Factorising is the "reverse" of expanding brackets.
Usually, the first step of 'factorising' is ‘taking out’/extracting the common factors
Factorising Linear Algebraic Expression
Extracting Common Factor/s
Rule: ab + ac = a(b+ c)
Example
Factorise a2 + 3a
= a(a + 3). (Step : Look for Common factors -> a, a2 = a x a)
Example
Simplify 3a + 6b
= 3(a + 2b) (Step : Look for common factor -> 3 for 3 and 6)
Example:
Simplify 2jk + 6j
= 2(jk + 3j) (Step : Look for common factor -> 2 for 2 and 6)
= 2j(k + 3) (Step : Look for common factor -> j)
Example
Factorise 5y2 + 10
= 5y2 + 10y (Step 1 : Look for Common factors -> 5)
= 5(y2 + 2y)
= 5y (y + 2) (Step 2 : Repeat looking for common factor -> y)
Factorisation of a Quadratic Expression x 2 +b x + c
To factorise a quadratic expression => 'putting' it into the format
x2 + bx + c = (x + P)(x + Q)
for x2 quadratic expression, and P and Q are numbers
Let's first expand (x + P)(x + Q) using the rainbow method
(x + P) (x + Q) = x x x + Q x x + P x x + P x Q
= x + (P + Q)x + PQ
=> x2 +bx + c = x2 + (P + Q)x + PQ
=> P + Q = b
=> PQ = c
There are a few methods to factorise x2+ b x + c.
The Cross Method X
Example
Factorise x2 + 3x + 2
=> x2 + 3x + 2 = (x + P)(x + Q)
=> P + Q = 3
P x Q = 2
Draw The cross X:
A P
\ /
/ \
B Q
What are A, B,?
(1). A and B = Product of x2 => A=x, B = x
=> x P
\ /
/ \
x Q
(2a) 3x = (P x x) + (Q x x) = (P + Q)x
(2b) 2 = P x Q
=> The factor of 2
2 = 1 x 2
Since there is only 1 set of factor => P = 1, Q = 2
=> x 1
\ /
/ \
x 2
Using the X to cross-multiply to confirm,
x 1
\ /
/ \
x 2
x x 1 = x, 2 x x = 2x
=> x + 2 x = 3 x
(3) Then the factorisation of x2 + 3x + 2 is as bracket:
=> (x +1)
\ /
/ \
( x +2)
= (x + 1) (x + 2)
Thus, x2 + 3x + 2 = (x + 1) (x + 2)
Note: P = 2 and Q = 1 is also correct.
Example (x 2 + bx + c)
Factorise x 2 + 5x + 6
=> To find P and Q where x 2 + bx + c = x 2 + 5x + 6 = (x + P)(x + Q)
=> b = 5 = P + Q, c = 6 = PQ
Step 1 : Draw the cross X
<< Since it is x2 , fill in x2 >>
x [What is P?]
\ /
/ \
x [What is Q?]
Step 2 : Find factors of c (P X Q) for b (P + Q)
When there is more than 1 set of factors,
Find the factor of c = 6 (P x Q)
6 = 1 x 6,
= 2 x 3
Using the X to cross-multiply,
x 2 x 1
\ / \ /
/ \ / \
x 3 x 6
x x 3 = 3 x, 2 x x = 2 x x x 1 = 3 x, x x 6= 2 x
2 x + 3 x = 5 x x + 6 x = 7 x
P = 2 , Q = 3
Step 3 : Factorise ( )
=> ( x +2 )
\ /
/ \
( x +3)
Thus,
x 2 + 5x + 6 = (x + 2) (x + 3)
Example (x 2 - bx + c)
Factorise x 2 - 5x + 6
<< Since it is x 2 , fill in x 2 >>
x [What is P?] [Step 1 : Draw the cross X]
\ /
/ \
x [What is Q?]
P + Q = -5. << it is -5 >>. [Step 2 : Find factors of c (P X Q) for b (P + Q)]
Factor of 6 = 1 x 6
= 2 x 3
For +6 (positive 6) to get a negative sum (P + Q),
=> -ve x -ve = +ve
6 = -1 x -6,
= -2 x -3
Using the X to cross-multiply,
x -2 x -1
\ / \ /
/ \ / \
x -3 x -6
x x 3 = -3x, -2 x x = -2x x x -1 = - x, x x -6= -6x
-2 x + (-3 x) = -5 x - x + (-6 x) = -7x
=> P = -2 , Q = -3
Step 3 : Factorise ( )
=> ( x -2 )
\ /
/ \
( x -3 )
Thus,
x 2 - 5x + 6 = (x - 2) (x - 3)
Example (x 2 - bx - c)
Factorise x2 - 5x - 6
<< Since it is x 2 , fill in x 2 >>
x [What is P?] [Step 1 : Draw the cross X]
\ /
/ \
x [What is Q?]
P + Q = -5. [Step 2 : Find factors of c (P X Q) for b = -6 (P + Q)]
With a -6 (negative 6), and to obtain a -5 (negative 5),
=> -ve x +ve = -ve
The factor for 6 = 1 x 6
= 2 x 3.
We can "quickly" filter the probable factor by:
=> -6 = -2 x 3 (=> -2 + 3 = 1)
= 2 x -3 ( 2 + (-3) = -1)
= 1 x -6 ( 1 + (-6) = -5) <=
= -1 x 6 ( -1 + 6) = 5
Using the X to cross-multiply to confirm,
x 1
\ /
/ \
x -6
x x 1 = x, -6 x x = -6x
x + (-6 x) = -5 x
=> P = 1 , Q = -6
Step 3 : Factorise
=> ( x +1 )
\ /
/ \
( x -6 )
Thus,
x 2 - 5x - 6 = (x + 1) (x - 6)
Example (x 2 + bx - c)
Factorise x 2 + 5x - 6
<< Since it is x 2 , fill in x 2 >>
x [What is P?] [Step 1 : Draw the cross X]
\ /
/ \
x [What is Q?]
P + Q = 5. [Step 2 : Find factors of c (P X Q) for b = -6 (P + Q)]
With a -6 (negative 6) => -ve x +ve = -ve
The factor for 6 = 1 x 6
= 2 x 3.
We can "quickly" filter the probable factor by:
=>. -6 = -2 x 3 (=> -2 + 3 = 1)
= 2 x -3 ( 2 + (-3) = -1)
= 1 x -6 ( 1 + (-6) = -5)
= -1 x 6 ( -1 + 6 = 5 ) <=
Using the X to cross-multiply to confirm,
x -1
\ /
/ \
x +6
x x -1 = -x, 6 x x = 6x
-x + 6x = -5 x
=> P = -1 , Q = 6
Step 3 : Factorise ( )
=> ( x -1 )
\ /
/ \
( x 6 )
Thus,
x 2 + 5x - 6 = (x - 1) (x + 6)
Addition and Subtraction of Quadratic Expressions
Number Operations (+ and -) can be performed on Quadratic Expressions.
Example
Simplify (2x2 + 2) + (3x2 + x + 1)
2x2 + 2 + 3x2 + x + 1 (Step 1 : Open bracket)
= 2x2 + 3x2 + x + 1 + 2 (Step2: Group x2, x , numbers)
= 5x2 + x + 3
—— Level 2 More challenging Question —-
Example
Simplify (2x2– 3x + 8) + (-3x2– x -3)
2x2– 3x + 8 – 3x2– x - 3 (Step 1 : Open bracket)
= 2x2– 3x2– 3x – x + 8 – 3 (Step2: Group x2, x , numbers)
= -x2– 4x + 5
Practice
1. Factorise 6x + 12y (S17/I/7/1)
2. Factorise x2 - 5x + 6 (S17/I/7/2)
3. Factorise x2 - 3x - 4
4. Factorise a2 - 3a
5. Factorise x2 + 5x + 4
