Normally, when an expression has a multiplication and an addition or subtraction, for example 5+8\times9, we evaluate the multiplication first. The exception is when the addition or subtraction is in brackets, for example, (5+8)\times9.

It will help to visualise a rectangle with a height of 9 cm and a width of 5+8 cm.

A rectangle with a height of 9 centimetres and a length of 5 plus 8 centimetres. Ask your teacher for more information.

The rectangle has an area of \left(5+8\right)\times9 \text{ cm}^{2}. We can work this out as follows.

\displaystyle \left(5+8\right)\times9	\displaystyle =	\displaystyle 13\times9	Evaluate the addition in the brackets
	\displaystyle =	\displaystyle 117 \text{ cm}^2	Evaluate the multiplication

However, we can see that the rectangle is made up of two smaller rectangles, one with area 5\times 9 \text{ cm}^2 and the other with area 8\times 9 \text{ cm}^2. So we can also work out the total area like this.

\displaystyle 5\times 9+8\times 9	\displaystyle =	\displaystyle 45+72	Evaluate the multiplication
	\displaystyle =	\displaystyle 117 \text{ cm}^2	Evaluate the addition

So \left(5+8\right)\times 9=5\times 9+8\times 9. This can be extended to any other numbers.

If A, B, and C are any numbers then A\left(B+C\right)=AB+AC. This is known as the distributive law.

The distributive law is particularly useful for algebraic expressions where we can't evaluate the expression in the brackets.

Exploration

The following applet explores the distributive law using algebra tiles.

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The total number of +x tiles corresponds to the coefficient of the variable and the number of +1 tiles corresponds to the constant after multiplication.

We can use the distributive law to expand an algebraic expression brackets like so: \\ A\left(B+C\right)=AB+AC and if the second term in the brackets is negative: A\left(B-C\right)=AB-AC where A,B and C are any numbers.

Examples

Example 1

Expand the expression 4\left(n-2\right).

Worked Solution

Create a strategy

Use the distributive law A\left(B-C\right)=AB-AC to expand the expression.

Apply the idea

\displaystyle 4\left(n-2\right)	\displaystyle =	\displaystyle 4\times n - 4\times 2	Expand the expression
	\displaystyle =	\displaystyle 4n-8	Simplify

Reflect and check

Because of the distributive law we know that both sides of the equation are equal. But now we have a way to write an equal expression without brackets.

We had to be careful of the negative sign here. Because A is positive and C is negative, AC is negative. We could also use a slightly different version of the rule that accounts for the negative sign: A\left(B-C\right)=AB-AC. Notice that in this case we are assuming C is positive, but we are taking away AC.

Example 2

Expand the expression 6s\left(9t+8\right).

Worked Solution

Create a strategy

Use the distributive law A\left(B+C\right)=AB+AC to expand the expression.

Apply the idea

\displaystyle 6s\left(9t+8\right)	\displaystyle =	\displaystyle 6s\times9t + 6s\times 8	Expand the expression
	\displaystyle =	\displaystyle 54st + 48s	Simplify

Example 3

Expand and simplify the expression: x\left(x+6\right) + 8\left(x+7\right).

Worked Solution

Create a strategy

Use the distributive law A\left(B+C\right)=AB+AC to expand the expression.

Apply the idea

\displaystyle x\left(x+6\right) + 8\left(x+7\right)	\displaystyle =	\displaystyle x\times x +x\times 6 +8 \times x+8 \times 7	Expand both brackets
	\displaystyle =	\displaystyle x^{2} +6x +8x+56	Simplify each term
	\displaystyle =	\displaystyle x^{2}+14x+56	Add like terms

Idea summary

We can use the distributive law to expand an algebraic expression brackets like so:A\left(B+C\right)=AB+AC

and if the second term in the brackets is negative:A\left(B-C\right)=AB-AC

where A,B and C are any numbers.

Outcomes

MA5.2-6NA

simplifies algebraic fractions, and expands and factorises quadratic expressions

2.02 The distributive law

Ideas

Distributive law

Exploration

Examples

Example 1

Example 2

Example 3

Outcomes

MA5.2-6NA

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