NCERT Solutions: Polynomials (Exercise 2.2)

# NCERT Solutions for Class 10 Maths Chapter 2 - Polynomials (Exercise 2.2)

Q1: Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients.
(i) x2–2x –8
Sol: ⇒ 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
Sol: ⇒ 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
Sol: ⇒ 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
Sol: ⇒ 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
Sol: ⇒ 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
Sol: ⇒ 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
Sol:
From the formulas of sum and product of zeroes, we know,
Sum of zeroes = α+β
Product of zeroes = αβ
Sum of zeroes = α+β = 1/4
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–(1/4)x +(-1) = 0
4x2–x-4 = 0
Thus,4x2–x–4 is the quadratic polynomial.

(ii)√2, 1/3
Sol:
Sum of zeroes = α + β =√2
Product of zeroes = αβ = 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
Sol:
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
Sol:
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+1 is the quadratic polynomial.

(v) -1/4, 1/4
Sol:
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
Sol:
Given,
Sum of zeroes = α+β = 4
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.

Check out the NCERT Solutions of all the exercises of Polynomials:

Exercise 2.1. NCERT Solutions: Polynomials

Exercise 2.3 NCERT Solutions: Polynomials

Exercise 2.4 NCERT Solutions: Polynomials

The document NCERT Solutions for Class 10 Maths Chapter 2 - Polynomials (Exercise 2.2) is a part of the Class 10 Course NCERT Textbooks & Solutions for Class 10.
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## FAQs on NCERT Solutions for Class 10 Maths Chapter 2 - Polynomials (Exercise 2.2)

 1. What is a polynomial?
Ans. A polynomial is an algebraic expression consisting of variables, coefficients, and exponents. It is made up of terms that are combined using addition, subtraction, and multiplication operations. For example, 2x^3 - 5x^2 + 3x - 7 is a polynomial.
 2. What are the different types of polynomials?
Ans. There are various types of polynomials based on the number of terms they have. Some common types include monomials (1 term), binomials (2 terms), trinomials (3 terms), and polynomials with more than 3 terms.
 3. How can I identify the degree of a polynomial?
Ans. The degree of a polynomial is the highest power of the variable in the polynomial. To identify the degree, look for the term with the highest exponent. For example, in the polynomial 3x^2 + 4x - 1, the degree is 2.
 4. What is the zero of a polynomial?
Ans. The zero of a polynomial is a value of the variable that makes the polynomial equal to zero. In other words, it is the solution to the equation obtained by setting the polynomial equal to zero. For example, if we have the polynomial 2x^2 - 5x + 3, the zeros would be the values of x that satisfy the equation 2x^2 - 5x + 3 = 0.
 5. How can I divide polynomials using long division?
Ans. To divide polynomials using long division, follow these steps: 1. Arrange the polynomials in descending order of the variable's exponent. 2. Divide the term with the highest exponent of the dividend by the term with the highest exponent of the divisor. 3. Multiply the quotient obtained by the entire divisor. 4. Subtract the product obtained in step 3 from the dividend. 5. Repeat steps 2-4 until the degree of the remaining polynomial is less than the degree of the divisor. The final result will be the quotient and any remaining terms will be the remainder.

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