Class 10 NCERT Solutions- Chapter 2 Polynomials – Exercise 2.3
Question 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
(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
R = 7x-9
Q = x-3
ii) p(x) = x4 – 3x2 + 4x + 5, g(x) = x2 + 1 – x
R = 8
Q = x2+x-3
iii) p(x) = x4– 5x + 6, g(x) = 2 – x2
Q = -x2-2
R = -5x+10
Question 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
(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
Q = 2t3+3t+4
R = 0
Yes 1st polynomial is factor of 2nd polynomial.
ii) x2 + 3x + 1, 3x4 + 5x3 – 7x2 + 2x + 2
R = 0
Q = 3x2-4x+2
iii) x3 – 3x + 1, x5 – 4x3 + x2 + 3x + 1
R = x2-1
Q = 2
Question 3. Obtain all other zeroes of 3x4 + 6x3 – 2x2 – 10x – 5, if two of its zeroes are and √(5/3) and -√(5/3).
Solution:
R = 0
Q = 3x2+6x+3
∴ we are factorizing
3x2+6x+3
x2+2x+1
(x+1)2
(x+1) (x+1) = 0
∴ x = -1 and x = -1
Question 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:
Dividend = Divisor * Quotient + Remainder
x3-3x2+3x-2/x-2
R = 0
Q = x2 -x +1
Answer: g(x)=x2-x+1
Question 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)
(ii) deg q(x) = deg r(x)
(iii) deg r(x) = 0
Solution:
i) deg p(x) = deg q(x)
p(x)=2x2-2x+14, g(x)=2
p(x)/g(x)=2x2-2x+14/2=(x2-x+7)
=x2-x+7=q(x)
=q(x)=x2-x+7
r(x)=0
ii) deg q(x)=deg r(x)
p(x)=4x2+4x+4, g(x)=x2+x+1
q(x) = 4
r(x) = 0
∴Here deg q(x)=deg r(x)
iii) deg r(x)=0
p(x)=x3+2x2-x+2 ,g(x)=x2-1
q(x) = x+2
r(x) = 4
deg of r(x) = 0
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