HCF AND LCM – EXERCISE 3.3.5 – Class 9

  1. find the HCF h(x) and the LCM m(x) for the polynomials p(x) = x6 – 1, q(x) = x4 + x2 + 1 and prove that p(x) q(x) = h(x) m(x)

p(x) = x6 – 1 = (x2)3 – 13

= (x2 – 1) [(x2)2 + x2.1 + 12]

= (x2 – 1) (x4 + x2 + 1)

= (x + 1) (x – 1) (x4 + x2 + 1)

q(x) = x4 + x2 + 1

h(x) = x4 + x2 + 1

m(x) = (x + 1) (x – 1) (x4 + x2 + 1)

p(x).q(x)

= (x + 1) (x – 1) (x4 + x2 + 1) x (x4 + x2 + 1)

= (x2 – 1) (x4 + x2 + 1) ……. (1)

h(x).m(x)

= (x4 + x2 + 1) x (x + 1) (x – 1) (x4 + x2 + 1)

= (x2 – 1) (x4 + x2 + 1) ……. (2)

From (1) and (2) we get

p(x) q(x) = h(x) m(x)


  1. The HCF h(y) = 2y (y + 2) and the LCM m(y) = 24y (y + 2)2 (y – 2) for two polynomials p(y) and q(y) is known. If p(y) = 8y3 + 32y2 + 32y and if b(y) m(y) = p(y) q(y) find g(y).

p(y) = 8y3 + 32y2 + 32y

= 8y (y2 + 4y + 4)

= 8y (y + 2)2 [(a + b)2 = a2 + 2ab + b2]

We have h(y) m(y) = – p(y) q(y)

2y (y + 2) x 24y (y + 2)2 (y – 2)

= – 8y (y + 2)2 = q(y)

q(y) = [−2 x 24y2(y+2)2(y – 2)]/[8y(y+2)2]

= – 6y (y + 2) (y – 2)

= – 6y (y2 – 4)

q(y) = –[6y3 – 24y]

= 24y – 6y3


  1. Find the HCF h(x) of the following polynomials and using HCF h(x) find the LCM m(x).

(i) 3x2 + 5x – 2, 3x2 – 7x +2

p(x) = 3x2 + 5x – 2

= 3x2 + 6x – x – 2

= 3x (x + 2) – 1 (x + 2)

= (3x – 1) (x + 2)

q(x) = 3x2 – 7x +2

= 3x2 – 6x – x + 2

=3x (x – 2) – 1(x – 2)

= (3x – 1) (x – 2)

h(x) = (3x – 1)

We know that

h(x) m(x) = p(x) q(x)

m(x) = [p(x) q(x)]/h(x)

= [(3x – 1)( x + 2) (x – 2) (3x – 1)]/ (3x – 1)

= (3x – 1) (x + 2) (x – 2) = (3x-1)(x2-4)

 

(ii) 3(x2 – 7x + 12), 24(x2 – 9x + 20)

p(x) = 3(x2 – 7x + 12)

= 3(x2 – 4x – 3x + 12)

= 3 [x (x – 4) – 3 (x – 4)

= 3 (x – 4) (x – 3)

q(x) = 24(x2 – 9x + 20)

= 23 x 3 (x2 – 4x – 5x + 20)

= 23 x 3 [x(x – 4) – 5 (x – 4)

= 23 x 3 (x – 4) (x – 5)

h(x) = 3 (x – 4)

We know that m(x) = [p(x) q(x)]/h(x)

= [3 (x – 4)( x – 3) . 23. 3 (x – 4) (x – 5)]/3 (x – 4)

= 23 x 3 (x – 3) (x – 4) (x – 5)

= 24 (x – 3) (x – 4) (x – 5)

 

(iii) 16 – 4x2 , x2 + x – 6

p(x) = 16 – 4x2

= 4 (4 – x2)

= 4 (22 – x2)

= – 4 (x2 – 22)

= – 4 (x + 2) (x – 2)

q(x) = x2 + x – 6

= x2 + 3x – 2x – 6

= x (x + 3) – 2 (x + 3)

= (x + 3) (x – 2)

h(x) = (x – 2)

We know that

m(x) = p(x) q(x)/h(x)

= [–[−4 (x + 2)( x – 2) ] (x + 3) (x – 2)]/(x – 2)

= 4 (x + 2) (x – 2) (x + 3)

= 4 (x2 – 4) (x + 3)

 

(iv) 8(x3 – x2 + x), 28(x3 + 1)

p(x) = 8(x3 – x2 + x)

= 23 x (x2 – x + 1)

q(x) = 28(x3 + 1)

= 22 x 7 (x + 1) (x2 – x + 1)

h(x) = 22 (x2 – x + 1)

We know that

m(x) = p(x) q(x)/h(x)

=[23 x (x2 – x + 1) 22 x 7 (x + 1) (x2 – x + 1)]/[22(x2 – x + 1) ]

= 23 x 7 (x + 1) (x2 – x + 1)

= 56x (x3 + 1)


  1. The HCL of two polynomials is h(a) = a – 7 and their LCM is m(a) = a3 – 10 a2 +11a + 70. If one of the polynomials is p(a) = a2 – 12a + 35, and if the leading coefficient q(a) is positive, find the other polynomial q(a).

p(a) = a2 – 12a + 35

= a2 – 7a – 5a + 35

= a (a – 7) – 5 (a – 7)

= (a – 7) (a – 5)

We know that

p(a) q(a) = h(a) m(a)

q(a) = h(a) m(a)/p(a)

= [(a – 7) (a3−10a2 + 11a+ 70)]/[(a – 7)(a – 5)]

=(a3−10a2 + 11a+ 70)/(a-5)

= a2 – 5a – 14


  1. The HCF of the two expressions is h(x) = (x + 3) and their LCM is

m(x) = x3 – 7x + 6. If one of the polynomial is q(x) = x2 + x – 6 and the other polynomial p(x) has negative leading coefficient. Find the other polynomial p(x)

[(x+3)(x3− 7x +6 )]/(x3+ x – 6)

By dividing

HCF AND LCM - EXERCISE 3.3.5  – Class 9

p(x) = (x + 3) (x – 2)

p(x) = x2 + 2x – 3

p(x) = – x2 + 2x – 3

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