C1.6; E1.6 - Signed products and order of operations | Cambridge IGCSE Mathematics

Multiplication and division with negative numbers follow a consistent pattern. You will use that pattern in directed calculations, work through brackets in the correct order, and see why a minus sign outside a square changes its value. These skills are shared by Core and Extended.

Multiplying signed numbers

A freezer temperature falls by $3\,^{\circ}\mathrm{C}$ each hour. What is its signed change over five hours forwards in time, and what happens if you look five hours backwards instead? A negative sign can describe a direction as well as a number below zero.

First, $3\times(-4)$ means three copies of $-4$ :

(-4)+(-4)+(-4)=-12.

Changing the order of the two factors does not change a product, so $(-4)\times3=-12$ too. Multiplying one positive and one negative number gives a negative answer.

Why do two negatives give a positive product? Keep multiplying by $-4$ , and reduce the other factor by $1$ each time. Each product must increase by $4$ : you have removed one copy of $-4$ .

First factor	$3$	$2$	$1$	$0$	$-1$	$-2$
Product with $-4$	$-12$	$-8$	$-4$	$0$	$4$	$8$

Continuing the same pattern through zero makes $(-1)\times(-4)=4$ , then $(-2)\times(-4)=8$ . The sign rule preserves ordinary multiplication patterns.

Key idea

For two non-zero factors, same signs give a positive product; different signs give a negative product. Multiply the sizes, then decide the sign. Any number multiplied by zero gives zero.

Use the rule to describe repeated directed changes, rather than dropping the minus sign halfway through.

Worked example 1: repeating a fall

Question. A temperature falls by $4\,^{\circ}\mathrm{C}$ per hour for $6$ hours. Work out its signed change.

Plan. Multiply the number of hours by the signed change per hour.

\begin{aligned} 6\times(-4)&=-(6\times4) && \text{different signs}\\ &=-24 && \text{six falls of four} \end{aligned}

Answer. The signed change is $-24\,^{\circ}\mathrm{C}$ .

Check. Two three-hour periods each change the temperature by $-12\,^{\circ}\mathrm{C}$ ; $-12+(-12)=-24$ .

Now change only the sign of the first factor. The size stays $24$ , but the product changes sign.

Worked example 2: multiplying two negatives

Question. Work out $(-6)\times(-4)$ .

Plan. Multiply the sizes, then use the rule for two equal signs.

\begin{aligned} 6\times4&=24 && \text{find the size}\\ (-6)\times(-4)&=24 && \text{same signs give positive} \end{aligned}

Answer. $24$ .

Check. The factors $6$ and $-6$ cancel to zero. Their products with $-4$ , namely $-24$ and $24$ , must cancel too.

Watch out

$(-6)\times(-4)=-24$ uses the rule for different signs. Two negative factors give $+24$ . But $-6+(-4)=-10$ : the multiplication sign rule is not an addition rule.

Dividing signed numbers

Division reverses multiplication. To work out $-20\div5$ , ask “what number multiplied by $5$ gives $-20$ ?” The answer is $-4$ , because $(-4)\times5=-20$ .

The same sign pattern follows from multiplication: a division of two non-zero numbers is positive when the signs match and negative when they differ. This applies to a horizontal division sign or to a fraction bar, which also means division.

Worked example 3: dividing different signs

Question. Work out $-42\div6$ .

Plan. Divide the sizes, then choose the sign that reverses multiplication by $6$ .

\begin{aligned} 42\div6&=7 && \text{divide the sizes}\\ -42\div6&=-7 && \text{different signs give negative} \end{aligned}

Answer. $-7$ .

Check. $(-7)\times6=-42$ , recovering the dividend.

Changing only the divisor's sign changes the sign of the quotient.

Worked example 4: dividing matching signs

Question. Work out $-42\div(-6)$ .

Plan. Use the same size calculation and the rule for matching signs.

\begin{aligned} 42\div6&=7 && \text{same size calculation}\\ -42\div(-6)&=7 && \text{same signs give positive} \end{aligned}

Answer. $7$ .

Check. $7\times(-6)=-42$ . A negative answer would multiply by $-6$ to give a positive result instead.

Zero needs special care. $0\div5=0$ , because $0\times5=0$ . But $5\div0$ is undefined: no number multiplied by zero gives $5$ . Also, $0\div0$ is undefined, because every number multiplied by zero gives zero, so there is no unique quotient.

Watch out

$-42\div(-6)=-7$ fails the inverse check: $(-7)\times(-6)=42$ , not $-42$ . Check a division by multiplying the answer by the divisor.

Brackets and the order of operations

An expression needs an agreed order so that it has one value. In $10-3\times2$ , the product $3\times2$ is one quantity being subtracted. Its value is $6$ , so the expression equals $4$ , not $14$ .

Work through these stages:

Evaluate brackets, working from the innermost pair outwards.
Evaluate powers and roots.
Do multiplication and division in their left-to-right order.
Do addition and subtraction in their left-to-right order.

Square brackets group a calculation in the same way as round brackets; using both shapes makes nested groups easier to read. Multiplication and division have equal priority: $8\div2\times3=4\times3=12$ . Likewise, $20-8+3=12+3=15$ . An abbreviation such as BIDMAS must not make you divide before every multiplication, or add before every subtraction.

Worked example 5: a bracket before division

Question. Work out $24\div(8-2)\times3$ .

Plan. Complete the bracket, then divide and multiply from left to right.

\begin{aligned} 24\div(8-2)\times3&=24\div6\times3 && \text{bracket first}\\ &=4\times3 && \text{leftmost division}\\ &=12 && \text{then multiply} \end{aligned}

Answer. $12$ .

Check. Undo the last multiplication: $12\div3=4$ . Multiplying this by the bracket value $6$ recovers $24$ .

Replace the $2$ being subtracted with the whole group $(2+2)$ . That extra inner bracket must be completed before the subtraction.

Worked example 6: nested brackets

Question. Work out $24\div[8-(2+2)]\times3$ .

Plan. Move from the innermost bracket outwards, then finish left to right.

\begin{aligned} 24\div[8-(2+2)]\times3&=24\div[8-4]\times3 && \text{inner bracket}\\ &=24\div4\times3 && \text{outer bracket}\\ &=6\times3 && \text{division before later product}\\ &=18 && \text{multiply} \end{aligned}

Answer. $18$ .

Check. The divisor is now $4$ , smaller than the previous $6$ , so the result should increase. Also, $18\div3\times4=24$ .

When a power occurs, evaluate it before an outside addition or subtraction: $2+3^2=2+9=11$ . Brackets can change that: $(2+3)^2=5^2=25$ .

Watch out

$24\div6\times3=24\div18$ is wrong. There is no bracket around $6\times3$ , so divide first as you read left to right: $4\times3=12$ .

Negative numbers and powers

Brackets tell you whether a minus sign belongs to the base of a power. In $(-5)^2$ , the entire negative number is squared. In $-5^2$ , the square applies only to $5$ ; the minus sign remains outside.

Writing the factors makes the difference visible:

(-5)^2=(-5)\times(-5)=25, \qquad -5^2=-(5\times5)=-25.

The principal root follows the same sign care: $\sqrt{25}=5$ , even when $25$ came from squaring $-5$ .

Worked example 7: a minus sign outside the square

Question. Work out $-5^2+2$ .

Plan. Square $5$ , apply the outside minus, then add $2$ .

\begin{aligned} -5^2+2&=-(5\times5)+2 && \text{power before addition}\\ &=-25+2 && \text{minus stays outside}\\ &=-23 && \text{move two right} \end{aligned}

Answer. $-23$ .

Check. Undo the addition: $-23-2=-25$ , the negative of the square $25$ .

Now add brackets around the negative base. Only the reach of the square changes.

Worked example 8: squaring the negative base

Question. Work out $(-5)^2+2$ .

Plan. Multiply two copies of $-5$ , then add $2$ .

\begin{aligned} (-5)^2+2&=(-5)\times(-5)+2 && \text{the base includes the minus}\\ &=25+2 && \text{two negative factors}\\ &=27 && \text{add last} \end{aligned}

Answer. $27$ .

Check. $27-2=25$ , the non-negative square of $-5$ . This is $50$ above the previous result because the square term changed from $-25$ to $25$ .

With an odd power, a negative base stays negative: $(-2)^3=(-2)\times(-2)\times(-2)=4\times(-2)=-8$ . Here $-2^3$ also equals $-8$ , but that agreement does not extend to even powers. It also shows why $\sqrt[3]{-8}=-2$ : a cube root can be negative.

Return to the freezer. Five forward hours give $5\times(-3)=-15\,^{\circ}\mathrm{C}$ . Looking five hours backwards reverses the falls, so the signed change from now to then is $(-5)\times(-3)=15\,^{\circ}\mathrm{C}$ . These are changes, not the actual temperatures; an actual temperature also needs a starting value.

Watch out

$-5^2=(-5)^2$ is false. The values are $-25$ and $25$ . If the number being squared is negative, include its sign inside brackets.

Quick Recap

Two non-zero factors or a division with matching signs give a positive result; different signs give a negative result.
Multiplication by zero gives zero. Division by zero is undefined.
Brackets come first, then powers/roots, then multiplication/division left to right, then addition/subtraction left to right.
$(-5)^2$ squares a negative base; $-5^2$ takes the negative of a square.

The opening freezer changes by $-15\,^{\circ}\mathrm{C}$ five hours forwards and $+15\,^{\circ}\mathrm{C}$ five hours backwards.

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C1.6; E1.6 - Signed products and order of operations

Multiplying signed numbers

Dividing signed numbers

Brackets and the order of operations

Negative numbers and powers

Quick Recap

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