# HCF AND LCM – EXERCISE 3.3.3 – Class 9

1. Find the HCF of each of the following pairs of polynomials

(i) x2 + 2x – 15 and x2 – 7x + 12

Solution:

x2 + 2x – 15                     &                x2 – 7x + 12

= x2 + 5x – 3x – 15                            = x2 – 4x – 3x – 12

= x (x + 5) – 3 (x + 5)                       = x (x – 4) – 3 (x – 4)

=(x + 5) (x – 3)                                 = (x – 4) (x – 3)

HCF = (x – 3)

(ii) m2 – 3m + 2 and m2 + m – 6

Solution:

m2 – 3m + 2

= m2 – 2m – m + 2

= m (m – 2) – 1 (m – 2)

= (m – 2) (m – 1)

m2 + m – 6

= m2 + 3m – 2m – 6

= m (m + 3) – 2 (m + 3)

= (m + 3) (m – 2)

HCF = (m – 2)

(iii) x2 – xy – 2y2 and x2 + 3xy + 2y2

x2 – xy – 2y2 = x2 – 2xy + xy – 2y2

= x (x – 2y) + y (x – 2y)

= (x – 2y) (x + y)

x2 + 3xy + 2y2

= x2 + 3xy + xy + 2y2

= x (x + 2y) + y (x + 2y)

= (x + 2y) (x + y)

HCF = (x + y)

(iv) x2 – 2x + 1 and x2 – 3x + 2

x2 – 2x + 1 = (x – 1)2

x2 – 3x + 2 = x2 – 2x – x + 2

= x (x – 2) – 1 (x – 2)

= (x – 2) (x – 1)

HCF = (X – 1)

1. Find the HCF of p(x) = (x3 -27) (x2 – 3x +2) and

q(x) =(x2 + 3x + 9) (x2 – 5x +6)

Solution:

p(x) = (x3 -27) (x2 – 3x +2)

(x3 -27) = x3 – 93 = (x – 3) (x2 + 3x + 32)

= (x – 3) (x2 + 3x + 9)

(x2 – 3x +2) = x2 – 2x – x – 2

= x (x – 2) – 1 (x – 2)

= (x – 2) (x – 1)

p(x) = (x – 3) (x2 + 3x + 9) (x – 2) (x – 1)

q(x) = (x2 + 3x + 9) (x2 – 5x +6)

(x2 – 5x +6) = x – 3x – 2 x + 6

=x (x – 3) – 2 (x – 3)

= (x – 3) (x – 2)

q(x) = (x – 3) (x – 2) (x2 + 3x + 9)

HCF = (x – 3) (x – 2) (x2 + 3x + 9)

1. Find the HCF of f(x) = x3 + x2 – x – 1 and g(x) = x3 + x2 + x + 1

Solution:

f(x) = x3 + x2 – x – 1

= x3 (x + 1) – 1 (x + 1)

= (x2 – 1) (x + 1)

= (x + 1) (x – 1) (x + 1)

= (x + 1)2 (x – 1)

g(x) = x3 + x2 + x + 1

= x2 (x + 1) + 1 (x + 1)

= (x2 – 1) (x + 1)

HCF = (x + 1)

1. Find the HCF of p(x) = x– 2x– 15x2and q(x) = x3 – 9x

p(x) = x4 – 2x3 – 15x2

= x2 (x2 – 2x – 15)

= x2(x2 -5x + 3x – 15)

= x2 [x (x – 5) + 3(x – 5)]

= x2 (x + 3) (x – 5)

q(x) = x3 – 9x

= x (x2 – 9)

= x (x2 – 32)

= x (x – 3) (x + 3)

HCF of p(x) and q(x) = x (x + 3)

1. Find the HCF of each of the following triples:

(i) a4 b – ab4 , a4b2 – a2b4 and a2 b2 (a4 – b4)

a4b – ab4 = ab (a3 – b3)

= ab (a – b) (a2 + 2ab +b2)

a4b2 – a2b4 = a2 b2 (a2 – b2)

= a2 b2 (a – b) (a + b)

a2 b2 (a4 – b4) = a2 b2 (a2 + b2) (a2 – b2)

= a2 b2 (a2 + b2) (a – b) (a + b)

HCF = ab (a – b)

(ii) 6(x2 + 10x + 24), 4(x2 – x – 20) and 8(x2 + 3x – 4)

Solution:

6(x2 + 10x + 24) = 6(x2 + 6x + 4 x + 247)

= 6 [x (x + 6) + 4 (x + 6)]

= 2 x 3 (x + 6) (x + 4)

4(x2 – x – 20) = 4 (x – 5x + 4x – 20)

= 4 [x (x – 5) + 4 (x – 5)]

= 22 (x – 5) (x + 4)

8(x2 + 3x – 4) = 23 (x2 + 4x – x – 4)

= 23 [x (x + 4) – 1 (x + 4)

= 23 (x + 4) (x – 1)

HCF = 2 (x + 4)

(iii) a2 (2x2 + 5x + 2), a2b (3x2 + 8x +4)and ab2 (2x2 + 3x – 2)

Solution:

a2 (2x2 + 5x + 2) = a2 (2x2 + 4x + x + 2)

= a2 [2x (x + 2) + 1 (x + 2)]

= a2 (2x + 1) (x + 2)

a2b (3x2 + 8x +4) = a2b (3x2 + 6x + 2x +4)

= a2b [3x(x + 2) + 2 (x + 2)]

= a2b (x + 2) (3x + 2)

ab2 (2x2 + 3x – 2) = ab2 (2x2 + 4x – x – 2)

= ab2 [2x (x + 2) – 1 (x + 2)]

= ab2 (2x – 1) (x + 2)

HCF is a (x + 2) HCF = a (x + 2)

1. If the HCF of the polynomials p(x) = (x + 1) (x2 + ax + 4) and

q(x) = (x + 4) (x² + bx +2) is h(x) = x2 + 5x + 4, find the values of a and b.

Solution:

p(x) = (x + 1) (x2 + ax + 4)

q(x) = (x + 4) (x2 + bx +2)

h(x) = x2 + 5x + 4

= x2 + 4x + x + 4

= x (x + 4) + 1(x + 4)

= (x + 4) (x +1)

Hence (x + 1) divides x2 + bx + 2

x2 + bx + 2 = (x + 1) f(x)

Let x = –1

(–1)2 + b (–1) + 2 = 0

–b + 3 = 0

b = 3

Also (x + 4) divides x2 + ax + 4

x2 + ax + 4 = (x + 4) g(x)

Let x = – 4

(–4)2 + a (–4) + 4 = 0

16 – 4a + 4 = 0

4a = 20

a = 5

1. If the HCF of the polynomials p(x) = (x – 3) (2x2 – ax + 2) and

q(x) = (x + 4) (x2 – bx – 6) is h(x) = x2 – 5x + 6, find the values of a and b.

Solution:

h(x) = x2 – 5x + 6

= x2 – 3x – 2x + 6

= x (x – 3) – 2 (x – 3)

= (x – 3) (x – 2)

p(x) = (x – 3) (2x2 – ax + 2)

Hence (x – 2) divides 2x2 – ax + 2

2x2 – ax + 2 = (x – 2) f(x)

Let x = 2

2(2)2 – a (2) + 2 = 0

8 – 2a + 2 = 0

2a = 10

a = 5

q(x) = (x + 4) (x2 – bx – 6)

Hence (x – 3) divides (x2 – bx – 6)

(x2 – bx – 6)= (x – 3) g(x)

Let x = 3

33 – b (3) – 6 = 0

9 – 3b – 6 = 0

3b = 3

b = 1

1. For what values of a and b the polynomials p(x) = (x2 +5x+6) (x2+2x – a) and q(x) = (x2 – x – 2) (x2 + 7x + b) have (x + 2) (x – 2) as their HCF.

Solution:

p(x) = (x2 + 5x + 6) (x2 + 2x – a)

= (x2 + 3x + 2x + 6) (x2 + 2x – a)

= x (x + 3) + 2 (x + 3) (x2 + 2x – a)

= (x + 2) (x + 3) (x2 + 2x – a)

Hence (x – 2) divides x2 + 2x – a

x2 + 2x – a = (x – 2) f(x)

Let x = 2

22 + 2(2) – a = 0

4 + 4 – a = 0

a = 8

q(x) = (x2 – x – 2) (x2 + 7x + b)

= (x2 – 2x + x – 2) (x2 + 7x + b)

= x (x – 2) + 1 (x – 2) (x2 + 7x + b)

= (x – 2) (x + 1) (x2 + 7x + b)

Hence (x + 2) divides x2 + 7x + b

x2 + 7x + b = (x + 2) g(x)

Let x = -2

(-2)2 + 7(-2) – b = 0

14 – 14 + b = 0

b = 10

1. If the HCF of x2 + x – 12 and 2x2 – kx – 9 is (x – 1) find the value of k.

Solution:

Since x – k is the HCF of p(x) = x2 + x – 12 and 2×2 – kx – 9, it is a common factor of p(x) = 0 and q(x) = 0.

Put x = k

p(x) = k2 + k – 12 = 0

q(x) = 2k2 – k2 – 9 = 0

k2 – 9 = 0

k2 = 9

k = ± 3

Given p(x) = 0, k ≠ -3. Taking k = 3 we verify

p(x) = 3 + 3 – 12

= 9 + 3 – 12

= 12 – 12 =0

k = 3