RD Sharma Class 10 Ex 2.3 Solutions Chapter 2 Polynomials

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Chapter 2: Polynomials Exercise – 2.3

Question: 1

Apply division algorithm to find the quotient q(x) and remainder r(x) on dividing f(x) by g(x) in each of the following:

(i) f(x) = x³ – 6x² + 11x – 6, g(x) = x² + x + 1

(ii) f(x) = 10x⁴ + 17x³ – 62x² + 30x – 105, g(x) = 2x² + 7x + 1

(iii) f(x) = 4x³ + 8x² + 8x + 7, g(x) = 2x² – x + 1

(iv) f(x) = 15x³ – 20x² + 13x – 12, g(x) = x² – 2x + 2

Solution: 

(i) f(x) = x³ – 6x² + 11x – 6 and g(x) = x² + x + 1

(ii) f(x) = 10x⁴ + 17x³ – 62x² + 30x – 105, g(x) = 2x² + 7x + 1

(iii) f(x) = 4x³ + 8x² + 8x + 7, g(x) = 2x² – x + 1

(iv) f(x) = 15x³ – 20x² + 13x – 12, g(x) = x² – 2x + 2

Question: 2

Check whether the first polynomial is a factor of the second polynomial by applying the division algorithm:

(i) g(t) = t² – 3; f(t) = 2t⁴ + 3t³ – 2t² – 9t – 12

(ii) g(x) = x² – 3x + 1; f(x) = x⁵ – 4x³ + x² + 3x + 1

(iii) g(x) = 2x² – x + 3; f(x) = 6x⁵ − x⁴ + 4x³ – 5x² – x – 15

Solution:

(i) g(t) = t² – 3; f(t) = 2t⁴ + 3t³ – 2t² – 9t – 12

g(t) = t² – 3g(t) = t² – 3 f(t) = 2t⁴ + 3t³ – 2t² – 9t Therefore, g(t) is the factor of f(t).

(ii) g(x) = x² – 3x + 1; f(x) = x⁵ – 4x³ + x² + 3x + 1

g(x) = x² – 3x + 1 g(x) = x² – 3x + 1 f(x) = x⁵ – 4x³ + x² + 3x + 1.Therefore, g(x) is not the factor of f(x).

(iii) g(x) = 2x² – x + 3; f(x) = 6x⁵ − x⁴ + 4x³ – 5x² – x – 15

g(x) = 2x² – x + 3 g(x) = 2x² – x + 3 f(x) = 6x⁵ − x⁴ + 4x³ – 5x² – x – 15

Question: 3

Obtain all zeroes of the polynomial f(x) = f(x) = 2x⁴ + x³ – 14x² – 19x – 6, if two of its zeroes are -2 and -1.

Solution:

f(x) = 2x⁴ + x³ – 14x² – 19x – 6

If the two zeroes of the polynomial are -2 and -1, then its factors are (x + 2) and (x + 1)

(x + 2)(x + 1) = x² + x + 2x + 2 = x² + 3x + 2

f(x) = 2x⁴ + x³ – 14x² – 19x – 6 = (2x² – 5x – 3)(x² + 3x + 2)

= (2x + 1)(x – 3)(x + 2)(x + 1)

Therefore, zeroes of the polynomial = – 1/2, 3, -2 , -1

Question: 4

Obtain all zeroes of f(x) = x³ + 13x² + 32x + 20, if one of its zeroes is -2.

Solution:

f(x) = x³ + 13x² + 32x + 20

Since, the zero of the polynomial is -2 so, it means its factor is (x + 2).

So, f(x) = x³ + 13x² + 32x + 20 =  (x² + 11x + 10)(x + 2)

= (x² + 10x + x + 10)(x + 2)

= (x + 10)(x + 1)(x + 2)

Therefore, the zeroes of the polynomial are – 1, – 10, – 2.

Question: 5

Obtain all zeroes of the polynomial f(x) = x⁴ – 3x³ – x² + 9x – 6, if the two of its zeroes are – √3 and √3.

Solution:

f(x) = x⁴ – 3x³ – x² + 9x – 6 Since, two of the zeroes of polynomial are -√3 and √3 so,(x + √3)(x – √3) = x²– 3x² – 3

So, f(x) = x⁴ – 3x² – x² + 9x – 6 = (x² – 3)(x² – 3x + 2)

= (x + √3)(x – √3)(x²– 2x – 2 + 2)

= (x + √3)(x – √3)(x – 1)(x – 2)

Therefore, the zeroes of the polynomial are – √3, √3, 1, 2.

Question: 6

Obtain all zeroes of the polynomial f(x) = 2x⁴ – 2x³ – 7x² + x – 1, if the two of its zeroes are – √(3/2) and √(3/2).

Solution:

f(x) = 2x⁴ – 2x³ – 7x² + x – 1 Since, – √(3/2)  and √(3/2) are the zeroes of the polynomial, so the factors are

So, f(x) = 2x⁴– 2x³–7x²+ x – 1 

Therefore, the zeroes of the polynomial = x = -1, 2, -√(3/2) and √(3/2).

Question: 7

Find all the zeroes of the polynomial x⁴ + x³ – 34x² – 4x + 120, if the two of its zeroes are 2 and – 2.

Solution:

x⁴ + x³ – 34x² – 4x + 120 Since, the two zeroes of the polynomial given is 2 and – 2 So, factors are (x + 2)(x – 2) =  x² + 2x – 2x – 4 = x² – 4x² – 4

So, x⁴ + x³ – 34x² – 4x + 120 = (x² – 4)(x² + x – 30)

= (x – 2)(x + 2)(x² + 6x – 5x – 30)

=  (x – 2)(x + 2)(x + 6)(x – 5)

Therefore, the zeroes of the polynomial = x = 2, – 2, – 6, 5

Question: 8

Find all the zeroes of the polynomial 2x⁴ + 7x³ – 19x² – 14x + 30, if the two of its zeroes are √2 and – √2.

Solution:

2x⁴ + 7x³ – 19x² – 14x + 30² Since, √2 and-√2 are the zeroes of the polynomial given. So, factors are (x + √2) and (x – √2)

= x²+√2x – √2x – 2 = x² – 2

So, 2x⁴ + 7x³ – 19x² – 14x + 30 = (x² – 2)(2x² + 7x – 15)

= (2x²+ 10x – 3x – 15)(x + √2)(x – √2)

= (2x – 3)(x + 5)(x + √2)(x – √2)

Therefore, the zeroes of the polynomial is √2,- √2,-5, 3/2.

Question: 9

Find all the zeroes of the polynomial f(x) = 2x³ + x² – 6x – 3, if two of its zeroes are – √3 and √3.

Solution: 

f(x) = 2x³ + x² – 6x – 3 Since, -√3 and √3 are the zeroes of the given polynomial So, factors are (x – √3) and (x + √3)

= (x²– √3x + √3x – 3) = (x² – 3)

So, f(x) = 2x³ + x² – 6x – 3 = (x² – 3)(2x + 1)

= (x – √3)(x + √3)(2x + 1)

Therefore, set of zeroes for the given polynomial = √3,– √3, –1/2

Question: 10

Find all the zeroes of the polynomial f(x) = x³ + 3x² – 2x – 6, if the two of its zeroes are √2 and – √2.

Solution:

f(x) = x³ + 3x² – 2x – 6 Since, √2 and -√2 are the two zeroes of the given polynomial. So, factors are (x + √2) and (x – √2)

= x²+ √2x – √2x – 2 = x² – 2

By division algorithm, we have:

f(x) = x³ + 3x² – 2x – 6 = (x² – 2)(x + 3)

= (x – √2)(x + √2)(x + 3)

Therefore, the zeroes of the given polynomial is – √2, √2 and – 3.

Question: 11

What must be added to the polynomial f(x) = x⁴ + 2x³ – 2x² + x − 1 so that the resulting polynomial is exactly divisible by g(x) = x² + 2x − 3.

Solution:

f(x) = x⁴ + 2x³ – 2x² + x − 1

We must add (x – 2) in order to get the resulting polynomial exactly divisible by g(x) = x² + 2x − 3.

Question: 12

What must be subtracted from the polynomial f(x) = x⁴ + 2x³ – 13x² –12x + 21 so that the resulting polynomial is exactly divisible by g(x) = x² – 4x + 3.

Solution:

f(x) = x⁴ + 2x³ – 13x² – 12x + 21

We must subtract (2x – 3) in order to get the resulting polynomial exactly divisible by g(x) = x² – 4x + 3.

Question: 13

Given that √2 is a zero of the cubic polynomial f(x) = 6x³+ √2x²– 10x – 4√2, find its other two zeroes.

Solution:

f(x) = 6x³+√2x² – 10x – 4√2 Since, √2 is a zero of the cubic polynomial So, factor is (x–√2)

Question: 14

Given that x – √5 is a factor of the cubic polynomialfind all the zeroes of the polynomial.

Solution: 

In the question, it’s given that x – √5 is a factor of the cubic polynomial.

So, the zeroes of the polynomial

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