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Answers to NCERT Class 10 Maths Chapter 2 – Polynomials


Exercise 2.1 Page: 28
1. The graphs of y = p(x) are given in Fig. 2.10 below, for some
polynomials p(x). Find the number of zeroes of p(x), in each case.
Solutions:
Graphical method to find zeroes:-
Total number of zeroes in any polynomial equation is equal to the total number of
times the curve intersects x-axis.
(i) In the graph, the number of zeroes of p(x) is 0 because the graph that is parallel to
x-axis does not cut it at any point.
(ii) In the graph, the number of zeroes of p(x) is 1 because the graph intersected the
x-axis at only one point.
(iii) In the graph, the number of zeroes of p(x) is 3 because the graph intersected the
x-axis at any three points.
(iv) In the graph, the number of zeroes of p(x) is 2 because the graph intersected the
x-axis at two points.
(v) In the given graph, the number of zeroes of p(x) is 4 because the graph
intersected the x-axis at four points.
(vi) In the given graph, the number of zeroes of p(x) is 3 because the graph
intersected the x-axis at three points.

Exercise 2.2 Page: 33
1. Find the zeroes of the following quadratic polynomials and verify
the relationship between the zeroes and the coefficients.
Solutions:
(i) x2–2x –8
⇒x2– 4x+2x–8 = x(x–4) +2(x–4) = (x-4)(x+2)
Therefore, zeroes of polynomial equation x2–2x–8 are (4, -2)
Sum of zeroes = 4 – 2 = 2 = - (-2)/1 = -(Coefficient of x)/(Coefficient of x 2)
Product of zeroes = 4 × (-2) = -8 = - 8/1 = (Constant term)/(Coefficient of x 2)
(ii) 4s2–4s+1
⇒4s2–2s–2s+1 = 2s (2s–1) –1(2s-1) = (2s–1)(2s–1)
Therefore, zeroes of polynomial equation 4s 2– 4s+1 are (1/2, 1/2)
Sum of zeroes = (½) + (1/2) = 1 = - (-4)/4 = - (Coefficient of s)/(Coefficient of s 2)
Product of zeros = (1/2) × (1/2) = 1/4 = (Constant term)/(Coefficient of s 2 )

, Answers to NCERT Class 10 Maths Chapter 2 – Polynomials

(iii) 6x2–3–7x
⇒6x2– 7x–3 = 6x2 – 9x + 2x – 3 = 3x(2x – 3) +1(2x – 3) = (3x+1)(2x-3)
Therefore, zeroes of polynomial equation 6x 2– 3 – 7x are (-1/3, 3/2)
Sum of zeroes = - (1/3) + (3/2) = (7/6) = -(Coefficient of x)/(Coefficient of x 2)
Product of zeroes = - (1/3) × (3/2) = -(3/6) = (Constant term) /(Coefficient of x 2 )
(iv) 4u2+8u
⇒ 4u (u+2)
Therefore, zeroes of polynomial equation 4u 2 + 8u are (0, -2).
Sum of zeroes = 0 + (-2) = -2 = - (8/4) = = -(Coefficient of u)/(Coefficient of u 2)
Product of zeroes = 0×-2 = 0 = 0/4 = (Constant term)/(Coefficient of u 2 )
(v) t2–15
⇒ t2 = 15 or t = ±√15
Therefore, zeroes of polynomial equation t 2 –15 are (√15, -√15)
Sum of zeroes =√15 + (-√15) = 0= - (0/1)= -(Coefficient of t) / (Coefficient of t 2)
Product of zeroes = √15 × (-√15) = -15 = -15/1 = (Constant term) / (Coefficient of t 2 )
(vi) 3x2–x–4
⇒ 3x2– 4x+3x– 4 = x (3x-4) +1(3x-4) = (3x – 4)(x + 1)
Therefore, zeroes of polynomial equation 3x 2 – x – 4 are (4/3, -1)
Sum of zeroes = (4/3) + (-1) = (1/3) = - (-1/3) = - (Coefficient of x) / (Coefficient of x 2)
Product of zeroes= (4/3) × (-1) = (-4/3) = (Constant term) /(Coefficient of x 2 )
2. Find a quadratic polynomial each with the given numbers as the sum and
product of its zeroes, respectively.
(i) 1/4 , -1
Solution:
From the formulas of sum and product of zeroes, we know,
Sum of zeroes = α + β = 1/4
Product of zeroes = α β = -1
Product of zeroes = α β ∴ If α and β are zeroes of any quadratic polynomial, then the
quadratic polynomial equation can be written directly as:-
x2 – (α+β)x + αβ = 0
x2 – (1/4)x + (-1) = 0
4x2– x- 4 = 0