Leaving Cert Factors
FACTORSType 1 Common Factor (Grouping)
2Example 1. Factorise 3 3x x xy y
2 3 3 Group like termsx x xy y 2 ( 3 ) ( 3 )x x xy y
( 3) ( 3)x x y x
( 3)( )x x y
2 2 = ( )( )x y x y x y 2Example 1. Factorise 16 25x
216 25x 2 2(4 ) (5)x
(4 5)(4 5)x x
2Example 2. Factorise 1 16a
21 16a2 2(1) (4 )a
(1 4 )(1 4 )a a
Difference of 2 squares
Factorise the following
6x + 24y 5ab+ 15bc 7x² - 28x
6( x + 4y) 5b( a + 3c) 7x( x – 4)
4x² -6xy +8xz 5xy² - 20x²y 2a²b -4ab² +12abc
2x( 2x -3y + 4z) 5xy( y – 4x) 2ab( a – 2b+ 6c)
Factorising
x² + 4x 2x² + 4x x ( x + 4 ) 2x (x + 2)
X² - 36 4x² - 100
(x – 6) (x + 6) (2x + 10) ( 2x – 10)
Type 2 Quadratic Factors
2Example 1. Factorise 3 2x x
Method 1 Brackets Method 2 Big X Method 3 Guide Number
2 3 2x x
( )( )x x2 1
( 2)( 1)x x
2 3 2x x x
x
2
1
( 2)( 1)x x
21 3 2x x
Guide Number ( )( )1 2 = 2Factors for Guide Number
2 1 21 3 2x x
21 2 1 2x x x 2(1 2 ) (1 2)x x x ( 2) 1( 2)x x x ( 2)( 1)x x
x² + 6x + 8 x² +3x ‑ 10 x² -2x ‑ 24
(x + 2) (x + 4) (x ‑ 2) (x + 5) (x – 6) (x + 4)
3x² + 13x + 4
(3x + 1) (x + 4) check (3x)(4) + (1)(x) = 12x + 1x = 13x 8x² +10x - 3
(8x + 1) (x - 3) check (8x)(-3)+ (1)(x) = -24x + 1x = -23x .......
wrong, try again
(4x ‑ 1) (2x + 3) check (4x )(3) + (‑1)(2x) = 12x ‑ 2x = 10x .......
correct
Factorising when there is a number in front of the x²
2Example 2. Factorise 12 7 10x x
Method 1 Brackets
212 7 10x x
( )( )4x 3x5 2
(4 5)(3 2)x x
4x
3x
5
2
Guide Number ( )( ) =12 10 120
Factors for Guide Number
120 1, 60 2, 40 3
30 4, 24 5, 2
15 8
0 6
, 12 10.
2(12 15 ) (8 10)x x x 3 (4 5) 2(4 5)x x x
( 2)( 1)x x
Method 2 Big X
212 7 10x x
(4 5)(3 2)x x
Method 3 Guide Number
2 712 10x x
212 15 8 10x x x
212 7 10x x
Simplify each of the following, using factors where necessary
8x + 8y x² + 8x+7 a² - 16 8 x + 1 3a - 12
8( x + y) (x + 7) ( x + 1) (a + 4) (a – 4 ) 8 x + 1 3( a – 4)
= x + y = x + 7 a + 4 3
x2 – 4x – 3 = 0
a = 1 b = - 4 c = - 3
Factorising, using the quadratic formula
X = 9.291 or x = - 1.291 2 2
X = 4.645 or x = - 0.645
X = 4.6 or x = - 0.6
Quadratic EquationsQuadratic equations have two solutions (roots).
2Example 1. Solve for if 3 2 0x x x Method 1 Using Factors
2 3 2 0x x ( 2)( 1) 0x x 2 0 or 1 0x x
2 or 1x x
Method 2 Using Quadratic Formula2 3 2 0x x
2 4
2
b b acx
a
1a 3b 2c 2( ) ( ) 4( )( )
2( )x
13 23
1
3 9 8 3 1
2 2x
3 1 3 1 or
2 2x x
1 or 2x x
2Example 2. Solve for if 4 0x x x
Method 1 Using Factors2 4 0x x ( 4) 0x x
0 or 4 0x x
0 or 4x x
Method 2 Using Quadratic Formula2 4 0x x
2 4
2
b b acx
a
1a 4b 0c 2( ) ( ) 4( )( )
2( )x
14 04
1
4 16 0 4 4
2 2x
4 4 4 4 or
2 2x x
4 or 0x x
2
1 2Example 3. Solve for if 4, 1.
1 1
Give answer in the form of , , .
x xx x
a ba b c N
c
1 2 4
1 ( 1)( 1) 1x x x
When dealing with fractions always
get a common denominator.
1( 1) 2 4( 1)( 1) (1)( 1)( 1) (1)( 1)( 1) (1)( 1)( 1)
x x x
x x x x x x
1( 1) 2 4( 1)( 1) x x x 21 2 4( 1) x x
24 4 3x x 24 1 0x x
4a 1b 1c
2 4Applying we get.
2
b b acx
a
1 17
8x