3.04 Expansion
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(x−12).
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(B−C)=AB−AC. Notice that in this case we are assuming $C$C is positive, but we are taking away $AC$AC.
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(B−C)=AB−AC
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(c−5).