NCERT Solutions for Class 10 Maths Chapter 2 Polynomials Exercise 2.4 are provided here to aid students in their studies. These solutions are prepared by our subject expert in Maths to let the students in Class 10 first term exam prepare well. These experts create NCERT Solutions for Maths so that it would help students to solve the NCERT problems easily. They also pay attention to the ease of understanding the concept by the students and also make sure that students can learn from this in a quick way. Exercise 2.4 is optional and is not given from the examination point of view.

It consists of extra questions to practice from the chapter. Our experts provide a detailed solution of each answer to the questions given in the exercises 2.4 in the NCERT textbook for Class 10. The NCERT Solutions for Class 10 Maths Chapter 2 Polynomials are prepared by following NCERT guidelines keeping in mind so that it should cover the whole syllabus accordingly. These are very helpful in scoring well in examinations.

Access Other Exercise Solutions of Class 10 Maths Chapter 2- Polynomials

Exercise 2.1 Solutions 1 Question
Exercise 2.2 Solutions 2 Question (2 short)
Exercise 2.3 Solutions 5 Question (3 short, 2 long)

Access Answers to NCERT Class 10 Maths Chapter 2 – 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:

Given, p(x) 2x3+x2-5x+2

And zeroes for p(x) are = 1/2, 1, -2

 

∴ p(1/2) = 2(1/2)3+(1/2)2-5(1/2)+2 = (1/4)+(1/4)-(5/2)+2 = 0

p(1) = 2(1)3+(1)2-5(1)+2 = 0

p(-2) = 2(-2)3+(-2)2-5(-2)+2 = 0

Hence, proved 1/2, 1, -2 are the zeroes of 2x3+x2-5x+2.

Now, comparing the given polynomial with general expression, we get;

∴ ax3+bx2+cx+d = 2x3+x2-5x+2

a=2, b=1, c= -5 and d = 2

As we know, if α, β, γ are the zeroes of the cubic polynomial ax3+bx2+cx+d , then;

α +β+γ = –b/a

αβ+βγ+γα = c/a

α βγ = – d/a.

Therefore, putting the values of zeroes of the polynomial,

α+β+γ = ½+1+(-2) = -1/2 = –b/a

αβ+βγ+γα = (1/2×1)+(1 ×-2)+(-2×1/2) = -5/2 = c/a

α β γ = ½×1×(-2) = -2/2 = -d/a

Hence, the relationship between the zeroes and the coefficients are satisfied.

(ii) x3-4x2+5x-2 ;2, 1, 1

Solution:

Given, p(x) = x3-4x2+5x-2

And zeroes for p(x) are 2,1,1.

∴ p(2)= 23-4(2)2+5(2)-2 = 0

p(1) = 13-(4×1)+(5×1)-2 = 0

Hence proved, 2, 1, 1 are the zeroes of x3-4x2+5x-2

Now, comparing the given polynomial with general expression, we get;

∴ ax3+bx2+cx+d = x3-4x2+5x-2

a = 1, b = -4, c = 5 and d = -2

As we know, if α, β, γ are the zeroes of the cubic polynomial ax3+bx2+cx+d , then;

α + β + γ = –b/a

αβ + βγ + γα = c/a

α β γ = – d/a.

Therefore, putting the values of zeroes of the polynomial,

α +β+γ = 2+1+1 = 4 = -(-4)/1 = –b/a

αβ+βγ+γα = 2×1+1×1+1×2 = 5 = 5/1= c/a

αβγ = 2×1×1 = 2 = -(-2)/1 = -d/a

Hence, the relationship between the zeroes and the coefficients are satisfied.

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:

Let us consider the cubic polynomial is ax3+bx2+cx+d and the values of the zeroes of the polynomials be α, β, γ.

As per the given question,

α+β+γ = -b/a = 2/1

αβ +βγ+γα = c/a = -7/1

α βγ = -d/a = -14/1

Thus, from above three expressions we get the values of coefficient of polynomial.

a = 1, b = -2, c = -7, d = 14

Hence, the cubic polynomial is x3-2x2-7x+14

3. If the zeroes of the polynomial x3-3x2+x+1 are a – b, a, a + b, find a and b.

Solution:

We are given with the polynomial here,

p(x) = x3-3x2+x+1

And zeroes are given as a – b, a, a + b

Now, comparing the given polynomial with general expression, we get;

∴px3+qx2+rx+s = x3-3x2+x+1

p = 1, q = -3, r = 1 and s = 1

Sum of zeroes = a – b + a + a + b

-q/p = 3a

Putting the values q and p.

-(-3)/1 = 3a

a=1

Thus, the zeroes are 1-b, 1, 1+b.

Now, product of zeroes = 1(1-b)(1+b)

-s/p = 1-b2

-1/1 = 1-b2

b2 = 1+1 = 2

b = ±√2

Hence,1-√2, 1 ,1+√2 are the zeroes of x3-3x2+x+1.

4. If two zeroes of the polynomial x4-6x3-26x2+138x-35 are 2 ±3, find other zeroes.

Solution:

Since this is a polynomial equation of degree 4, hence there will be total 4 roots.

Let f(x) = x4-6x3-26x2+138x-35

Since 2 +√and 2-√are zeroes of given polynomial f(x).

∴ [x−(2+√3)] [x−(2-√3)] = 0

(x−2−√3)(x−2+√3) = 0

On multiplying the above equation we get,

x2-4x+1, this is a factor of a given polynomial f(x).

Now, if we will divide f(x) by g(x), the quotient will also be a factor of f(x) and the remainder will be 0.

So, x4-6x3-26x2+138x-35 = (x2-4x+1)(x2 –2x−35)

Now, on further factorizing (x2–2x−35) we get,

x2–(7−5)x −35 = x2– 7x+5x+35 = 0

x(x −7)+5(x−7) = 0

(x+5)(x−7) = 0

So, its zeroes are given by:

x= −5 and x = 7.

Therefore, all four zeroes of given polynomial equation are: 2+√3 , 2-√3−5 and 7.

 


NCERT solutions for class 10 Maths Chapter 2 – Polynomials Exercise 2.4 is the fourth exercise of Chapter 2 of Class 10 Maths. Polynomials are introduced in Class 9 and this is discussed more in details in Class 10.

  • This exercise is not from the examination point of view.

Key benefits of NCERT Solutions for Class 10 Maths Chapter 2- Polynomials Exercise 2.4

  • After going through the stepwise solutions given by our subject expert teachers, you will be able to score more marks.
  • It follows NCERT guidelines which help in preparing the students accordingly.
  • It contains all the important questions from the examination point of view.
  • It helps in scoring well in Maths first term exams.
  • These NCERT Solutions let you solve and revise all questions of exercise 2.4.

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