FA4, HA4 - Simplifying powers in algebra | Eduqas GCSE Mathematics

The index laws count equal factors whether the base is a number or a letter. Learn to simplify products, quotients and powers of algebraic terms, then distinguish those calculations from adding like terms. This develops the sums, products, powers and index-law scope within FA4 and HA4.

Products and quotients

Are $2x^3\times3x^2$ and $2x^3+3x^2$ the same expression? The symbols between the terms matter just as much as the indices.

You already know that $x^3=x\times x\times x$ . Thus $x^3x^2=x^5$ : there are five equal factors. A letter with no written index has index $1$ , so $x^2x=x^3$ .

Index laws for algebraic terms

x^m x^n=x^{m+n},\qquad \frac{x^m}{x^n}=x^{m-n}.

These laws use integer indices wherever the expressions are defined. A base in a denominator, or raised to zero or a negative integer power, must be nonzero.

Not in the formula sheets: learn these laws.

Work with numerical coefficients separately from letters. For example, $2x^2\times3x=(2\times3)x^{2+1}=6x^3$ . If there are two letters, combine each letter only with matching bases.

Worked example 1: multiplying algebraic terms

Question. Simplify $12x^5y^3\times4x^2y^5$ .

Plan. Multiply the coefficients, then add the indices separately for $x$ and $y$ .

\begin{aligned} 12x^5y^3\times4x^2y^5 &=(12\times4)x^{5+2}y^{3+5} &&\text{group matching factors}\\ &=48x^7y^8 &&\text{simplify coefficients and indices} \end{aligned}

Answer. $48x^7y^8$ .

Check. Put $x=1,y=2$ . The original gives $(12\times8)(4\times32)=96\times128=12288$ . The answer gives $48\times1\times256=12288$ . This checks the calculation, although one substitution alone is not a proof of an identity.

Keep both terms, but change multiplication to division. Subtract indices and divide coefficients instead.

Worked example 2: dividing algebraic terms

Question. Simplify $\dfrac{12x^5y^3}{4x^2y^5}$ , where $x\ne0$ and $y\ne0$ .

Plan. Divide the coefficients and subtract each denominator index from the matching numerator index.

\begin{aligned} \frac{12x^5y^3}{4x^2y^5} &=3x^{5-2}y^{3-5} &&\text{divide matching factors}\\ &=3x^3y^{-2} &&\text{two }y\text{ factors remain below}\\ &=\frac{3x^3}{y^2} &&\text{write the reciprocal power} \end{aligned}

Answer. $3x^3/y^2$ , with the original restrictions $x\ne0,y\ne0$ .

Check. With $x=1,y=2$ , the original is $96/128=3/4$ . The answer is $3/2^2=3/4$ .

The original restrictions still matter after cancellation. For example, $x/x=1$ only for $x\ne0$ ; substituting zero into the original would give $0/0$ , which is undefined. Likewise, $x^0=1$ and $x^{-2}=1/x^2$ require $x\ne0$ .

Watch out

In $4x^3y\times7x^2y^5$ , multiply $4$ by $7$ , but add the indices of each matching letter. The result is $28x^5y^6$ , not $11x^5y^6$ . The unwritten index of the first $y$ is $1$ .

Apply the same division method with a different coefficient and powers.

Powers of expressions

An outside power repeats the whole expression in brackets, including its coefficient. For example, $(3x)^2=(3x)(3x)=9x^2$ .

The numerical rules transfer directly:

(x^m)^n=x^{mn},\qquad (xy)^n=x^ny^n,\qquad \left(\frac xy\right)^n=\frac{x^n}{y^n}.

These statements use integer indices wherever all expressions are defined; a denominator must be nonzero. The outside power multiplies an existing index and applies to every factor.

A negative coefficient is part of the repeated expression: $(-2x)^3=-8x^3$ , but $(-2x)^2=4x^2$ . Brackets show whether the negative sign is included in the power.

Worked example 3: a power of a product

Question. Simplify $(3x^2y)^2$ .

Plan. Square each factor, multiplying the letter indices by $2$ .

\begin{aligned} (3x^2y)^2 &=3^2(x^2)^2y^2 &&\text{square every factor}\\ &=9x^4y^2 &&\text{multiply the }x\text{ indices} \end{aligned}

Answer. $9x^4y^2$ .

Check. With $x=2,y=3$ , the original is $(3\times4\times3)^2=36^2=1296$ . The answer is $9\times16\times9=1296$ .

Change the product with $y$ to division by $y$ . The denominator must also be squared.

Worked example 4: a power of a quotient

Question. Simplify $\left(\dfrac{3x^2}{y}\right)^2$ , where $y\ne0$ .

Plan. Square numerator and denominator, including the numerical coefficient.

\begin{aligned} \left(\frac{3x^2}{y}\right)^2 &=\frac{3^2(x^2)^2}{y^2} &&\text{square both parts of the fraction}\\ &=\frac{9x^4}{y^2} &&\text{evaluate and multiply indices} \end{aligned}

Answer. $9x^4/y^2$ , for $y\ne0$ .

Check. With $x=2,y=3$ , the original is $(12/3)^2=16$ . The answer is $9\times16/9=16$ .

A negative outside index first takes the reciprocal. For example, for nonzero $x,y$ ,

\left(\frac{2x}{y}\right)^{-2} =\left(\frac{y}{2x}\right)^2 =\frac{y^2}{4x^2}.

The nonzero conditions come from the original denominator and the reciprocal; do not drop them because a letter moves.

Watch out

$(3x^2)^2=9x^4$ , not $3x^4$ or $9x^2$ . The coefficient is squared and the existing index is multiplied by $2$ .

Use the same quotient method with a cube rather than a square.

Combined calculations

In an expression with products and sums, simplify powers and products first. Then collect only terms with exactly the same letter factors and indices.

For example, $2x^3+5x^3=7x^3$ , because there are two lots and five lots of the same quantity. But $2x^3+5x^2$ has unlike terms. It cannot become $7x^5$ : addition does not join factors.

Like terms

Like terms have the same letter part, including the indices. Their coefficients can be added or subtracted. For example, $4a^2b$ and $-3a^2b$ are like terms; $4a^2b$ and $4ab^2$ are not.

The operation order gives a reliable route: simplify each product, identify like terms, then combine their coefficients.

Worked example 5: products before sums

Question. Simplify $2x^2\times3x+4x^3-x^2$ .

Plan. Simplify the product first, then collect the cubic terms.

\begin{aligned} 2x^2\times3x+4x^3-x^2 &=6x^3+4x^3-x^2 &&\text{multiply the first two terms}\\ &=10x^3-x^2 &&\text{collect only the cubic terms} \end{aligned}

Answer. $10x^3-x^2$ .

Check. With $x=2$ , the original gives $2\times4\times6+4\times8-4=76$ . The answer gives $10\times8-4=76$ .

Change the power in the second factor from $x$ to $x^2$ . The resulting product is now a different power from both other terms.

Worked example 6: unlike powers stay separate

Question. Simplify $2x^2\times3x^2+4x^3-x^2$ .

Plan. Use the product law, then check whether any like terms remain.

\begin{aligned} 2x^2\times3x^2+4x^3-x^2 &=6x^{2+2}+4x^3-x^2 &&\text{add indices in the product}\\ &=6x^4+4x^3-x^2 &&\text{the three powers are unlike} \end{aligned}

Answer. $6x^4+4x^3-x^2$ .

Check. With $x=2$ , the original gives $2\times4\times12+32-4=124$ . The answer gives $6\times16+32-4=124$ .

We can now settle the opening comparison. The product $2x^3\times3x^2$ simplifies to $6x^5$ , while the sum $2x^3+3x^2$ contains unlike terms. At $x=1$ their values are $6$ and $5$ , so they are not the same expression.

Watch out

$x^2+x^3\ne x^5$ . If unsure, try $x=2$ : the left side is $4+8=12$ , while the right side is $32$ . A single counterexample disproves a proposed identity; the index laws provide the general valid simplification.

Simplify the product and leave unlike powers separate.

Quick Recap

Multiply coefficients while adding indices of matching bases; for division, divide coefficients and subtract indices.
An outside power acts on every factor and multiplies each existing index.
Preserve nonzero restrictions from the original expression.
Simplify products before collecting like terms; unlike powers cannot be added into a single power.

The opening product and sum are different: multiplication joins factors, while addition combines only like terms.

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FA4, HA4 - Simplifying powers in algebra

Products and quotients

Powers of expressions

Combined calculations

Quick Recap

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