- 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)
- 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
- 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)
- 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
- 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
p(x) = (x + 3) (x – 2)
p(x) = x2 + 2x – 3
p(x) = – x2 + 2x – 3
1 thought on “HCF AND LCM – EXERCISE 3.3.5 – Class 9”
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