leaving cert factors

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Leaving Cert Factors

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Leaving Cert Factors. FACTORS. Difference of 2 squares. Factorise the following 6x + 24y 5ab + 15bc 7x² - 28x 6( x + 4y) 5b( a + 3c) 7x( x – 4) - PowerPoint PPT Presentation

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Page 1: Leaving Cert Factors

Leaving Cert Factors

Page 2: 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

Page 3: Leaving Cert Factors

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

Page 4: Leaving Cert Factors

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)

Page 5: Leaving Cert Factors

Factorising

x² + 4x 2x² + 4x x ( x + 4 ) 2x (x + 2)

X² - 36 4x² - 100

(x – 6) (x + 6) (2x + 10) ( 2x – 10)

Page 6: Leaving Cert Factors

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

Page 7: Leaving Cert Factors

x² + 6x + 8 x² +3x ‑ 10 x² -2x ‑ 24

(x + 2) (x + 4) (x ‑ 2) (x + 5) (x – 6) (x + 4)

Page 8: Leaving Cert Factors

  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

Page 9: Leaving Cert Factors

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

Page 10: Leaving Cert Factors

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

Page 11: Leaving Cert Factors

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

Page 12: Leaving Cert Factors

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

Page 13: Leaving Cert Factors

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

Page 14: Leaving Cert Factors

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