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NCERT Solutions Class 10 Maths Chapter 2 Polynomials are provided here to help the students in learning efficiently for their exams. The subject experts of Maths have prepared these solutions to help students prepare well for their exams. They solve these solutions in such a way that it becomes easier for students to practise the questions of Chapter 2 Polynomials using the Solutions of NCERT. This makes it simple for the students to learn by adding step-wise explanations to these Maths NCERT Class 10 Solutions.
NCERT Solutions for Class 10 Maths is an extremely important study resource for students. Solving these Polynomials NCERT solutions of Class 10 Maths would help the students fetch good marks in board exams. Also, following the NCERT guidelines is focused on while preparing these solutions.
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-(4×12 )+(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 +√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
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.
