NCERT Solutions For Class 10 Math Chapter – 2 Polynomial Exercise – 2.3
Q1. Divide the polynomial p(x) by the polynomial g(x) and find the quotient and remainder in each of the following:
- P(x) = x3 – 3x2 + 5x – 3 g(x) = x2 – x
- P(x) = x4– 3x2 + 4x + 5 g(x) = x2 + 1 – x
- P(x) = x4– 5x + 6 g(x) = 2 – x2
Solution:-
- Here p(x) = x3– 3x2 + 5x – 3 and g(x) = x2 – 2
Dividing p(x) by g(x) =
Quotient = x – 3 Remainder = 7x – 9
- P(x) = x4– 3x2 + 4x + 5 and g(x) = x2 + 1 – x
Dividing p(x) by g(x)
Quotient = x2 + x – 3 Remainder = 8
- P(x) = x4– 5x + 6 g(x) = 2 – x2
Quotient = -x2 – 2
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.
- T2– 3 , 2t4 + 3t3 – 2t2 – 9t – 12
- X2+ 3x + 1, 3x4 + 5x3 – 7x2 + 2x + 12
- X2+ 3x + 1, x5 – 4x3 + x2 + 3x + 1
Solution:-
- First polynomial = t2– 3
Second polynomial = 2t4 + 3t3 – 2t2 – 9t – 12
Dividing second polynomial by first polynomial.
Remainder is zero.
First polynomial is a factor of second polynomial.
- First polynomial = x2+ 3x + 1
Second polynomial = 3x4 + 5x3 – 7x2 + 2x + 2
Dividing second polynomial by first polynomial =
Remainder is zero.
First polynomial is a factor of second polynomial.
- First polynomial = x3– 3x + 1
Second polynomial = x5 – 4x3 + x2 + 3x + 1.
Remainder ≠ 0
First polynomial is not a factor of second polynomial.
Q3. Obtain all zeroes of 3x4 + 6x3 – 2x2 – 10x – 5 , If two of its zeroes are and √5/3 and –√5/3
Solution:-
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:-
Q5. Give examples of polynomial p(x) , q(x) and r(x) which satisfy the division algorithm and:
- deg p(x) = deg q(x)
- Deg q(x) = deg r(x)
- Deg r(x) = 0
Solution:-
- P(x) = 2x2+ 2x + 8
Q(x) = x2 + x + 4
G(x) = 2 and r(x) = 0
- P(x) = x3– x2 + 2x + 3
Q(x) = x + 1
G(x) = x2 – 1 and r(x) = 2x +2
- P(x) = x3– x2 + 2x + 3
G(x) = x2 + 2
Q(x) = x – 1 and r(x) = 5