Chapter 2 : Polynomials
Ncert maths solutions for class 10 chapter 2 Polynomials , serve as an invaluable companion for students embarking on their mathematical journey. This chapter introduces students to the world of polynomials. NCERT solutions for Chapter 2 are designed to simplify the learning process and foster a genuine appreciation for the elegance of polynomials in mathematics.
Exercise 2.1 Polynomials
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.
Solution:
(i) The graph is parallel to x-axis and not cut it at any point hence the number of zeroes of p(x) is 0 .
(ii) The graph intersects the x-axis at only one point hence the number of zeroes of p(x) is 1 .
(iii) The graph intersects the x-axis at three points hence the number of zeroes of p(x) is 3 .
(iv) The graph intersects the x-axis at two points hence the number of zeroes of p(x) is 2 .
(v) The graph intersects the x-axis at four points hence the number of zeroes of p(x) is 4 .
(vi) The graph intersects the x-axis at three points hence the number of zeroes of p(x) is 3 .
Exercise 2.2 Polynomials
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)
x – 4 = 0 or x + 2 = 0
⇒ x = 4 or x = -2
Therefore, zeroes of polynomial = 4, -2
Sum of zeroes = 4–2 = 2 =
Again sum of the zeroes using formula = -(Coefficient of x)/(Coefficient of x2)
= -(-2)/1 = 2
Product of zeroes = 4×(-2) = -8
Again product of zeroes using formula = (Constant term)/(Coefficient of x2)
=-(8)/1 = -8
verify
(ii) 4s2–4s+1
⇒4s2–2s–2s+1
= 2s(2s–1)–1(2s-1)
= (2s–1)(2s–1)
Therefore, zeroes of polynomial = 1/2, 1/2
Sum of zeroes = (½)+(1/2) = 1
Again sum of the zeroes using formula = -(Coefficient of s)/(Coefficient of s2)
= -(-4)/4 = 1
Product of zeros = (1/2)×(1/2) = 1/4
Again product of the zeroes using formula = (Constant term)/(Coefficient of s2 )
= 1/4
verify
(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 = -1/3, 3/2
Sum of zeroes = -(1/3)+(3/2) = (7/6)
Again sum of the zeroes using formula = -(Coefficient of x)/(Coefficient of x2)
= -(-7)/6 = 7/6
Product of zeroes = -(1/3)×(3/2) = -3/6
Again product of the zeroes using formula = (Constant term) /(Coefficient of x2 )
= -3/6
(iv) 4u2+8u
⇒ 4u(u+2)
Therefore, zeroes of polynomial = 0, -2
Sum of zeroes = 0+(-2) = -2
Again sum of the zeroes using formula = -(Coefficient of u)/(Coefficient of u2)
= -(8/4) = -2
Product of zeroes = 0×-2 = 0
Again product of the zeroes using formula = (Constant term)/(Coefficient of u2 )
= 0/4
verify
(v) t2–15
⇒ t2 = 15 or t = ±√15
Therefore, zeroes of polynomial equation t2 –15 are (√15, -√15)
Sum of zeroes =√15+(-√15) = 0= -(0/1)= -(Coefficient of t) / (Coefficient of t2)
Product of zeroes = √15×(-√15) = -15 = -15/1 = (Constant term) / (Coefficient of t2 )
verify
(vi) 3x2–x–4
⇒ 3x2–4x+3x–4
= x(3x-4)+1(3x-4)
= (3x – 4)(x + 1)
Therefore, zeroes of polynomial = (4/3, -1)
Sum of zeroes = (4/3)+(-1) = (1/3)
Again sum of the zeroes using formula = -(Coefficient of x) / (Coefficient of x2)
= -(-1/3) = 1/3
Product of zeroes=(4/3)×(-1) = (-4/3)
Again product of the zeroes using formula = (Constant term) /(Coefficient of x2 )
= (-4/3)
verify
2. Find a quadratic polynomial each with the given numbers as the sum and product of its zeroes, respectively.
(i) 1/4 , -1
Solution:
Let zeroes of the quadratic polynomial
Sum of zeroes = α+β = 1/4
Product of zeroes = α β = -1
The quadratic polynomial equation can be written as:-
x2–(α+β)x +αβ = 0
⇒ x2–(1/4)x +(-1) = 0
⇒ 4x2–x-4 = 0
Thus, 4x2–x–4 is the quadratic polynomial.
(ii)√2, 1/3
Solution:
Sum of zeroes = α + β =√2
Product of zeroes = α β = 1/3
The quadratic polynomial equation can be written as:-
x2–(α+β)x +αβ = 0
⇒ x2 –(√2)x + (1/3) = 0
⇒ 3x2-3√2x+1 = 0
Thus, 3x2-3√2x+1 is the quadratic polynomial.
(iii) 0, √5
Solution:
Given,
Sum of zeroes = α+β = 0
Product of zeroes = α β = √5
The quadratic polynomial equation can be written as:-
x2–(α+β)x +αβ = 0
⇒ x2–(0)x +√5= 0
Thus, x2+√5 is the quadratic polynomial.
(iv) 1, 1
Solution:
Given,
Sum of zeroes = α+β = 1
Product of zeroes = α β = 1
The quadratic polynomial equation can be written as:-
x2–(α+β)x +αβ = 0
⇒ x2–x+1 = 0
Thus, x2–x+1 is the quadratic polynomial.
(v) -1/4, 1/4
Solution:
Given,
Sum of zeroes = α+β = -1/4
Product of zeroes = α β = 1/4
Then the quadratic polynomial equation can be written directly as:-
x2–(α+β)x +αβ = 0
⇒ x2–(-1/4)x +(1/4) = 0
⇒ 4x2+x+1 = 0
Thus, 4x2+x+1 is the quadratic polynomial.
(vi) 4, 1
Solution:
Given,
Sum of zeroes = α+β =4
Product of zeroes = αβ = 1
Then the quadratic polynomial equation can be written directly as:-
x2–(α+β)x+αβ = 0
⇒ x2–4x+1 = 0
Thus, x2–4x+1 is the quadratic polynomial.
Ncert solution Class 10
Chapter 1: Real Number
Ncert maths solution class 10 chapter 1.1
Ncert maths solution class 10 chapter 1.2
Chapter 4:Quadratic Equation
NCERT Solutions Maths for Class 10 Chapter 4