EXERCISE 2.3
Ques 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) = x^{3}  3x^{2} + 5x  3, g(x) = x^{2}  2
(ii) p(x) = x^{4}  3x^{2} + 4x + 5, g(x) = x^{2} + 1  x
(iii) p(x) = x^{4}  5x + 6, g(x) = 2  x^{2}
Sol: Here, dividend p(x) = x^{3}  3x^{2} + 5x  3
divisor g(x) = x^{2}  2
∴ We have
Thus, the quotient = (x  3) and remainder = (7x  9)
(ii) Here, dividend p(x) = x^{4 } 3x^{2} + 4x + 5
and divisor g(x) = x^{2} + 1  x
= x^{2}  x + 1
∴ We have
Thus, the quotient is (x^{2} + x  3) and remainder = 8
(iii) Here, dividend, p(x) = x^{4}  5x + 6
and divisor, g(x) = 2  x^{2}
=  x^{2} + 2
∴ We have
Thus, the quotient = –x^{2} – 2 and remainder = –5x + 10.
Ques 2: Check whether the first polynomial is a factor of the second polynomial by dividing the second polynomial by the first polynomial:
Sol. (i) Dividing 2t^{4} + 3t^{3}  2t^{2}  9t  12 by t^{2}  3, we have:
∵ Remainder = 0
∴ (t^{2}  3) is a factor of 2t^{4} + 3t^{3}  2t^{2}  9t  12.
(ii) Dividing 3x^{4} + 5x^{3}  7x^{2} + 2x + 2 by x^{2} + 3x + 1, we have:
∵ Remainder = 0.
∴ x^{2} + 3x + 1 is a factor of 3x^{4} + 5x^{3}  7x^{2 }+ 2x + 2.
(iii) Dividing x^{5}  4x^{3} + x^{2} + 3x + 1 by x^{3}  3x + 1, we get:
∴ The remainder = 2, i.e., remainder ≠ 0
∴ x^{3}  3x + 1 is not a factor of x^{5 } 4x^{3} + x^{2} + 3x + 1.
Ques 3: Obtain all other zeroes of 3x^{4} + 6x^{3}  2x^{2}  10x  5, if two of its zeroes are  .
Sol: We have p(x) = 3x^{4} + 6x^{3 } 2x^{2}  10x  5
α = and β = 
∴ is a factor of p (x).
Now, let us divide 3x^{4} + 6x^{3}  2x^{2}  10x  5 by .
For p (x) = 0, we have
i.e.,
or 3x + 3 = 0 ⇒ x =  1
or x + 1 = 0 ⇒ x =  1
Thus, all the other zeroes of the given polynomial are  1 and  1.
Ques 4: On dividing x^{3 } 3x^{2} + x + 2 by a polynomial g(x), the quotient and remainder were x  2 and 2x + 4 respectively. Find g(x).
Sol: Here,
Dividend p(x) = x^{3}  3x^{2} + x + 2
Divisor = g(x)
Quotient = (x  2)
Remainder = (2x + 4)
Since,
(Quotient × Divisor) + Remainder = Dividend
∴ [(x − 2) × g(x)] + [(−2x + 4)] = x^{3} − 3x^{2} + x + 2
⇒ (x − 2) × g(x)= x^{3} − 3x^{2} + x + 2 − (−2x + 4)
= x^{3} − 3x^{2} + x + 2 + 2x −4
= x^{3}  3x^{2} + 3x  2
∴
Now, dividing x^{3}  3x^{2 }+ 3x  2 by x  2, we have
∴
Thus, the required divisor g(x) = x^{2 } x + 1.
Ques 5: Give example 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
Sol:
(i) p(x), g(x), q(x), r(x)
deg p(x) = deg q(x)
∴ both g(x) and r(x) are constant terms.
p(x) = 2x^{2} 2x + 14
g(x) = 2
q(x) = x^{2}  x + 7
r(x) = 0
(ii) deg q(x) = deg r(x)
∴ this is possible when deg of both q(x) and r(x) should be less than p(x) and g(x).
p(x) = x^{3} + x^{2} + x + 1
g(x) = x^{2}  1
q(x) = x + 1, r(x) = 2x + 2
(iii) deg r(x) is 0.
This is possible when product of q(x) and g(x) form a polynomial whose degree is equal to degree of p(x) and constant term.
NOTE: We have given one example for each of the above cases, however, there can be several examples for them.
NCERT Solutions: Polynomials (Exercise 2.4) Doc  5 pages 
1. What is a polynomial? 
2. How do you identify the degree of a polynomial? 
3. What is the difference between a monomial and a binomial? 
4. How do you add or subtract polynomials? 
5. Can a polynomial have a negative degree? 
NCERT Solutions: Polynomials (Exercise 2.4) Doc  5 pages 

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