Chapter 2 Polynomials NCERT Exemplar Solutions Exercise 2.2 Class 10 Maths
Chapter Name  NCERT Maths Exemplar Solutions for Chapter 2 Polynomials Exercise 2.2 
Book Name  NCERT Exemplar for Class 10 Maths 
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Related Study  NCERT Solutions for Class 10 Maths 
Exercise 2.2 Solutions
Short Answer Questions with Reasoning:
1. Answer the following and justify:
(i) Can x^{2} – 1 be the quotient on division of x^{6} + 2x^{3} + x  1 by a polynomial in x of degree 5 ?
(ii) What will the quotient and remainder be on division of ax^{2} + bx + c by px^{3} + qx^{2} + rx + s, p ≠ 0 ?
(iii) If on division of a polynomial p(x) by a polynomial g(x), the quotient is zero, what is the relation between the degrees of p (x) and g (x)?
(iv) If on division of a nonzero polynomial p (x) by a polynomial g (x), the remainder is zero, what is the relation between the degrees of p (x) and g (x)?
(v) Can the quadratic polynomial x^{2} + kx + k have equal zeroes for some odd integer k ? 1 ?
Solution
(i) No, x^{2}  1 cannot be the quotient on division of x^{6} + 2x^{3} + x  1 by a polynomial in x of degree 5.
Explanation :
When a polynomial with degree 6 is divided by degree 5 polynomial, the quotient will be of degree 1.
Assuming that (x^{2}  1) divides the degree 6 polynomial and the quotient obtained is degree 5 polynomial.
As a = bq + r, so,
(Degree 6 polynomial) = (x^{2}  1) (degree 5 polynomial) + r(x)
= (degree 7 polynomial) + r(x)
[As, (x^{2} term × x^{5} term = x^{7} term)]
= (degree 7 polynomial)
So, this contradict our assumption.
Hence, x^{2}  1 cannot be the quotient on division of x^{6} + 2x^{3} + x  1 by a polynomial in x of degree 5.
(ii) Degree of the polynomial px^{3} + qx^{2} + rx + s = 3
Degree of the polynomial ax^{2} + bx + c = 2
Here, Degree of px^{3} + qx^{2} + rx + s is greater than degree of the ax^{2} + bx + c
Hence, the quotient would be zero,
Therefore, the remainder would be the dividend = ax^{2} + bx + c.
(iii) We have, p(x) = g(x) × q(x) + r(x)
As given in the question,
q(x) = 0
When q(x) = 0,
r(x) is also = 0
Here, when we divide p(x) by g(x),
Then, p(x) should be = 0
Therefore, the relation between the degrees of p(x) and g(x) is the degree p(x) < degree g(x).
(iv) To divide p(x) by g(x)
We have,
Degree of p(x) > degree of g(x)
or
Degree of p(x) = degree of g(x)
Hence, the relation between the degrees of p(x) and g(x) is degree of p(x) > degree of g(x)
(v) A Quadratic Equation has equal roots when :
b^{2}  4ac = 0
Given,
x^{2} + kx + k = 0
a = 1,
b = k,
x = k
Putting values in the equation we get,
k^{2}  4(1)(k) = 0
⇒ k^{3}  4k = 0
⇒ k(k  4) = 0
⇒ k = 0,
⇒ k = 4
Here, it is given that k is greater than I, So, the value of k is 4 if the equation has common roots.
If k = 4, then the equation (x^{2} + kx + k) will have equal roots.
2. Are the following statements ‘True’ or ‘False’? Justify your answers.
(i) If the zeroes of a quadratic polynomial ax^{2} + bx + c are both positive, then a, b and c all have the same sign.
(ii) If the graph of a polynomial intersects the x  axis at only one point, it cannot be a quadratic polynomial.
(iii) If the graph of a polynomial intersects the x  axis at exactly two points, it need not be a quadratic polynomial.
(iv) If two of the zeroes of a cubic polynomial are zero, then it does not have linear and constant terms.
(v) If all the zeroes of a cubic polynomial are negative, then all the coefficients and the constant term of the polynomial have the same sign.
(vi) If all three zeroes of a cubic polynomial x^{3} + ax^{2}  bx + c are positive, then at least one of a, b and c is non  negative.
(vii) The only value of k for which the quadratic polynomial kx^{2} + x + k has equal zeros is 1/2.
Solution
(i) False
Taking Î± and Î² as the roots of he quadratic polynomial. If Î± and Î² are positive then Î± + Î² = b/a. For making sum of roots positive either b or a must be negative.
(ii) False
The statement is false, because when two zeroes of a quadratic polynomial are equal, then two intersecting points coincide to become one point.
(iii) True
If a polynomial of degree more than two has two zeroes and other zeroes are not real or are imaginary, then graph of the polynomial will intersect at two points on xaxis.
(iv) True
Taking,
Î² = 0,
Î³ = 0
f(x) = (x  Î±)(x  Î²)(x  Î³)
= (x  Î±) x . x
f(x) = x^{3}  ax^{2}
So, it has no linear (coefficient of x) and constant terms.
(v) True
Î±, Î² and Î³ are all negative for cubic polynomial ax^{3} + bx^{2} + cx + d.
Here, d and a will have same sign.
So, sign of b, c, d are same as of a.
Signs of a, b, c, d will be same either positive or negative.
(vi) False: As all zeroes of cubic polynomial are positive
let f(x) = x^{3} + ax^{2}  bx + c
Î± + Î² + Î³ = positive, (say + x)
a and d have opposite signs.
Therefore, we can conclude that,
From (i) if a is positive, then b is negative.
From (ii) if a is positive, then c is also positive.
From (iii) if a is positive, then d is negative.
So, if zeroes Î±, Î², Î³ of cubic polynomial are positive then out of a, b, c at least one is negative.
(vii) False
So, the values of k are ± 1/2 so that the given equation has equals roots.