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 experts in Maths to help the students prepare well for the Class 10 board exam. These experts create NCERT Solutions for Maths which would help students to solve the NCERT problems easily. They also make sure the concepts are easy to understand and that students can learn quickly. 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 for each answer to the questions given in Exercise 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 and syllabus. These are very helpful in scoring well in the examinations.
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Exercise 2.1 Solutions 1 Question
Exercise 2.2 Solutions 2 Question (2 short)
Exercise 2.3 Solutions 5 Questions (2 short, 3 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
Â
∴ 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 the 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 is 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-(4×12 )+(5×1)-2 = 0
Hence proved, 2, 1, 1 are the zeroes of x3-4x2+5x-2
Now, comparing the given polynomial with the 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 is 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 are 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 the above three expressions, we get the values of the coefficients of the 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 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 the 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, there will be total of 4 roots.
Let f(x) = x4-6x3-26x2+138x-35
Since 2 +√3 and 2-√3 are zeroes of given polynomial f(x).
∴ [x−(2+√3)] [x−(2-√3)] = 0
(x−2−√3)(x−2+√3) = 0
After multiplication, 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 the given polynomial equation are 2+√3 , 2-√3, −5 and 7.
Q.5: If the polynomial x4 – 6x3 + 16x2 – 25x + 10 is divided by another polynomial x2 – 2x + k, the remainder comes out to be x + a, find k and a.
Solution:
Let’s divide x4 – 6x3 + 16x2 – 25x + 10 by x2 – 2x + k.
Given that the remainder of the polynomial division is x + a.
(4k – 25 + 16 – 2k)x + [10 – k(8 – k)] = x + a
(2k – 9)x + (10 – 8k + k2) = x + a
Comparing the coefficients of the above equation, we get;
2k – 9 = 1
2k = 9 + 1 = 10
k = 10/2 = 5
And
10 – 8k + k2 = a
10 – 8(5) + (5)2 = a [since k = 5]
10 – 40 + 25 = a
a = -5
Therefore, k = 5 and a = -5.
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 in more detail 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
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