Class10 NCRT Polynomials Exercise – 2.3 pdf || UP Board

Polynomials
Exercise – 2.3

 

1. 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

Solution:

Quotient = x – 2 and remainder = 7x – 9

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

Solution:

Quotient = x² + x - 3 and remainder = 8

(iii) p(x) = x⁴-5x+6, g(x) = 2 – x²

Solution:

Quotient = -x² -2 and remainder = -5x + 10

 

2. 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

Solution:

Since, remainder is 0, therefore the polynomial t² – 3 is a factor of the polynomial 2t⁴ + 3t³ – 2t² – 9t – 12.

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

Solution: 

Since, remainder is 0, therefore the polynomial x² + 3x + 1 is a factor of the polynomial 3x⁴ + 5x³ – 7x² +2x +2.

(iii) x³ -3x + 1, x⁵ -4x³ +x² +3x +1

Solution: 

The polynomial x³ -3x + 1 is not a factor of the polynomial x⁵ -4x³ +x² +3x +1 because the remainder is 0 not 2.

3. Obtain all other zeroes of 3x⁴ +6x³ -2x² -10x -5, if two of its zeroes are √(5/3) and -√(5/3).

Solution:

Let p(x) = 3x⁴ +6x³ -2x² -10x -5

Therefore, x =√(5/3), x =-√(5/3)

Or

3x² – 5 = 0

3x² – 5 is a factor of p(x),

 by division method,

p(x) = (3x² -5) (x² + 2x +1)

Now, on factoring x² + 2x +1,

x² + 2x +1 = 0

x² + x + x +1 = 0

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

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

x = -1, -1

Hence two other zeroes are -1, -1.

4. 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) = ?

We know that,

p(x) = g(x)× q(x) + r(x)

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

 by division method,

Hence, g(x) = x² - x +1

 

5. 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)

Solution:

The degree of the dividend and the quotient can be equal only if the divisor is a constant (to the degree 0) number.

Let p(x) = 3x² -6x +5

Let g(x) = 3

Therefore, q(x) = x² -2x +1 and r(x) = 2

 

(ii) deg q(x) = deg r(x)

Solution:

Let p(x) = 2x² -4x +3

Let g(x) = x² -2x +1

Therefore, q(x) = 2 and r(x) = 1

 

(iii) deg r(x) = 0

Solution:

Let p(x) = 2x² -4x +3

Let g(x) = x² -2x +1

Therefore, q(x) = 2 and r(x) = 1 

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