CLASS 8 MATHS 14 Factorisation SOLUTIONS

 CLASS 8 MATHS  14 Factorisation SOLUTIONS

Example 1: Factorise 12a2b + 15ab2 

Solution: We have 12a2b = 2 × 2 × 3 × a × a × b 

15ab2 = 3 × 5 × a × b × b

The two terms have 3, a and b as common factors. 

Therefore,

12a2b + 15ab2

= (3 × a × b × 2 × 2 × a) + (3 × a × b × 5 × b)

= 3 × a × b × [(2 × 2 × a) + (5 × b)]

 = 3ab × (4a + 5b) 

= 3ab (4a + 5b) (required factor form)

Example 2:Factorise 10x2 – 18x3 + 14x4 

Solution: 10x2 = 2 × 5 × x × x 

 18x3 = 2 × 3 × 3 × x × x × x 

 14x4 = 2 × 7 × x × x × x × x 

The common factors of the three terms are 2, x and x. 

Therefore, 10x2 – 18x3 + 14x4 = (2 × x × x × 5) –(2 × x × x × 3 × 3 × x) + (2 × x × x × 7 × x × x)

(2 × x × x) - [5-(3x3x3xx) + 7xxxx)] = 2x2 × (5 – 9x + 7x2) =2x2(7x2 -9x+5)         

(combining the three terms

Example 3: Factorise 6xy – 4y + 6 – 9x. 

Solution: 

6xy – 4y + 6 – 9x = 6xy – 4y – 9x + 6 

= 2y (3x – 2) – 3 (3x – 2) 

= (3x – 2) (2y – 3) 

The factors of (6xy – 4y + 6 – 9 x) are (3x – 2) and (2y – 3).

Example 4: Factorise x2 + 8x + 16 

Solution:

a2 + 2ab + b2 = x2 + 2 (x) (4) + 42 

= x2 + 8x + 16 

Since a2 + 2ab + b2 = (a + b)2, 

by comparison x2 + 8x + 16 = ( x + 4)2

Example 5: Factorise 4y2 – 12y + 9

Solution: 4y2 = (2y)2, 9 = 32 and 12y = 2 × 3 × (2y)

Therefore, 4y2 – 12y + 9 = (2y)2 – 2 × 3 × (2y) + (3)2

= ( 2y – 3)2

Example 6: Factorise 49p2 – 36 

Solution: (a2 – b2).

 49p2 – 36 = (7p)2 – ( 6 )2 

= (7p – 6 ) ( 7p + 6)

Example 7: Factorise a2 – 2ab + b2 – c2 

Solution:

a2 – 2ab + b2 – c2 = (a – b)2– c2

= [(a – b) – c) ((a – b) + c)] 

= (a – b – c) (a – b + c)

Example 8: Factorise m4 – 256

Solution: We note m4 = (m2)2 and 256 = (16) 2

m2–16 = m2 – 42 

= (m – 4) (m + 4) 

Therefore, m4 – 256 = (m – 4) (m + 4) (m2 +16)

Example 9: Factorise x2 + 5x + 6

Solution:

(x +2) (x + 3)

a = 2, b = 3. For this a + b = 5 and ab = 6

 x2 + (a + b) x + ab

Example 10: Find the factors of y2 –7y +12. 

Solution: 12 = 3 × 4 and 3 + 4 = 7.

Therefore, y2 – 7y+ 12 = y2 – 3y – 4y + 12 

 = y (y –3) – 4 (y –3) = (y –3) (y – 4)

Example 11: Obtain the factors of z2 – 4z – 12. 

Solution: z2 – 4z –12 = z2 – 6z + 2z –12 

= z(z – 6) + 2(z – 6 ) 

= (z – 6) (z + 2) 

Example 12: Find the

factors of 3m2 + 9m + 6. 

Solution:

3m2 + 9m + 6 = 3(m2 + 3m + 2) 

Now, m 2 + 3m + 2 = m2 + m + 2m + 2 (as 2 = 1 × 2)

= m(m + 1)+ 2( m + 1) 

= (m + 1) (m + 2) 

Therefore, 3m2 + 9m + 6

= 3(m + 1) (m + 2)

Example 13: Do the following divisions. 

(i) –20x4 ⎟ 10x2(ii) 7x2y2z2 ⎟ 14xyz 

Solution: (i) –20x4 = –2 × 2 × 5 × x × x × x × x 

10x2 = 2 × 5 × x × x

Solution= –2x2

Example 14: Divide 24(x2yz + xy2z + xyz2) by 8xyz

Solution: 24 (x2yz + xy2z + xyz2) 

= 2 × 2 × 2 × 3 × [(x × x × y × z) + (x × y × y × z) + (x × y × z × z)]

= 2 × 2 × 2 × 3 × x × y × z × (x + y + z) = 8 × 3 × xyz × (x + y + z)

Therefore, 24 (x2yz + xy2z + xyz2) ➗ 8xyz

= 3x + 3y + 3z = 3(x + y + z)

Example 15: Divide 44(x4 – 5x3 – 24x2) by 11x (x – 8) 

Solution: Factorising 44(x4 – 5x3 – 24x2), we get 

44(x4 – 5x3 – 24x2) = 2 × 2 × 11 × x2(x2 – 5x – 24)

= 2 × 2 × 11 × x2(x2 – 8x + 3x – 24)

= 2 × 2 × 11 × x2 [x (x – 8) + 3(x – 8)] 

= 2 × 2 × 11 × x2 (x + 3) (x – 8)

= 2 × 2 × x (x + 3) = 4x(x + 3)

Example 16: Divide z(5z2 – 80) by 5z(z + 4)

Solution: Dividend = z(5z2 – 80)

= z × 5 × (z2 – 16)

= z[(5 × z2) – (5 × 16)]

= 5z × (z + 4) (z – 4) [using the identity a2 – b2 = (a + b) (a – b)]

(z – 4)