RBSE Solutions for Class 10 Maths Chapter 3 Polynomials Ex 3.6

RBSE Solutions for Class 10 Maths Chapter 3 Polynomials Ex 3.6 is part of RBSE Solutions for Class 10 Maths. Here we have given Rajasthan Board RBSE Class 10 Maths Chapter 3 Polynomials Exercise 3.6.

Board	RBSE
Te×tbook	SIERT, Rajasthan
Class	Class 10
Subject	Maths
Chapter	Chapter 3
Chapter Name	Polynomials
E×ercise	Exercise 3.6
Number of Questions Solved	5
Category	RBSE Solutions

Rajasthan Board RBSE Class 10 Maths Chapter 3 Polynomials Ex 3.6

Question 1.
Find the L.C.M. of following (RBSESolutions.com) expressions
(i) 24x²yz and 27x⁴y²z²
(ii) x² – 3x + 2 and x⁴ + x³ – 6x²
(iii) 2x² – 8 and x² – 5x + 6
(iv) x² – 1, (x² + 1)(x + 1) and x² + x – 1
(v) 18(6x⁴ + x³ – x²) and 45(25x⁶ + 3x⁵ – x⁴)
Solution

Question 2.
Find the H.C.F. of following (RBSESolutions.com) expressions.
(i) a³b⁴, ab⁵, a²b⁸
(ii) 16x²y², 48x⁴z
(iii) x² – 7x + 12; x² – 10x + 21 and x² + 2x – 15
(iv) (x + 3)²(x – 2) and (x + 3)(x – 2)²
(v) 24(6x⁴ – x³ – 2x²) and 20(6x⁶ + 5x⁵ + x⁴)
Solution

Question 3.
If u(x) = (x – 1)² and v(x) = (x² – 1) then, (RBSESolutions.com) verify the relation L.C.M. × H.C.F. = u(x) × v(x).
Solution
u(x) = (x – 1)²
⇒ u(x) = (x – 1)(x – 1)
and v(x) = (x² – 1) = (x – 1)(x + 1)
Product of common least powers = (x – 1)
H.C.F. = (x – 1)
Product of highest power of prime factors = (x – 1)²(x + 1)
L.C.M. = (x – 1)²(x + 1)
Test:
u(x) × v(x) = (x – 1)² × (x² – 1) = (x – 1)²(x² – 1)
and H.C.F. × L.C.M. = (x – 1)(x – 1)²(x + 1) = (x – 1)²(x² – 1)
So, L.C.M. × H.C.F. = w(x) × v(x).
Hence proved.

Question 4.
The product of two (RBSESolutions.com) expression are (x – 7)(x² + 8x + 12) if their H.C.F. is (x + 6), then find their L.C.M.
Solution
Given-Product of two expressions = (x – 7)(x² + 8x + 12)
= (x – 7)[x² + 6x + 2x + 12]
= (x – 7)[x(x + 6) + 2(x + 6)]
= (x – 7)(x + 2)(x + 6)
and H.C.F. = x + 6
L.C.M. = ?
we know that
H.C.F. × L.C.M. = Product of expressions

Question 5.
H.C. F. and L.C.M of two quadratic (RBSESolutions.com) expressions are respectively (x – 5) and x³ – 19x – 30, then find both expressions.
Solution
we know that H.C.F. = (x – 5)
and L.C.M. = x³ – 19x – 30
factorize x³ – 19x – 30
then putting x = 1 = (1)³ – 19(1) – 30 ≠ 0
then putting x = -1 = (-1)³ – 19(-1) – 30 ≠ 0
then putting x = 2 = (2)³ – 19(2) – 30 = 8 – 38 – 30 ≠ 0
then putting x = -2 = (-2)³ – 19(-2) – 30 = -8 + 38 – 30 = 0
Putting x = – 2 in expression (RBSESolutions.com) we get zero.
(x + 2) is a factor of expression

x³ – 19x – 30 = (x + 2)(x² – 2x – 15) = (x + 2)(x – 5)(x + 3)
Now H.C.F. = (x – 5) and L.C.M. = (x + 2)(x – 5)(x + 3)
⇒ (x – 5) is common factor
⇒ (x – 5) will exists in second (RBSESolutions.com) expressions
⇒ First expression = (x – 5)(x + 2) = x² – 3x – 10
and second expression = (x – 5)(x + 3) = x² – 2x – 15
Hence, x² – 3x – 10 and x² – 2x – 15 are required expressions.

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