Factorise ax + bx + kay + by (Group Common Factors)
Example
Factorise x2 – 3x + 2xy – 6y
Step 1 : Look for "paired" common factors
x2 – 3x + 2xy – 6y
Step 2: Extract Common Factors
= x(x - 3) + 2yx - 6y
Step3: Repeat for other common factors (if any)
= x(x - 3) + 2y (x – 3) (Extract y as common factor)
Example
Factorise 2yz + 8z + 3xy + 12x
2yz + 8z + 3xy + 12x (Step 1 : Look for "paired" common factors)
= 2z(y + 4) + 3x(y + 4) (Step 2 : Extract common factor 2z , 3x)
= (y+4)(2z + 3x) : Extract common factor (y + 4)
Recap - Use of:
1. (a + b)2 = (a + b) x (a + b) = a x a + a x b + a x b + b x b = a2+ 2ab + b2
=> (a + b)2 = a2 + 2ab + b2 = (b + a)2
2. (a - b)2 = (a - b) x (a - b) = a x a + a x (-b) + (-b) x a + (-b x -b)
= a2 -ab -ab + b2 = a2 - 2ab + b2
=> (a - b)2 = a2 - 2ab + b2
3. a2 – b2 = (a + b) x (a - b) = a2 -ab + ab + b2 = a2 - b2
=> a2 – b2 = (a + b) x (a - b)
To remember:
(a + b)2 = a2 + 2ab + b2 = (b + a)2
(a - b)2 = a2 - 2ab + b2
a2 – b2 = (a + b) x (a - b)
22 = 4 32 = 9 42 = 16 52 = 25 62 = 36
72 = 49 82 = 64 92 = 81 102 = 100 112 = 121
Factorisation of ax2 +bx + c
There are a few methods to factorise ax2+ bx + c
Example: Factorise 2x2+ 5x + 3 => ax2 +bx + c
Step 1: Draw The cross X and list a, b, c
a=2, b = 5, c = 2
[J]x [P]
\ /
/ \
[K]x [Q]
Step 2: Write Factor of a , J and K
a=2 = 1 x 2
2 x [What is P?]
\ /
/ \
x [What is Q?]
2b: Write factor of c
c = 1 x 3
Using the X to cross-multiply to find b ,
2x 1 2x 3
\ / \ /
/ \ / \
x 3 x 1
2x x 3 = 6x, x x 1= x 2x x1 = 2x, x x 3= 3x
6x + x = 7x 2x + 3x = 5x. <=
(3) Then the factorisation of 2x2 + 5x + 3 is as bracket:
=> (2x +3)
\ /
/ \
( x +1)
= (2x + 3) (x + 2)
Thus, 2x 2 + 5x + 3 = (2x + 3) (x + 2)
Example
The + Method
Example: Factorise 2x2+ 5x + 3 => ax2 +bx + c
Step 1 : Draw + and fill put in the expression
[J] x [P] |
[K] x [Q] |
—————————-
x 2 + 6 | 5 x
|
We need to find the values of [J], [K], [P], [Q]
J x K are factors of 2(a), and P x Q are factors or 3(b)
Step 2 : Find in the values
2 = 1 x 2 (J, K) b = 3 = 1 x 3
1x 1 | x x 3 | 3x
2x 3 | 6x 2x 1 | 2x
—————————- ——————
2x 2 + 3 | 7 x x 2 + 6 | 5 x <===
Step 3 Factorise
( x +3) | 3x
(2x +1) | 2x
—————————-
x 2 + 3 | 5x
2x 2 + 5x + 3 = ( x + 3) (x + 2)
Example: Factorise 2x2+ 5x– 3
Factorise 2x2+ 5x - 3 => ax2 +bx + c
Step 1 : Draw The cross X and list a, b, c
a=2, b = 5, c = -3
[J]x [P]
\ /
/ \
[K]x [Q]
Step 2 : Write Factor of a , J and K
a=2 = 1 x 2
2 x [What is P?]
\ /
/ \
x [What is Q?]
2b: Write factor of c
c = 1 x 3
Since c is negative, => c = 1 x -3 or -1 x 3
Using the X to cross-multiply to find b ,
2x 1 2x -1
\ / \ /
/ \ / \
x -3 x 3
2x x 3 = -6x, x x 1= x 2x x -3 = 6, x -1 = -x
-6x + x = -5x 6x - x = 5x <==
Step 3 : The factorisation of 2x2 + 5x + 3 is as bracket:
=> (2x -1)
\ /
/ \
( x 3)
= (2x - 3) (x + 3)
Thus, 2x 2 + 5x - 3 = (2x - 1) (x + 3)
Example
The + Method
Factorise x 2 + 5x + 6 => ax 2 + bx + c
Step 1 : Draw + and fill put in the expression
[J] x [P] |
[K] x [Q] |
—————————-
x 2 + 6 | 5 x
|
We need to find the values of [J], [K], [P], [Q]
J x K are factors of 1(a), and P x Q are factors or 6(b)
Step 2 : Find in the values
[1] x 1 | 1 x x [1] x 3 | 3 x x
[1] x 6 | 6 x x [1] x 2 | 2 x x
—————————- —————————-
x 2 + 6 | 7 x x 2 + 6 | 5 x <===
a= 1 x 1 (J, K) b = 6 = 1 x 6
= 3 x 2
Step 3 : Factorise
( x +3) | 3x
(x +2) | 2x
—————————-
x 2 + 6 | 5x
x 2 + 5x + 6 = ( x + 3) (x + 2)
Practice
1. Factorise x2 - 6x + 9 [2010/P1/7/1]
2. Factorise 4x2 - 9
3. Factorise 3ax - 6ay + bx -2by [2010/P2/8a/3]
4. Factorise 6x2 - 12y - 3 [11/P1/18a/1]
5. Factorise 2x2 - 7x + 3 [11/P2/6/2]
Solve for 2x2 - 7x + 3 = 0
