FACTORISATION AND FACTORISING EXPRESSIONS
There are five topics discussed in this post;
(i) Factorizing expressions with a common factor bracket
(ii) Switch around
(iii) Factorizing by grouping
(iv) Difference of squares
(v) Trinomials with coefficients
FACTORISING EXPRESSION WITH A COMMON FACTOR BRACKET
Sometimes, there is a bracket in the expression that is common to all the terms. This common bracket is the common factor. Consider the following expression:
2x(x + 1) + y(x + 1) {do you see that (x + 1) is common}
We can look at it in another way.
● Let us replace the bracket (x + 1) with P:
● Factorise the expression by taking P out as the common factor:
= P(2x + y)
● Replace the bracket:
= (x +1)(2x + y)
We can only take out a bracket as a common factoe if the brackets are exactly the same. For example, if the above expression were the 2x(1 + x) + y(x + 1), (x +1) is still a common bracket because the brackets are the same.
If the expression were 2x(1 - x) + y(x + 1), it is not possible to factorize it.
There are more to factorization than replacing brackets. Let's consider this;
SWITCH AROUND
Consider the following expressions:
3x(x - 3) + 2(3 - x)
● Notice that the brackets look very similar, but they are not identical. As a result, it cannot be factorized. Can you think of a way that we can change the order of the second bracket to match the first bracket?
To change the second bracket to match the first one, we multiply the second bracket by -1:
= 3x(x -3) - 2(-3 + x)
That is, we multiplied +2(3 - x) with -1 which gave us -2(-3 + x)
This is called a switch around. We perform this operation whenever we want to change brackets to factorise expression
● The switch around method changes the signs of the terms inside the bracket, as well as the sign of the term outside the bracket. Putting the term in the second bracket in the right order gives us;
= 3x(x -3) - 2(x -3)
● The brackets are now identical, and we can take out (x -3) as a common factor:
= (3x - 2)(x - 3)
EXAMPLE
Factorize the following expression completely: 2x(a + 2b) - (a + 2b)
Solution
(a + 2b) is a common factor bracket. You can replace the bracket by a single variable to make it easier to factorize:Let's
let's replace (a + 2b) with P
The expression will now be:
= 2xP - P
= P(2x - 1)
Now replace (a + 2b) with P
= (a + 2b)(2x - 1)
EXAMPLES ON SWITCH AROUND METHOD
(a) Factorize the following expression completely: p(m - 2) - (2 - m)
solution
In order to take out a common factor, we need to make the brackets look exactly the same. So, we perform switch around on the second bracket:
= p(m - 2) + (-2 + m)
= p(m - 2) + (m - 2)
= (p + 1)(m - 2)
Notice that the sign in front of the second bracket changed to +. That is because the signs in the second term was multiplied by -1.
The switch around can be performed on the first bracket:
= -p(-m + 2) - (2 - m)
= -p(2 - m) - (2 - m)
= (-p - 1)(2 - m)
FACTORISE BY GROUPING
Sometimes, we have a polynomial where there are no common factors to any of the terms in the expression. However, we can break up the expression into smaller groups that can be factorised.
EXAMPLE 1
Factorise the following expression:
ax + ay + 3x + 3y
solution
We divide ax + ay + 3x + 3y into two pairs:
(ax + ay)(3x + 3y)
In (ax + ay)(3x + 3y), "a" is the common factor of the first pair and 3 is the common factor of the second pair.
We then remove the common factors:
= a(x + y) + 3(x + y)
We notice the common factor in both is x + y
a(x + y) + 3(x + y)
= (x + y)(a + 3)
EXAMPLE 2
Factorise the following expression: 6ax +2bx + 3ay + by
solution
The first pair of terms with a common factor is 6ax + 2bx and the second pair with a common factor is 3ay + by.
So, lets group them;
(6ax + 2bx) + (3ay + by)
= 2x(3a + b) + y(3a + b)
= (3a + b)(2x + y)
DIFFERENCE OF SQUARES
Recall that:
(a + b)(a - b)----{a2-ab +ab - b2}
= (a2 - b2)
The result is the difference of squares. This means that both terms of the expression are squares, and we are subtracting them.
Remember that factorising means to work backwards. So, the factors of a2 - b2 are;
(a + b)(a - b).
EXAMPLE 1
Factorise 9x2 - 4
solution
a2 - b2 = (a + b)(a - b)
Here, a2 is the term 9x2 = (3x)2
Here, b2 is the term 4 = 22
So, a = 3x
So, b = 2
Replace a with 3x and b with 2;
(3x - 2)(3x + 2)
EXAMPLE 2
Factorise t2 - (u - v)2
solution
Here, a2 = t2
Here, b2 = (u - v)
we replace a with "t"
we also replace b with "(u - v)"
= [t - (u - v)][t + (u - v)]
Open the inner brackets
(t - u + v)(t + u - v)
EXAMPLE 3
Factorise 18p3 - 2pq2
solution
Look closely, you notice p is common in both terms; 2 is also common.
18p3 - 2pq2
= 2p(9p2 - q2)
The factor (9p2 - q2) is a difference of two squares that can be factorised.
Here a2 = 9p2
Here b2 = q2
We replace a with 3p
We replace b2 with q
= (3p - q)(3p + q)
Remember the common term "2p"
2p(3p - q)(3p +q)
Can you solve these?
(i) 4 - 25b2
(ii) 1212 - 1202
TRINOMIALS WITH COEFFICIENTS
EXAMPLE
(i) x2 + 3x + 2
(ii) x2 + 10x + 24
solution
(i) x2 + 3x + 2
Here, it's quite different from what we've done. It's difficult to factorise three terms. The way we can do that is to split it into four terms.
x2 + 3x + 2
Let's start with giving each term a representative
a b c
x2 + 3x + 2 ---(1)
Now, multiply the coefficient of a and c together
In this case, coefficient of a = +1 while c = +2
a x c
= 1 x 2 ----(2)
= 2
The next step is to find the factors of 2 i.e all numbers we can multiply together to give us 2.
+1 x +2 ----(3)
Basically, the only number we can multiply together to give us 2 is 1 x 2
The next step is to manipulate by adding signs to 1 x 2 to give the coefficient of b. Which is +3
That will be +1 x +2 ----(4)
Now that we have +1 and +2, adding them together gives us +3 (coefficient of b)
The next step is to replace +3 with +1+2
x2 +1x + 2x +2 ----(5)
Notice the above expression, 3x was substituted with +1x + 2x. What is left is to factorise.
[x2 + 1x + 2x + 2]
x(x +1) +2(x + 1)
= (x + 1)(x + 2)
(ii) x2 + 10x + 24
a b c
x2 + 10x + 24
a x c
1 x 24
= 24
Factors of 24
1 x 24
2 x 12
6 x 4
8 x 3
Here, there are multiple factors. We choose the best factor to manipulate to give us the coefficient of b which is +10
6 x 4
Manipulate the factors
+6 +4
= +10
Substitute +10 with +6 +4
x2 + 6x + 4x + 24
x(x + 6) + 4(x + 6)
= (x + 6)(x + 4)
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