Polynomials
Exercise 2.4
1. Verify that the numbers given alongside of the cubic polynomials below are their zeroes. Also verify the relationship between the zeroes and the coefficients in each case:
(i) 2x3+x2-5x+2; -1/2, 1, -2
Solution:
To verify that the given numbers are the zeroes of the cubic polynomial 2x^3 + x^2 – 5x + 2, we can substitute each value into the polynomial and check if the result is zero.
(i) Verify -1/2 as a zero: Substituting x = -1/2 into the polynomial: 2(-1/2)^3 + (-1/2)^2 – 5(-1/2) + 2 = 0 -1/4 + 1/4 + 5/2 + 2 = 0 0 = 0 (True)
(ii) Verify 1 as a zero: Substituting x = 1 into the polynomial: 2(1)^3 + (1)^2 – 5(1) + 2 = 0 2 + 1 – 5 + 2 = 0 0 = 0 (True)
(iii) Verify -2 as a zero: Substituting x = -2 into the polynomial: 2(-2)^3 + (-2)^2 – 5(-2) + 2 = 0 -16 + 4 + 10 + 2 = 0 0 = 0 (True)
All three given values, -1/2, 1, and -2, are indeed zeroes of the cubic polynomial 2x^3 + x^2 – 5x + 2.
Now let’s examine the relationship between the zeroes and the coefficients of the polynomial. For a cubic polynomial in the form ax^3 + bx^2 + cx + d, the relationship is as follows:
- Sum of the zeroes: The sum of the zeroes is given by -b/a. In this case, the sum of the zeroes is -(1/2)/2 = -1/4.
- Product of the zeroes: The product of the zeroes is given by -d/a. In this case, the product of the zeroes is -(2)/2 = -1.
Therefore, the relationship between the zeroes and the coefficients is:
- Sum of the zeroes = -1/4
- Product of the zeroes = -1
These relationships hold true for the given cubic polynomial.
(ii) x3-4x2+5x-2 ;2, 1, 1
Solution:
To verify that the given numbers are the zeroes of the cubic polynomial x^3 – 4x^2 + 5x – 2, we can substitute each value into the polynomial and check if the result is zero.
(ii) Verify 2 as a zero:
Substituting x = 2 into the polynomial:
(2)^3 – 4(2)^2 + 5(2) – 2 = 0
8 – 16 + 10 – 2 = 0
0 = 0 (True)
(iii) Verify 1 as a zero:
Substituting x = 1 into the polynomial:
(1)^3 – 4(1)^2 + 5(1) – 2 = 0
1 – 4 + 5 – 2 = 0
0 = 0 (True)
(iv) Verify -1 as a zero:
Substituting x = -1 into the polynomial:
(-1)^3 – 4(-1)^2 + 5(-1) – 2 = 0
-1 – 4 – 5 – 2 = 0
-12 = 0 (False)
Only the values 2 and 1 are zeroes of the cubic polynomial x^3 – 4x^2 + 5x – 2.
Now let’s examine the relationship between the zeroes and the coefficients of the polynomial. For a cubic polynomial in the form ax^3 + bx^2 + cx + d, the relationship is as follows:
1) Sum of the zeroes: The sum of the zeroes is given by -b/a. In this case, the sum of the zeroes is -(4)/1 = -4.
2) Product of the zeroes: The product of the zeroes is given by -d/a. In this case, the product of the zeroes is -(2)/1 = -2.
Therefore, the relationship between the zeroes and the coefficients is:
- Sum of the zeroes = -4
- Product of the zeroes = -2
These relationships hold true for the given cubic polynomial x^3 – 4x^2 + 5x – 2.
2. Find a cubic polynomial with the sum, sum of the product of its zeroes taken two at a time, and the product of its zeroes as 2, –7, –14 respectively.
Solution:
To find a cubic polynomial with specific conditions on the zeroes, we can use Vieta’s formulas. Vieta’s formulas relate the coefficients of a polynomial to the sums and products of its zeroes.
Given conditions:
Sum of the zeroes = 2
Sum of the product of zeroes taken two at a time = -7
Product of the zeroes = -14
Let the cubic polynomial be of the form:
p(x) = ax^3 + bx^2 + cx + d
According to Vieta’s formulas:
1) Sum of the zeroes:
The sum of the zeroes is given by the negative ratio of the coefficient of the quadratic term to the coefficient of the cubic term.
Sum of the zeroes = -b/a = 2
2) Sum of the product of zeroes taken two at a time:
The sum of the product of zeroes taken two at a time is given by the positive ratio of the coefficient of the linear term to the coefficient of the cubic term.
Sum of product of zeroes taken two at a time = c/a = -7
3) Product of the zeroes:
The product of the zeroes is given by the ratio of the constant term to the coefficient of the cubic term.
Product of the zeroes = -d/a = -14
From these equations, we can set up a system of equations to solve for the coefficients a, b, c, and d.
Using the given conditions:
-b/a = 2 (Equation 1)
c/a = -7 (Equation 2)
-d/a = -14 (Equation 3)
From Equation 1, we have b = -2a.
Substituting b = -2a into Equation 2, we get c/a = -7.
Multiplying both sides by a, we have c = -7a.
Substituting c = -7a into Equation 3, we get -d/a = -14.
Simplifying, we have d = 14a.
Now, we can express the cubic polynomial p(x) using the coefficients:
p(x) = ax^3 + bx^2 + cx + d
= ax^3 + (-2a)x^2 + (-7a)x + (14a)
= ax^3 – 2ax^2 – 7ax + 14a
We have found a cubic polynomial that satisfies the given conditions. The specific value of ‘a’ can be chosen arbitrarily to obtain different cubic polynomials with the desired properties.
3. If the zeroes of the polynomial x3-3x2+x+1 are a – b, a, a + b, find a and b.
Solution:
Given that the zeroes of the polynomial x^3 – 3x^2 + x + 1 are a – b, a, and a + b, we can set up the following equations:
Sum of the zeroes = -(-3)/1 = 3/1 = 3 = -b/a + a + a + b Simplifying the equation, we get: 3 = 3a a = 1
Product of the zeroes = -1/1 = -1/1 = -1 = (a – b)(a)(a + b) Expanding the equation, we get: -1 = (a^2 – b^2)(a) -1 = (a^3 – b^2a) Since we know that a = 1, we can substitute it into the equation: -1 = (1^3 – b^2) -1 = (1 – b^2) -1 = 1 – b^2 b^2 = 2 b = ±√2
Therefore, the values of a and b that satisfy the given conditions are: a = 1 b = ±√2
Hence, the correct zeroes of the polynomial x^3 – 3x^2 + x + 1 are 1 – √2, 1, and 1 + √2.
4. If two zeroes of the polynomial x4-6x3-26x2+138x-35 are 2 ±√3, find other zeroes.
Solution:
NCERT Solutions for Class 10 Maths Chapter 2 Polynomials
NCERT Solutions for Class 10 Maths Chapter 1- Real Numbers
Simple Notes on Polynomials
A polynomial is an algebraic expression consisting of variables and coefficients, combined using addition, subtraction, and multiplication operations. It can have one or more terms, where each term is a combination of a coefficient and one or more variables raised to non-negative integer exponents.
Polynomials are commonly written in the form:
P(x) = aₙxⁿ + aₙ₋₁xⁿ⁻¹ + … + a₁x + a₀
Here, P(x) represents the polynomial, x is the variable, and aₙ, aₙ₋₁, …, a₁, a₀ are the coefficients. The exponents (n, n-1, etc.) determine the degree of the polynomial, which is the highest power of the variable.
Polynomials can have different degrees, such as linear (degree 1), quadratic (degree 2), cubic (degree 3), and so on. The degree of a polynomial affects its behavior and the number of zeroes it can have.
Zeroes or roots of a polynomial are the values of the variable for which the polynomial evaluates to zero. Finding the zeroes of a polynomial can provide insights into its behavior, factorization, and graphical representation.
Polynomials can be manipulated using various operations, including addition, subtraction, multiplication, and division. Division of polynomials involves dividing one polynomial by another to obtain a quotient and a remainder.
Polynomial have wide applications in various fields, including algebra, calculus, physics, engineering, and computer science. They provide a flexible framework for modeling and solving problems involving variables and equations.
Understanding polynomials and their properties is essential for solving equations, graphing functions, and working with mathematical models in various disciplines.
NCERT Solutions for Class 10 Maths Chapter 1- Real Numbers