6.01c - Multiplying and dividing algebraic terms | OCR GCSE Mathematics

Why is $x+x$ different from $x\times x$ ? Multiply and divide algebraic terms by treating coefficients and variable factors carefully, then combine powers without losing the restrictions on division.

Multiplying algebraic terms

Two packs each contain $x$ counters. Together they contain $2x$ . But $x$ rows of $x$ counters contain $x^2$ counters. Addition counts copies of a quantity; multiplication can build a new power.

For $3a\times4b$ , rearrange the factors: $3\times4\times a\times b=12ab$ . For repeated letters, count every factor: $a^2\times a^3$ contains five factors of $a$ , giving $a^5$ . Different letters stay separate, so $a\times b=ab$ , not $a^2$ .

Multiplying powers

For non-negative whole-number indices $m$ and $n$ ,

a^m\times a^n=a^{m+n}

The base must be the same. If a zero index occurs, use a non-zero base, so that $a^0=1$ is defined. This rule is not supplied on the 2027 formula sheets: learn it.

Multiply the numerical coefficients using the signed-number rules, then handle each letter separately. Writing letters in alphabetical order makes equivalent products easier to recognise.

Worked example 1: multiplying two terms

Question. Simplify $3a^2b\times4ab^2$ .

Plan. Multiply the coefficients and add the indices of each matching letter.

\begin{aligned} 3a^2b\times4ab^2&=(3\times4)(a^2\times a)(b\times b^2) &&\text{regroup factors}\\ &=12a^{2+1}b^{1+2} &&\text{count each letter}\\ &=12a^3b^3 &&\text{simplify powers} \end{aligned}

Answer. $12a^3b^3$ .

Check. At $a=2$ , $b=1$ , the original is $12\times8=96$ ; the answer gives $12\times8\times1=96$ .

Change only the sign of the second coefficient. The powers do not change, because the variable factors have not changed.

Worked example 2: a negative coefficient

Question. Simplify $3a^2b\times(-4ab^2)$ .

Plan. Determine the sign from the coefficients, then collect repeated letters.

\begin{aligned} 3a^2b\times(-4ab^2)&=(3\times(-4))a^{2+1}b^{1+2} &&\text{regroup and count}\\ &=-12a^3b^3 &&\text{one negative coefficient} \end{aligned}

Answer. $-12a^3b^3$ .

Check. At $a=2$ , $b=1$ , $12\times(-8)=-96$ , matching $-12\times8=-96$ .

The negative coefficient does not guarantee that the expression's value is negative for every substitution: the letters themselves may be negative.

Watch out

$3x\times4x=12x^2$ , not $7x$ . Multiply coefficients in a product. By contrast, $3x+4x=7x$ : add coefficients in a sum of like terms.

Apply the multiplication rule to new coefficients and powers.

Dividing algebraic terms

Division undoes multiplication. In $x^5\div x^2$ , writing out five factors above two factors lets you cancel two matching pairs. Three factors remain. Each cancellation uses $x/x=1$ , so it requires $x\ne0$ .

Dividing powers

For whole-number indices $m\ge n$ , and $a\ne0$ ,

\frac{a^m}{a^n}=a^{m-n}

When $m=n$ , the result is $a^0=1$ . The original divisor must remain non-zero, even if no letter remains in the simplified answer. This rule is not supplied on the 2027 formula sheets: learn it.

Divide coefficients as numbers and cancel matching variable factors. You can cancel multiplying factors, not arbitrary terms in a sum. Any letter in the original divisor contributes a restriction.

Worked example 3: leaving variable factors

Question. Simplify $18a^4b^2\div6a^2b$ , where $a\ne0$ and $b\ne0$ .

Plan. Write a fraction, divide coefficients and subtract matching indices.

\begin{aligned} \frac{18a^4b^2}{6a^2b}&=\frac{18}{6}a^{4-2}b^{2-1} &&\text{cancel common factors}\\ &=3a^2b &&\text{simplify each part} \end{aligned}

Answer. $3a^2b$ , for the stated non-zero values.

Check. Multiplying back gives $6a^2b\times3a^2b=18a^4b^2$ , the original dividend.

Change only the power of $b$ in the dividend from two to one. Now every $b$ -factor cancels, but the original non-zero restriction still matters.

Worked example 4: a letter disappears

Question. Simplify $18a^4b\div6a^2b$ , where $a\ne0$ and $b\ne0$ .

Plan. Cancel common factors and replace the zero power with one.

\begin{aligned} \frac{18a^4b}{6a^2b}&=3a^{4-2}b^{1-1} &&\text{divide each factor}\\ &=3a^2\times1 &&\text{non-zero }b\text{ gives }b^0=1\\ &=3a^2 &&\text{remove the factor 1} \end{aligned}

Answer. $3a^2$ , still with $a\ne0$ and $b\ne0$ .

Check. $6a^2b\times3a^2=18a^4b$ . At $b=0$ , the original division would be by zero, so that substitution is not allowed.

There is the useful surprise: a letter can disappear from the answer while its restriction survives. Simplifying does not repair an undefined original calculation.

Watch out

$x^3\div x^3=1$ , not zero, when $x\ne0$ . All factors cancel to ones. Do not substitute $x=0$ : $0/0$ is undefined.

Keep the original divisor in mind when you simplify.

Combining index operations

An outside power repeats the whole term. For example, $(2x^3)^2=(2x^3)(2x^3)=4x^6$ . Both the coefficient and the variable factor are squared.

For positive whole-number $n$ , $(a^m)^n=a^{mn}$ : there are $n$ groups, each containing $m$ factors of $a$ . With $m=0$ , require $a\ne0$ . When the entire expression has a zero outside power, its base must also be non-zero. We will combine ordinary positive powers here.

You already know

Brackets and powers come before multiplication and division. Multiplication joins factors; division cancels factors only when the original divisor is non-zero.

Write the result of a bracket power first. Then combine matching letters across the remaining product or quotient.

Worked example 5: a powered term in a product

Question. Simplify $(2x^2)^3\times3x$ .

Plan. Cube every factor inside the bracket, then multiply by $3x$ .

\begin{aligned} (2x^2)^3\times3x&=2^3x^{2\times3}\times3x &&\text{cube both factors}\\ &=8x^6\times3x &&\text{evaluate the power}\\ &=24x^7 &&\text{multiply and add indices} \end{aligned}

Answer. $24x^7$ .

Check. At $x=1$ , the original is $2^3\times3=24$ , matching the answer. Counting the repeated letters also gives six inside the cube and one more outside: seven in total.

Change only the final operation from multiplication to division. Now the last $x$ -factor is cancelled, and $x=0$ becomes forbidden.

Worked example 6: a powered term in a quotient

Question. Simplify $(2x^2)^3\div3x$ , where $x\ne0$ .

Plan. Evaluate the bracket power before dividing coefficients and powers.

\begin{aligned} \frac{(2x^2)^3}{3x}&=\frac{8x^6}{3x} &&\text{cube both factors}\\ &=\frac83x^{6-1} &&\text{cancel a non-zero }x\\ &=\frac{8x^5}{3} &&\text{write an exact coefficient} \end{aligned}

Answer. $\frac{8x^5}{3}$ , for $x\ne0$ .

Check. Multiplying by the divisor gives $3x\times\frac{8x^5}{3}=8x^6=(2x^2)^3$ .

Integer coefficients can produce a fractional coefficient after division. Leave it exact; the fraction does not change the index rules.

Watch out

$(2x^3)^2=4x^6$ , not $2x^6$ . The outside power applies to the coefficient too. Expand the repeated product if you are unsure.

Try the quotient method with a negative coefficient inside the power.

Quick Recap

Multiply coefficients and add indices of the same base in a product.
Divide coefficients and subtract indices when cancelling powers.
A power outside brackets applies to every multiplying factor inside.
A simplified quotient keeps the restrictions of its original divisor.

$x+x=2x$ counts two copies of $x$ ; $x\times x=x^2$ multiplies two copies. The operation decides the rule.

Last Accessed

Number

Algebra

Ratio and proportion

Geometry and measures

Graphs and functions

Probability

Statistics

6.01c - Multiplying and dividing algebraic terms

Multiplying algebraic terms

Dividing algebraic terms

Combining index operations

Quick Recap

Learn faster with Nova