Class 10 Exam  >  Class 10 Notes  >  Mathematics (Maths) Class 10  >  NCERT Solutions: Polynomials (Exercise 2.1 & 2.2)

NCERT Solutions for Class 10 Maths Chapter 2 - Polynomials (Exercise 2.1)

 Exercise 2.1

Q1: 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.
NCERT Solutions for Class 10 Maths Chapter 2 - Polynomials (Exercise 2.1)

Sol:
Graphical method to find zeroes:
Total number of zeroes in any polynomial equation = total number of times the curve intersects x-axis.
(i) In the given graph, the number of zeroes of p(x) is 0 because the graph is parallel to x-axis does not cut it at any point.
(ii) In the given graph, the number of zeroes of p(x) is 1 because the graph intersects the x-axis at only one point.
(iii) In the given graph, the number of zeroes of p(x) is 3 because the graph intersects the x-axis at any three points.
(iv) In the given graph, the number of zeroes of p(x) is 2 because the graph intersects the x-axis at two points.
(v) In the given graph, the number of zeroes of p(x) is 4 because the graph intersects the x-axis at four points.
(vi) In the given graph, the number of zeroes of p(x) is 3 because the graph intersects the x-axis at three points.


(Exercise 2.2)

Q1. 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 x2)
Product of zeroes = 4×(-2) = -8 =-(8)/1 = (Constant term)/(Coefficient of x2)

(ii) 4s2–4s+1
⇒ 4s2–2s–2s+1 = 2s(2s–1)–1(2s-1) = (2s–1)(2s–1)
Therefore, zeroes of polynomial equation 4s2–4s+1 are (1/2, 1/2)
Sum of zeroes = (½)+(1/2) = 1 = -4/4 = -(Coefficient of s)/(Coefficient of s2)
Product of zeros = (1/2)×(1/2) = 1/4 = (Constant term)/(Coefficient of s)

(iii) 6x2–3–7x
⇒ 6x2–7x–3 = 6x– 9x + 2x – 3 = 3x(2x – 3) +1(2x – 3) = (3x+1)(2x-3)
Therefore, zeroes of polynomial equation 6x2–3–7x are (-1/3, 3/2)
Sum of zeroes = -(1/3)+(3/2) = (7/6) = -(Coefficient of x)/(Coefficient of x2)
Product of zeroes = -(1/3)×(3/2) = -(3/6) = (Constant term) /(Coefficient of x)

(iv) 4u2+8u
⇒ 4u(u+2)
Therefore, zeroes of polynomial equation 4u2 + 8u are (0, -2).
Sum of zeroes = 0+(-2) = -2 = -(8/4) = = -(Coefficient of u)/(Coefficient of u2)
Product of zeroes = 0×-2 = 0 = 0/4 = (Constant term)/(Coefficient of u)

(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 t)

(vi) 3x2–x–4
⇒ 3x2–4x+3x–4 = x(3x-4)+1(3x-4) = (3x – 4)(x + 1)
Therefore, zeroes of polynomial equation3x2 – x – 4 are (4/3, -1)
Sum of zeroes = (4/3)+(-1) = (1/3)= -(-1/3) = -(Coefficient of x) / (Coefficient of x2)
Product of zeroes=(4/3)×(-1) = (-4/3) = (Constant term) /(Coefficient of x)

Q2. 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
Product of zeroes = α β
Sum of zeroes = α+β = 1/4
Product of zeroes = α β = -1
∴ 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
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
∴ 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 –(√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
∴ 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–(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
∴ 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–x+1 = 0
Thus , x2–x+1is the quadratic polynomial.

(v) -1/4, 1/4
Solution:
Given,
Sum of zeroes = α+β = -1/4
Product of zeroes = α β = 1/4
∴ 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 +(1/4) = 0
4x2+x+1 = 0
Thus,4x2+x+1 is the quadratic polynomial.

(vi) 4, 1
Solution:
Given,
Sum of zeroes = α+β =
Product of zeroes = αβ = 1
∴ 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
x2–4x+1 = 0
Thus, x2–4x+1 is the quadratic polynomial.

The document NCERT Solutions for Class 10 Maths Chapter 2 - Polynomials (Exercise 2.1) is a part of the Class 10 Course Mathematics (Maths) Class 10.
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FAQs on NCERT Solutions for Class 10 Maths Chapter 2 - Polynomials (Exercise 2.1)

1. What is a polynomial?
Ans. A polynomial is an algebraic expression that consists of variables, coefficients, and exponents. It is a sum of terms, where each term is a product of a coefficient and one or more variables raised to non-negative integer powers.
2. How is the degree of a polynomial determined?
Ans. The degree of a polynomial is determined by the highest power of the variable in the polynomial expression. For example, if the highest power of the variable is 3, then the degree of the polynomial is 3.
3. What is the difference between a monomial, binomial, and trinomial?
Ans. A monomial is a polynomial expression with only one term, a binomial has two terms, and a trinomial has three terms. The number of terms in a polynomial determines its name.
4. What is the Remainder Theorem in polynomials?
Ans. The Remainder Theorem states that when a polynomial f(x) is divided by a linear divisor x-a, the remainder obtained is equal to f(a). This theorem helps in finding the remainder of a polynomial when divided by a linear factor.
5. How can polynomials be used in real-life applications?
Ans. Polynomials have numerous real-life applications, such as in physics, engineering, and finance. They can be used to model and solve problems related to motion, electrical circuits, financial forecasting, and many other fields.
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