**find the HCF h(x) and the LCM m(x) for the polynomials p(x) = x****6****– 1, q(x) = x**^{4}**+ x**^{2}**+ 1 and prove that p(x) q(x) = h(x) m(x)**

p(x) = x^{6} – 1 = (x^{2})^{3} – 1^{3}

= (x^{2} – 1) [(x^{2})^{2} + x^{2}.1 + 1^{2}]

= (x^{2} – 1) (x^{4} + x^{2} + 1)

= (x + 1) (x – 1) (x^{4} + x^{2} + 1)

q(x) = x^{4} + x^{2} + 1

h(x) = x^{4} + x^{2} + 1

m(x) = (x + 1) (x – 1) (x^{4} + x^{2} + 1)

p(x).q(x)

= (x + 1) (x – 1) (x^{4} + x^{2} + 1) x (x^{4} + x^{2} + 1)

= (x2 – 1) (x^{4} + x^{2} + 1) ……. (1)

h(x).m(x)

= (x^{4} + x^{2 }+ 1) x (x + 1) (x – 1) (x^{4} + x^{2} + 1)

= (x^{2} – 1) (x^{4} + x^{2} + 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) = 8y**^{3}**+ 32y**^{2}**+ 32y and if b(y) m(y) = p(y) q(y) find g(y).**

p(y) = 8y^{3} + 32y^{2} + 32y

= 8y (y^{2} + 4y + 4)

= 8y (y + 2)^{2} [(a + b)^{2} = a^{2} + 2ab + b^{2}]

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 24y^{2}(y+2)^{2}(y – 2)]/[8y(y+2)^{2}]

= – 6y (y + 2) (y – 2)

= – 6y (y^{2} – 4)

q(y) = –**[6y**^{3}**– 24y] **

**= 24y – 6y**^{3}

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

**(i) 3x**^{2}**+ 5x – 2, 3x**^{2}**– 7x +2 **

p(x) = 3x^{2} + 5x – 2

= 3x^{2 }+ 6x – x – 2

= 3x (x + 2) – 1 (x + 2)

= (3x – 1) (x + 2)

q(x) = 3x^{2} – 7x +2

= 3x^{2} – 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)(x ^{2}-4)**

**(ii) 3(x**^{2}**– 7x + 12), 24(x**^{2}**– 9x + 20) **

p(x) = 3(x^{2} – 7x + 12)

= 3(x^{2} – 4x – 3x + 12)

= 3 [x (x – 4) – 3 (x – 4)

**= 3 (x – 4) (x – 3) **

q(x) = 24(x^{2} – 9x + 20)

= 2^{3} x 3 (x^{2} – 4x – 5x + 20)

= 2^{3} x 3 [x(x – 4) – 5 (x – 4)

**= 2**^{3}**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) . 2^{3}. 3 (x – 4) (x – 5)]/_{3 (x – 4)}

= 2^{3} x 3 (x – 3) (x – 4) (x – 5)

**= 24 (x – 3) (x – 4) (x – 5) **

**(iii) 16 – 4x**^{2}**, x**^{2}**+ x – 6 **

p(x) = 16 – 4x^{2}

= 4 (4 – x^{2})

= 4 (2^{2} – x^{2})

= – 4 (x^{2} – 2^{2})

= – 4 (x + 2) (x – 2)

q(x) = x^{2 }+ x – 6

= x^{2} + 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 (x**^{2}**– 4) (x + 3)**

**(iv) 8(x**^{3}**– x**^{2}**+ x), 28(x**^{3}**+ 1) **

p(x) = 8(x^{3} – x^{2} + x)

= 2^{3} x (x^{2} – x + 1)

q(x) = 28(x^{3 }+ 1)

= 2^{2} x 7 (x + 1) (x^{2} – x + 1)

h(x) = 2^{2} (x^{2} – x + 1)

We know that

m(x) = ^{p(x) q(x)}/_{h(x)}

=[2^{3} x (x^{2} – x + 1) 2^{2} x 7 (x + 1) (x^{2} – x + 1)]/[2^{2}(x^{2} – x + 1) ]

= 2^{3 }x 7 (x + 1) (x^{2} – x + 1)

**= 56x (x**^{3}**+ 1)**

**The HCL of two polynomials is h(a) = a – 7 and their LCM is m(a) = a****3****– 10 a**^{2}**+11a + 70. If one of the polynomials is p(a) = a**^{2}**– 12a + 35, and if the leading coefficient q(a) is positive, find the other polynomial q(a).**

p(a) = a^{2} – 12a + 35

= a^{2} – 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) (a^{3}−10a^{2} + 11a+ 70)]/[(a – 7)(a – 5)]

=(a^{3}−10a^{2} + 11a+ 70)/(a-5)

**= a**^{2}**– 5a – 14**

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

**m(x) = x**^{3}**– 7x + 6. If one of the polynomial is q(x) = x**^{2}**+ x – 6 and the other polynomial p(x) has negative leading coefficient. Find the other polynomial p(x)**

[**(**x+3)(x^{3}− 7x +6 )]/(x^{3}+ x – 6)

By dividing

p(x) = (x + 3) (x – 2)

p(x) = x^{2} + 2x – 3

**p(x) = – x**^{2}**+ 2x – 3**

## One thought on “HCF AND LCM – EXERCISE 3.3.5 – Class 9”

Comments are closed.