Class 10 NCERT Solutions Chapter 2 Polynomials Exercise 2.2 (original) (raw)
Last Updated : 26 Aug, 2024
Exercise 2.2: Division Algorithm for Polynomials
In this exercise, you'll delve into the concept of dividing polynomials. Similar to how you divide numbers, you can divide polynomials to find a quotient and remainder. The key idea is the division algorithm, which states that for any polynomials p(x) and g(x) (where g(x) ≠ 0), there exist unique polynomials q(x) and r(x) such that:
p(x) = g(x) * q(x) + r(x)
Here, q(x) is the quotient, and r(x) is the remainder. The degree of r(x) is always less than the degree of g(x).
This exercise will guide you through the process of polynomial long division, helping you find quotients and remainders. You'll also apply the division algorithm to solve various polynomial-related problems.
Understanding the Division Algorithm
The division algorithm is a fundamental concept in polynomial algebra. It's similar to the division process you learned for numbers, but it operates on expressions with variables and exponents. Grasping this algorithm is crucial for further studies in algebra and its applications.
**Question 1. Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients.
****(i) x** 2 **– 2x – 8
x2 – 2x – 8 = x2 – 4x + 2x – 8
= x (x – 4) + 2(x – 4)
= (x - 4) (x + 2)
Therefore, zeroes of equation x2 – 2x – 8 are (4, -2)
Sum of zeroes is equal to [4 – 2]= 2 = -(-2)/1
i.e. = -(Coefficient of x) / (Coefficient of x2)
Product of zeroes is equal to 4 × (-2) = -8 =-(8)/1
i.e.= (Constant term) / (Coefficient of x2)
****(ii) 4s2 – 4s + 1**
4s2 – 4s + 1 = 4s2 – 2s – 2s +1
= 2s(2s – 1) – 1(2s - 1)
= (2s – 1) (2s – 1)
Therefore, zeroes of equation 4s2 – 4s +1 are (1/2, 1/2)
Sum of zeroes is equal to [(1/2) + (1/2)] = 1 = -4/4
i.e.= -(Coefficient of s) / (Coefficient of s2)
Product of zeros is equal to [(1/2) × (1/2)] = 1/4
i.e.= (Constant term) / (Coefficient of s2 )
****(iii) 6x** 2 **– 3 – 7x
6x2 – 3 – 7x = 6x2 – 7x – 3
= 6x2 – 9x + 2x – 3
= 3x(2x – 3) + 1(2x – 3)
= (3x + 1) (2x - 3)
Therefore, zeroes of equation 6x2 – 3 – 7x are (-1/3, 3/2)
Sum of zeroes is equal to -(1/3) + (3/2) = (7/6)
i.e.= -(Coefficient of x) / (Coefficient of x2)
Product of zeroes is equal to -(1/3) × (3/2) = -(3/6)
i.e.= (Constant term) / (Coefficient of x2 )
****(iv) 4u** 2 **+ 8u
4u2 + 8u = 4u(u + 2)
Therefore, zeroes of equation 4u2 + 8u are (0, -2).
Sum of zeroes is equal to [0 + (-2)] = -2 = -(8/4)
i.e. = -(Coefficient of u) / (Coefficient of u2)
Product of zeroes is equal to 0 × -2 = 0 = 0/4
i.e. = (Constant term) / (Coefficient of u2 )
****(v) t** 2 **– 15
t2 – 15
⇒ t2 = 15 or t = ±√15
Therefore, zeroes of equation t2 – 15 are (√15, -√15)
Sum of zeroes is equal to [√15 + (-√15)] = 0 = -(0/1)
i.e.= -(Coefficient of t) / (Coefficient of t2)
Product of zeroes is equal to √15 × (-√15) = -15 = -15/1
i.e. = (Constant term) / (Coefficient of t2 )
****(vi) 3x** 2 – x – 4
3x2 – x – 4 = 3x2 – 4x + 3x – 4
= x(3x - 4) + 1(3x - 4)
= (3x – 4) (x + 1)
Therefore, zeroes of equation 3x2 – x – 4 are (4/3, -1)
Sum of zeroes is equal to (4/3) + (-1) = (1/3) = -(-1/3)
i.e. = -(Coefficient of x) / (Coefficient of x2)
Product of zeroes is equal to (4/3) × (-1) = (-4/3)
i.e. = (Constant term) / (Coefficient of x2)
**Question 2. Find a quadratic polynomial each with the given numbers as the sum and product of its zeroes respectively.
****(i) 1/4, -1**
Let two zeroes be α, β
∴ Sum of zeroes = α + β
∴ Product of zeroes = αβ
Given, 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
∴ 4x2 – x – 4 is the quadratic polynomial.
****(ii) √2, 1/3**
Let two zeroes be α, β
∴ Sum of zeroes = α + β
∴ Product of zeroes = αβ
Given 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
∴ 3x2 - 3√2x + 1 is the quadratic polynomial.
****(iii) 0, √5**
Let two zeroes be α, β
∴ Sum of zeroes = α + β
∴ Product of zeroes = αβ
Given, Sum of zeroes = α + β = 0
Product of zeroes = αβ = √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
∴ x2 + √5 is the quadratic polynomial.
****(iv) 1, 1**
Let two zeroes be α, β
∴ Sum of zeroes = α + β
∴ Product of zeroes = αβ
Given, Sum of zeroes = α + β = 1
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 – x + 1 = 0
∴ x2 – x + 1 is the quadratic polynomial.
****(v) -1/4, 1/4**
Let two zeroes be α, β
∴ Sum of zeroes = α + β
∴ Product of zeroes = αβ
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
∴ 4x2 + x + 1 is the quadratic polynomial.
****(vi) 4, 1**
Let two zeroes be α, β
∴ Sum of zeroes = α + β
∴ Product of zeroes = αβ
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
∴ x2 – 4x + 1 is the quadratic polynomial.
Summary
Exercise 2.2 of NCERT Class 10 Chapter 2 - Polynomials focuses on the division algorithm for polynomials. It introduces students to the concept that for polynomials p(x) and g(x), we can find polynomials q(x) and r(x) such that p(x) = g(x)q(x) + r(x), where the degree of r(x) is less than the degree of g(x). The exercise covers long division of polynomials, finding quotients and remainders, and using the division algorithm to solve various polynomial-related problems.