 Middle Years

# 2.04 Expansion

Lesson

Normally, when an expression has a multiplication and an addition or subtraction, for example $5+8\times9$5+8×9, we evaluate the multiplication first. The exception is when the addition or subtraction is in brackets, for example, $\left(5+8\right)\times9$(5+8)×9.

It will help to visualise a rectangle with a height of $9$9 cm and a width of $5+8$5+8 cm. The rectangle has an area of $\left(5+8\right)\times9$(5+8)×9 cm2. We can work this out as follows.

 $\left(5+8\right)\times9$(5+8)×9 $=$= $13\times9$13×9 Evaluate the addition in the brackets first $=$= $117$117 cm2 Evaluate the multiplication

However, we can see that the rectangle is made up of two smaller rectangles, one with area $5\times9$5×9 cm2 and the other with area $8\times9$8×9 cm2. So we can also work out the total area like this.

 $5\times9+8\times9$5×9+8×9 $=$= $45+72$45+72 Evaluate the multiplications $=$= $117$117 cm2 Evaluate the addition

So $\left(5+8\right)\times9=5\times9+8\times9$(5+8)×9=5×9+8×9. This can be extended to any other numbers.

If $A,B$A,B, and $C$C are any numbers then $A\left(B+C\right)=AB+AC$A(B+C)=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.

#### Worked Example

Expand $7\left(x-12\right)$7(x12).

Think: Expand means to write an algebraic expression without brackets. We can expand this expression using the distributive law.

Do:

 $7\left(x-12\right)$7(x−12) $=$= $7\times x+7\times\left(-12\right)$7×x+7×(−12) Use the distributive law, $A\left(B+C\right)=AB+AC$A(B+C)=AB+AC. Here, $A=7,B=x,$A=7,B=x, and $C=-12$C=−12 $=$= $7x-84$7x−84 Evaluate the multiplication

Reflect: 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 $A$A is positive and $C$C is negative, $AC$AC is negative.
To solve the previous example we could also use a slightly different version of the rule that accounts for the negative sign: $A\left(B-C\right)=AB-AC$A(BC)=ABAC. Notice that in this case we are assuming $C$C is positive, but we are taking away $AC$AC.

Summary

We can use the distributive law to expand an algebraic expression brackets like so:

$A\left(B+C\right)=AB+AC$A(B+C)=AB+AC,

and if the second term in the brackets is negative:

$A\left(B-C\right)=AB-AC$A(BC)=ABAC

where $A,B$A,B, and $C$C are any numbers.

#### Practice questions

##### Question 1

Expand the expression $9\left(5+w\right)$9(5+w).

##### Question 2

Expand the expression $\left(y+8\right)\times7$(y+8)×7.

##### Question 3

Expand the expression $-8\left(c-5\right)$8(c5).