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) 24x^{2}yz and 27x^{4}y^{2}z^{2}

(ii) x^{2} – 3x + 2 and x^{4} + x^{3} – 6x^{2}

(iii) 2x^{2} – 8 and x^{2} – 5x + 6

(iv) x^{2} – 1, (x^{2} + 1)(x + 1) and x^{2} + x – 1

(v) 18(6x^{4} + x^{3} – x^{2}) and 45(25x^{6} + 3x^{5} – x^{4})

Solution

Question 2.

Find the H.C.F. of following (RBSESolutions.com) expressions.

(i) a^{3}b^{4}, ab^{5}, a^{2}b^{8}

(ii) 16x^{2}y^{2}, 48x^{4}z

(iii) x^{2} – 7x + 12; x^{2} – 10x + 21 and x^{2} + 2x – 15

(iv) (x + 3)^{2}(x – 2) and (x + 3)(x – 2)^{2}

(v) 24(6x^{4} – x^{3} – 2x^{2}) and 20(6x^{6} + 5x^{5} + x^{4})

Solution

Question 3.

If u(x) = (x – 1)^{2} and v(x) = (x^{2} – 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) = (x^{2} – 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} × (x^{2} – 1) = (x – 1)^{2}(x^{2} – 1)

and H.C.F. × L.C.M. = (x – 1)(x – 1)^{2}(x + 1) = (x – 1)^{2}(x^{2} – 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)(x^{2} + 8x + 12) if their H.C.F. is (x + 6), then find their L.C.M.

Solution

Given-Product of two expressions = (x – 7)(x^{2} + 8x + 12)

= (x – 7)[x^{2} + 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 x^{3} – 19x – 30, then find both expressions.

Solution

we know that H.C.F. = (x – 5)

and L.C.M. = x^{3} – 19x – 30

factorize x^{3} – 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

x^{3} – 19x – 30 = (x + 2)(x^{2} – 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) = x^{2} – 3x – 10

and second expression = (x – 5)(x + 3) = x^{2} – 2x – 15

Hence, x^{2} – 3x – 10 and x^{2} – 2x – 15 are required expressions.

We hope the given RBSE Solutions for Class 10 Maths Chapter 3 Polynomials Ex 3.6 will help you. If you have any query regarding Rajasthan Board RBSE Class 10 Maths Chapter 3 Polynomials Exercise 3.6, drop a comment below and we will get back to you at the earliest.

## Leave a Reply