Exercise 2.3 Class 10 maths 2. Polynomials - ncert solutions - Toppers Study

Exercise 2.3 Class 10 maths 2. Polynomials - ncert solutions - Toppers Study

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Exercise 2.3 class 10 maths chapter 2. Polynomials

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Q1. Divide the polynomial p(x) by the polynomial g(x) and find the quotient and remainder in each of the following:

(i) p(x) = x³ – 3x² + 5x – 3, g(x) = x² – 2

(ii) p(x) = x⁴ – 3x² + 4x + 5, g(x) = x² + 1 – x

(iii) p(x) = x⁴ – 5x + 6, g(x) = 2 – x²

Solution: (i) p(x) = x³ – 3x² + 5x – 3, g(x) = x² – 2

Quotients q(x) = x –  3 and Remainder = 7x –  9

Solution: (ii) p(x) = x⁴ – 3x² + 4x + 5, g(x) = x² + 1 – x

Quotients q(x) = x² + x –  3 and Remainder = 8

Solution: (iii) p(x) = x⁴ – 5x + 6, g(x) = 2 – x²

Quotients q(x) = – x² –  2 and Remainder = –  5x + 10  

Q2. Check whether the first polynomial is a factor of the second polynomial by dividing the second polynomial by the first polynomial:

(i) t² – 3, 2t⁴ + 3t³ – 2t² – 9t – 12

(ii) x² + 3x + 1, 3x⁴ + 5x³ – 7x² + 2x + 2

(iii) x³ – 3x + 1, x⁵ – 4x³ + x² + 3x + 1

Solution: (i) t² – 3, 2t⁴ + 3t³ – 2t² – 9t – 12

Hence Remainder r(x) is 0

Therefore, t² – 3 is the factor of 2t⁴ + 3t³ – 2t² – 9t – 12

Solution: (ii) x² + 3x + 1, 3x⁴ + 5x³ – 7x² + 2x + 2

Hence Remainder r(x) is 0

Therefore,  x² + 3x + 1 is the factor of 3x⁴ + 5x³ – 7x² + 2x + 2

Solution: (iii) x³ – 3x + 1, x⁵ – 4x³ + x² + 3x + 1

Hence Remainder r(x) = 2

Therefore, x3 – 3x + 1, is not a factor of x⁵ – 4x³ + x² + 3x + 1

Q3. Obtain all other zeroes of 3x⁴ + 6x³ – 2x² – 10x – 5, if two of its zeroes are

Solution:

Given that : p(x) = 3x⁴ + 6x³ – 2x² – 10x – 5

Or 3x²​ - 5 = 0

Therefore, 3x² - 5 is the factor of p(x)

Now Dividing 3x⁴ + 6x³ - 2x² - 10x - 5 by 3x² - 5

Therefore,  p(x) = (3x² –  5) (x² + 2x + 1)

Now, factorizing and getting zeroes x² + 2x + 1 -

= x² + x + x + 1 = 0

= x(x + 1) + 1(x + 1) = 0

= (x + 1) (x + 1) = 0 

Or x + 1 = 0, x + 1 = 0

Or x = – 1, x = – 1

Therefore, two zeroes are – 1 and – 1.

Q4. On dividing x3 – 3x2 + x + 2 by a polynomial g(x), the quotient and remainder were x – 2 and –2x + 4, respectively. Find g(x).

Solution:

Given that: Dividend p(x) = x³ – 3x² + x + 2

Quotient q(x) = x – 2,

Remainder r(x) = – 2x + 4

Divisor g(x) =?

Dividend = divisor × quotient + remainder

p(x) = g(x) × q(x) + r(x)

x³ – 3x² + x + 2 = g(x) (x – 2) + (– 2x + 4)

x³ – 3x² + x + 2 + 2x – 4 = g(x) (x – 2)

g(x) (x – 2) = x³ – 3x² + 3x – 2

Dividing x³ – 3x² + 3x – 2 by x - 2 we obtain g(x)-

Therefore, Divisor g(x) = x² – x + 1

Q5. Give examples of polynomials p(x), g(x), q(x) and r(x), which satisfy the division algorithm and

(i) deg p(x) = deg q(x)

(ii) deg q(x) = deg r(x)

(iii) deg r(x) = 0

Solution:

Using Euclid’s Division algorithm:

p(x) = g(x) × q(x) + r(x)  where q(x) – 0

(i) deg p(x) = deg q(x)

The deg of dividend p(x) and quotient q(x) can be equal when deg of divisor is 0 or any number.

Example : Let p(x) = 2x² - 6x + 3

And let g(x) = 2

On dividing  

p(x) = 2x² - 6x + 2 + 1

     = 2(x² - 3x + 1) + 1

Now comparing 2(x² - 3x + 1) + 1 by p(x) = g(x) × q(x) + r(x) we get:

So, q(x) = x² - 3x + 1and r(x) = 1

By which we obtain deg p(x) = deg q(x)

(ii) deg q(x) = deg r(x)

Solution: This situation comes when deg p(x) and deg g(x) is equal-

Let p(x) = 2x² + 6x + 7 and g(x) = x² + 3x + 2

On dividing: q(x) = 2 and r(x) = 3

Therefore, deg q(x) = deg r(x)

(iii) deg r(x) = 0

Solution:

r(x) = 0 is obtained when p(x) is completely divisible by g(x):

Let p(x) = x² – 1 and g(x) = x + 1

On dividing we obtain:

q(x) = x – 1 and r(x) = 0