Expanding and Factorising Algebraic expressions
What is an expansion?
The process of re writing an algebraic expression that contains brackets by expanding them is called an expansion.
To expand a bracket means to multiply each term in the bracket by the expression outside the bracket.
Be careful!
Watch out for negative numbers in front of brackets because they always require extra care. Remember that: + x + = + , + x – = – , – x – = +
To expand expressions that involve multiplication, you need to follow the rules of the distributive property which state that any number can be multiplied by any number. So numbers can be multiplied by a variable, by another number or by itself. When we expand terms by distribution we need to combine like terms to simplify. Like terms are numbers from the same group (1,0,5 or 56) or they share the same variable and exponent ( 2x and 3x)
Example 1: Expand and simplify the following expressions.
a) $\displaystyle -5(x+3)$
b) $\displaystyle 7(x+2x)$
c) $\displaystyle 4(y-7)-5(3y+5)$
d) $\displaystyle 8(p+4)-10(9p-6)$
Solution
a) $ \displaystyle -5(x+3)$
Using BODMAS we first start with the brackets. Since the numbers inside the brackets are not like terms we can´t add them so we multiply the numbers on the outside of the bracket by everything inside and that the negative sign is attached to the 5, because $\displaystyle -5\times x\text{ }=-5x$ and $\displaystyle -5\times 3=-15$.
$ \displaystyle -5(x+3)=-5x-15$
b) $\displaystyle 7(x+2x)$
Using BODMAS we see that the numbers in the brackets are like terms so we combine them firsts and the we multiply by number 7.
$\displaystyle 7(x+2x)=7\times 3x=21x$
c) $\displaystyle 4(y-7)-5(3y+5)$
Expand each bracket first and remember that the “-5“ must keep the negative sign when it is multiplied through the second bracket.
$\displaystyle 4(y-7)=4y-28$
$\displaystyle -5(3y+5)=-15y-25$
Collect like terms and simplify.
$\displaystyle 4(y-7)-5(3y+5)$
$\displaystyle =4y-28-15y-25$
$\displaystyle =-11y-53$
d) $\displaystyle 8(p+4)-10(9p-6)$
It is important to note that when you expand the second bracket `-10` will need to be multiplied by `-6`, giving a positive result for that term.
$ \displaystyle 8(p+4)-10(9p-6)=$
$ \displaystyle -10(9p-6)=-90p+60$
Collect like terms and simplify.
$\displaystyle 8(p+4)-10(9p-6)$
$\displaystyle =8p+32-90p+60$
$\displaystyle -82p+92$
Expansion of two sets of brackets
To expand two sets of brackets you need to multiply each term in the first bracket with each term of the second bracket. Then you need to combine the like terms and we careful with the signs.
a) $\displaystyle (x+3)(x-2)$
b) $ \displaystyle 2(x+1)(x+8)$
Solution
a) $\displaystyle (x+3)(x-2)=$
Multiply each term in the first bracket with each term of the second bracket.
$ \displaystyle x\cdot x+x\cdot (-2)+3\cdot x+3\cdot (-2)=$
Collect like terms and simplify.
$\displaystyle {{x}^{2}}-2x+3x-6=$
$\displaystyle {{x}^{2}}+x-6$
b) $ \displaystyle 2(x+1)(x+8)=$
We can multiply the first bracket with the number outside the first bracket or we can expansion the two brackets and then multiply each element with the number 2.
Expansion of two brackets by multiplying each term of the firsts bracket with each term of the second bracket.
$\displaystyle 2(x\cdot x+8x+1x+8\cdot 1)=$.
Collect like terms and then multiply with the number 2.
$\displaystyle 2({{x}^{2}}+9x+8)=$
$\displaystyle 2{{x}^{2}}+18x+16$
We have seen in detail expanding brackets. But sometimes it can be helpful to carry out the opposite process and put bracket back into an algebraic expression.
What is Factorisation?
Factorisation is the opposite process of expanding brackets. The first step of factorizing an expression is to take out any common factors which the terms have.
We consider the algebraic expression 4x+8. This expression is already simplified but notice that 4 and 8 have a common factor. In fact the HCF of 4 and 8 is 4.
Now, 8 = 4 x 2 and 4 = 4 x 1
So, 4x + 8 = 4 x 1x – 4 x 2
= 4(x + 2)
Notice that the HCF has been taken out of the bracket and written at the front. The terms inside are found by considering what you need to multiply by 4 to get 4x and 8.
As we explained in the definition above this process of writing an algebraic expression using brackets in this way is known as factorization. The expression 4x+8 has been factorized to give 4(x+2).
Example 3: Factorise each of the following expressions as fully as possible.
a)$ \displaystyle 15x+12y$
b)$ \displaystyle 18mn-30m$
c) $ \displaystyle 36{{p}^{2}}q-24p{{q}^{2}}$
d) $\displaystyle 15(x-2)-20{{(x-2)}^{3}}$
Solution
a) $\displaystyle 15x+12y=$
The HCF of 12 and 15 is 3, but x and y have no common factors, because $ \displaystyle 3\times 5x=15x$ and $\displaystyle 3\times 4y=12y$
$ \displaystyle 15x+12y=3(5x+4y)$
b) $\displaystyle 18mn-30m=$
The HCF of 18 and 30=6 and HCF of mn and m is m, because $\displaystyle 6m\times 3n=18mn$ and $\displaystyle 6m\times 5=30m$
$\displaystyle 18mn-30m=6m(3n-5)$
c) $ \displaystyle 36{{p}^{2}}q-24p{{q}^{2}}=$
The HCF of 36 and 24 is 12 and p2q and pq2 have common factor pq, because $\displaystyle 12pq\times
p=36{{p}^{2}}q$ and $\displaystyle 12pq\times -2q=-24p{{q}^{2}}$
$ \displaystyle 36{{p}^{2}}q-24p{{q}^{2}}=12pq(3p-2q)$
d) $\displaystyle 15(x-2)-20{{(x-2)}^{3}}=$
Sometimes the terms can have an expression in brackets that is common to both terms.
The HCF of 15 and 20 is 5 and the HCF of $l\displaystyle (x-2)$ and $ \displaystyle {{(x-2)}^{3}}$ is $ \displaystyle (x-2)$, because $ \displaystyle 5(x-2)\times 3=15(x-2)$ and $\displaystyle 5(x-2)\times 4{{(x-2)}^{2}}=20{{(x-2)}^{3}}$
$\displaystyle 15(x-2)-20{{(x-2)}^{3}}$
$\displaystyle =5(x-2)\left[ {3-4{{{(x-2)}}^{2}}} \right]$
