1.02 Expansion and factorisation
Expand brackets
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
example 1
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).
Factorise
The distributive law says that for any numbers $A,B$A,B, and $C$C, $A\left(B+C\right)=AB+AC$A(B+C)=AB+AC. We saw how to use this rule to expand algebraic terms, but we can also use the rule in reverse.
The reverse of expanding algebraic expressions is called factorising. Factorising an algebraic expression means writing the expression with any common factors between the terms taken outside of the brackets.
Worked examples
Example 2
Fully factorise $12x+20y$12x+20y.
Think: Before we can factorise the expression, we should look for the factors of each term. $12x$12x has the factors $12$12 and $x$x and $20y$20y has the factors $20$20 and $y$y.
We should check for any common factors in the coefficients. $12=4\times3$12=4×3 and $20=4\times5$20=4×5, so the highest common factor of the coefficients is $4$4.
Do:
| $12x+20y$12x+20y | $=$= | $4\times3x+4\times5y$4×3x+4×5y |
Write the coefficients as a product of the highest common factor. |
| $=$= | $4\left(3x+5y\right)$4(3x+5y) |
Factorise using the reverse of the distributive law, $AB+AC=A\left(B+C\right)$AB+AC=A(B+C). Here, $A=4,B=3x,$A=4,B=3x, and $C=5y$C=5y. |
Reflect: It is helpful to find the highest common factor of the coefficients before we factorise the expression. Otherwise we might end up having to factorise a second time. For example, if we factorised $2$2 instead, we would have $2\left(6x+10y\right)$2(6x+10y). Since $6$6 and $10$10 have a common factor of $2$2 we would still need to factorise this again.
Example 3
Fully factorise $-3xz-5yz$−3xz−5yz.
Think: In this case, there is no highest common factor of the coefficients. However, we can still factorise this expression.
First, notice that both terms have a pronumeral factor of $z$z. This means that we can factorise $z$z.
Also, both terms are negative. Being negative is the same as having a factor of $-1$−1. So we can also factorise $-1$−1.
Do:
| $-3xz-5yz$−3xz−5yz | $=$= | $-1\times3xz+\left(-1\right)\times5yz$−1×3xz+(−1)×5yz |
Write the terms as products of $-1$−1. Note that the sign of the terms changes when we do this. |
| $=$= | $-1\times z\times3x+\left(-1\right)\times z\times5y$−1×z×3x+(−1)×z×5y |
Write the terms as products of $z$z. |
|
| $=$= | $-1\times z\left(3x+5y\right)$−1×z(3x+5y) |
Factorise using the law, $AB+AC=A\left(B+C\right)$AB+AC=A(B+C). Here, $A=-1\times z,B=3x,$A=−1×z,B=3x, and $C=5y$C=5y. |
|
| $=$= | $-z\left(3x+5y\right)$−z(3x+5y) |
Simplify the factor $-1\times z$−1×z. |
Reflect: It is important to be careful about factorising $-1$−1, because the sign of each term will change.
We can use the reverse of distributive law to factorise an algebraic expression like so.
$AB+AC=A\left(B+C\right)$AB+AC=A(B+C)
This means means writing the expression with any common factors between the terms taken outside of the brackets. Factorising is the reverse of expanding.
Practice questions
Question 4
Factorise $3w+15$3w+15.
Question 5
Factorise $40c-35$40c−35.
Question 6
Factorise $10g-35h+15$10g−35h+15.