ is inside, so it gets the power.

Combined expressionUses product, quotient and power in one go.

Order of operationsPowers $\to$ multiplication $\to$ division.

Spot the Trap

heads-up

Wrong: “$(2 a^3)^4 = 2 a^{12}$” — ignoring the coefficient.

Right: $(2 a^3)^4 = 2^4 a^{12} = 16 a^{12}$.

Wrong: “$\dfrac{15 x^5}{3 x^2} = 5 x^{5/2}$” — dividing the indices instead of subtracting.

Right: $\dfrac{15 x^5}{3 x^2} = 5 x^{5-2} = 5 x^3$.

Quotient rule with algebra

+5 XP

Divide the coefficients normally; subtract the indices using the quotient rule. Multi-variable terms still get one column per variable.

$\dfrac{15 x^5}{3 x^2} = \dfrac{15}{3} \cdot x^{5-2} = 5 x^3$. $\dfrac{12 x^4 y^3}{4 x y^2} = \dfrac{12}{4} \cdot x^{4-1} \cdot y^{3-2} = 3 x^3 y$.

$\dfrac{12 x^4 y^3}{4 x y^2} = 3 x^3 y$

Power of a product

+5 XP

When a bracket is raised to a power, every factor inside gets that power — including the coefficient.

$(2x^3)^4 = 2^4 \cdot x^{3 \times 4} = 16 x^{12}$. $(3 x^2 y)^3 = 3^3 \cdot x^{6} \cdot y^3 = 27 x^6 y^3$. Coefficient $\to$ raised; each variable $\to$ index multiplied.

$(3 x^2 y)^3 = 27 x^6 y^3$

Watch Me Solve It · 3 examples

Watch Me Solve It · Quotient with two variables

+15 XP per step

PROBLEM

Simplify $\dfrac{12 x^4 y^3}{4 x y^2}$.

1
Divide the coefficients
$\dfrac{12}{4} = 3$
2
Subtract indices per variable
$x^{4-1} = x^3$; $y^{3-2} = y^1$
3
Combine
$3 x^3 y$

Answer$3 x^3 y$

Watch Me Solve It · Power of a product

+15 XP per step

PROBLEM

Expand $(2x^3)^4$.

1
Raise the coefficient
$2^4 = 16$
The $2$ is inside the bracket too.
2
Multiply the variable's index
$x^{3 \times 4} = x^{12}$
3
Combine
$16 x^{12}$

Answer$16 x^{12}$

Watch Me Solve It · All three laws together

+15 XP per step

PROBLEM

Simplify $\dfrac{(2x^2)^3 \cdot x^4}{4 x^3}$.

1
Apply the power first
$(2x^2)^3 = 2^3 x^{6} = 8 x^6$
Coefficient and variable both raised.
2
Multiply in the numerator
$8 x^6 \cdot x^4 = 8 x^{6+4} = 8 x^{10}$
3
Divide by the denominator
$\dfrac{8 x^{10}}{4 x^3} = \dfrac{8}{4} x^{10-3} = 2 x^7$

Answer$2 x^7$

Common Pitfalls

heads-up

Forgetting to raise the coefficient

$(2a^3)^4 = 2 a^{12}$ is wrong; the $2$ becomes $2^4 = 16$.

Fix: Underline the coefficient; remind yourself it's inside the bracket.

Dividing indices instead of subtracting

$\dfrac{x^5}{x^2}$ is $x^3$, not $x^{5/2}$.

Fix: Quotient rule subtracts. Division is for coefficients only.

Wrong order for combined expressions

Powers should be applied before product/quotient steps.

Fix: Expand brackets first, then multiply/divide.

Copy Into Your Books

Quotient with algebra

$\dfrac{15x^5}{3x^2} = 5x^3$
Divide coefficients
Subtract indices

Multi-variable quotient

$\dfrac{12 x^4 y^3}{4xy^2} = 3 x^3 y$
One column per variable

Power of a product

$(2x^3)^4 = 16 x^{12}$
$(3x^2 y)^3 = 27 x^6 y^3$
Coefficient gets the power

Combined

Powers $\to$ multiply $\to$ divide
$\dfrac{(2x^2)^3 x^4}{4x^3} = 2x^7$

How are you completing this lesson?

Brain Trainer · 4 problems

Brain Trainer · Quotient and power drills

4 problems

Mix quotients with powers, and one combined challenge.

1 Simplify $\dfrac{24 m^7}{6 m^3}$.
$24 \div 6 = 4$; $7 - 3 = 4$.$4 m^4$
2 Expand $(3 y^4)^2$.
$3^2 = 9$; $4 \times 2 = 8$.$9 y^8$
3 Expand $(2 a b^2)^3$.
$2^3 = 8$; $a^3$; $b^{2 \times 3} = b^6$.$8 a^3 b^6$
4 Simplify $\dfrac{(3 x^2)^2 \cdot x^3}{x^4}$.
$(3x^2)^2 = 9 x^4$; numerator $9 x^{4+3} = 9 x^7$; divide $9 x^{7-4} = 9 x^3$.$9 x^3$

Complete in your workbook.

Quick Check · 5 questions

Simplify $\dfrac{15 x^5}{3 x^2}$.

+10 XP

Expand $(2x^3)^4$.

+10 XP

Simplify $\dfrac{12 x^4 y^3}{4 x y^2}$.

+10 XP

Expand $(3 x^2 y)^3$.

+10 XP

Simplify $\dfrac{(2 x^2)^3 \cdot x^4}{4 x^3}$.

+10 XP

Show Your Working · 3 questions

Show Your Working

9 marks total

ApplyMedium3 MARKS

Q6. Simplify each: (a) $\dfrac{20 a^6}{5 a^2}$, (b) $\dfrac{18 p^7 q^2}{6 p^3 q}$, (c) $\dfrac{-14 x^5}{2 x^2}$.

Answer in your workbook.

ApplyMedium3 MARKS

Q7. Expand each: (a) $(4 b^2)^3$, (b) $(2 m n^3)^4$, (c) $(5 x^3 y)^2$.

Answer in your workbook.

ReasonHard3 MARKS

Q8. Simplify $\dfrac{(3 x^2)^3 \times 2 x}{6 x^4}$ as a single term, showing each rule used. Then evaluate when $x = 2$.

Answer in your workbook.

Comprehensive Answers

Quick Check

1. C — $5 x^3$.

2. A — $16 x^{12}$.

3. D — $3 x^3 y$.

4. B — $27 x^6 y^3$.

5. A — $2 x^7$.

Show Your Working Model Answers

Q6 (3 marks): (a) $20/5 = 4$, $a^{6-2} = a^4$, so $4 a^4$ [1]; (b) $18/6 = 3$, $p^{7-3} = p^4$, $q^{2-1} = q$, so $3 p^4 q$ [1]; (c) $-14/2 = -7$, $x^{5-2} = x^3$, so $-7 x^3$ [1].

Q7 (3 marks): (a) $4^3 \cdot b^{6} = 64 b^6$ [1]; (b) $2^4 \cdot m^4 \cdot n^{12} = 16 m^4 n^{12}$ [1]; (c) $5^2 \cdot x^6 \cdot y^2 = 25 x^6 y^2$ [1].

Q8 (3 marks): Power first: $(3x^2)^3 = 27 x^6$ [1]. Product in numerator: $27 x^6 \cdot 2x = 54 x^7$. Then quotient: $\dfrac{54 x^7}{6 x^4} = 9 x^3$ [1]. At $x = 2$: $9 \times 2^3 = 9 \times 8 = 72$ [1].

Stretch Challenge · +25 XP, +10 coins

Layered Laws

Simplify $\dfrac{(4 x^2 y)^2 \cdot 3 x y^3}{8 x^3 y^4}$ as a single term. State which index law you used at each step.

Reveal solution

Power: $(4 x^2 y)^2 = 16 x^4 y^2$. Product: $16 x^4 y^2 \cdot 3 x y^3 = 48 x^5 y^5$. Quotient: $\dfrac{48}{8} \cdot x^{5-3} \cdot y^{5-4} = 6 x^2 y$.

Quick Review

Quotient (algebra)

$\dfrac{15x^5}{3x^2} = 5x^3$

Multi-variable

$\dfrac{12 x^4 y^3}{4 x y^2} = 3 x^3 y$

Power of product

$(2x^3)^4 = 16 x^{12}$

Coefficient too

$2^4 = 16$, not just $2$

Order

Powers $\to$ $\times$ $\to$ $\div$

Combined

$\dfrac{(2x^2)^3 x^4}{4x^3} = 2x^7$

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