Class 10 NCERT Solutions – Chapter 2 Polynomials – Exercise 2.4
Last Updated :
01 May, 2024
Question 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:
p(x) = 2x3+x2-5x+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
Therefore, 1/2, 1, -2 are the zeroes of 2x3+x2-5x+2.
Now, comparing the given polynomial with general expression
ax3+bx2+cx+d = 2x3+x2-5x+2
a=2, b=1, c= -5 and d = 2
α, β, γ are the zeroes of the cubic polynomial ax3+bx2+cx+d
α +β+γ = –b/a
αβ+βγ+γα = c/a
α βγ = – d/a.
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:
p(x) = x3-4x2+5x-2
Zeroes are 2,1,1.
p(2)= 23-4(2)2+5(2)-2
= 0
p(1) = 13-(4)(12 )+(5)(1)-2 = 0
Therefore, proved, 2, 1, 1 are the zeroes of x3-4x2+5x-2
On comparing the given polynomial with general expression
ax3+bx2+cx+d = x3-4x2+5x-2
a = 1, b = -4, c = 5 and d = -2
Therefore,
α + β + γ = –b/a
= 2+1+1
= 4
–b/a = -(-4)/1
αβ + βγ + γα = c/a
= 2×1+1×1+1×2
= 5
c/a = 5/1
α β γ = – d/a.
= 2×1×1
= 2
-d/a = -(-2)/1
Hence, the relationship between the zeroes and the coefficients is satisfied.
Question 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 as ax3+bx2+cx+d and zeroes of the polynomials be α, β, γ.
α+β+γ = -b/a = 2/1
αβ +βγ+γα = c/a = -7/1
α βγ = -d/a = -14/1
On comparing
a = 1, b = -2, c = -7, d = 14
Therefore, the cubic polynomial is x3-2x2-7x+14
Question 3. If the zeroes of the polynomial x3 – 3x2 + x + 1 are a – b, a, a + b, find a and b.
Solution:
p(x) = x3-3x2+x+1
Zeroes are given as a – b, a, a + b
px3+qx2+rx+s = x3-3x2+x+1
On comparing
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
Therefore, zeroes are 1-b, 1, 1+b.
Product of zeroes = 1(1-b)(1+b)
-s/p = 1-b2
-1/1 = 1-b2
b2 = 1+1 = 2
b = √2
Therefore,1-√2, 1,1+√2 are the zeroes of x3-3x2+x+1.
Question 4. If two zeroes of the polynomial x4-6x3-26x2+138x-35 are 2 ±√3, find other zeroes.
Solution:
Degree of polynomial is 4
Therefore, it has four roots
f(x) = x4-6x3-26x2+138x-35
As 2 +√3 and 2-√3 are zeroes of given polynomial f(x).
Therefore, [x−(2+√3)] [x−(2-√3)] = 0
(x−2−√3)(x−2+√3) = 0
Therefore, x2-4x+1 is a factor of polynomial f(x).
Let it be g(x) = x2-4x+1
By dividing f(x) by g(x) we get another factor of f(x)
x4-6x3-26x2+138x-35 = (x2-4x+1)(x2 –2x−35)
On factorizing (x2–2x−35) by splitting the middle term
x2–(7−5)x −35 = x2– 7x+5x-35
=x(x −7)+5(x−7)
(x+5)(x−7) = 0
x= −5 and x = 7.
Therefore, all four zeroes of given polynomial equation are: 2+√3, 2-√3, −5 and 7.
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