NCERT Solution for Polynomial Chapter 2 Exercise 2.1
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Question 1:
Find the value of the polynomial $P(x)= 5x-4x^2 +3$ at
(i) x = 0 (ii) x = -1 (iii) x = 2
Solution:
(i)$P(x)= 5x-4x^2 +3$
P(0)= 0-0+3=3
(ii)$P(x)= 5x-4x^2 +3$
P(-1) =-5-4+3=-6
(iii)$P(x)= 5x-4x^2 +3$
P(2)= 10-16+3=-3
Question 2
Find p(0), p(1) and p(2) for each of the following polynomials:
(i) $p(y) = y^2 - y + 1$
(ii) $p(t) = 2 + t + 2t^2- t^3$
(iii) $p(x) = x^3$
(iv) $p(x) = (x - 1) (x + 1)$
Solution:
(i) $p(y) = y^2 - y + 1$
p(0) = (0)
^{2}- (0) + 1 = 1
p(1) = (1)
^{2}- (1) + 1 = 1
p(2) = (2)
^{2}- (2) + 1 = 3
(ii) $p(t) = 2 + t + 2t^2- t^3$>
p(0) = 2 + 0 + 2 (0)
^{2}- (0)3 = 2
p(1) = 2 + (1) + 2(1)
^{2}- (1)
^{3}
= 2 + 1 + 2 - 1 = 4
p(2) = 2 + 2 + 2(2)
^{2}- (2)
^{3}
= 2 + 2 + 8 - 8 = 4
(iii) $p(x) = x^3$
p(0) = (0)
^{3}= 0
p(1) = (1)
^{3}= 1
p(2) = (2)
^{3}= 8
(iv) $p(x) = (x - 1) (x + 1)$
p(0) = (0 - 1) (0 + 1) = (- 1) (1) = - 1
p(1) = (1 - 1) (1 + 1) = 0 (2) = 0
p(2) = (2 - 1 ) (2 + 1) = 1(3) = 3
Question 3
Verify whether the following are zeroes of the polynomial, indicated against them.
(i) $p(x) = 3x + 1$, x =- 1/3
(ii) $p(x) = 5x - \pi $, x =4/5
(iii) $p(x) = x^2 - 1$, x = 1, -1
(iv) $p(x) = (x + 1) (x - 2)$, x = - 1, 2
(v) $p(x) = x^2$ , x = 0
(vi) $p(x) = lx + m$, x = - m/l
(vii) $p(x) = 3x^2 - 1$, x =-1/√3 and 2/√3
(viii) $p(x) = 2x + 1$, x =1/2
Solution:
(i) $p(x) = 3x + 1$,x=-1/3
p(-1/3) = 3 (-1/3)+1=-1+1=0
p(-1/3) = 0 which means that-1 /3is zero of the polynomial p(x) = 3x+1.
(ii) $p(x) = 5x - \pi $, x =4/5
p(4/5) = 5(4/5)-π = 4-π
p(4/5) ≠ 0 which means that 4/5 is not zero of the polynomial $p(x) = 5x - \pi $.
(iii) $p(x) = x^2 - 1$,x = 1,- 1
p(1)=1
^{2}-1=1-1=0
p(-1)=(-1)
^{2}-1=1-1=0
Both p(1) and p(-1) are equal to 0. It means that 1 and -1 are zeroes of the polynomial $p(x) = x^2 - 1$.
(iv) $p(x) = (x + 1) (x - 2)$,x=-1,2
p(-1) = (-1+1)(-1-2) = 0×-3 = 0
p(2) = (2+1)(2-2) = 3×0 = 0
Both p(-1) and p(2) are equal to 0. It means that -1 and 2 are zeroes of the polynomial $p(x) = (x + 1) (x - 2)$.
(v) $p(x) = x^2$,x = 0
p(0) = 0
^{2}= 0
p(0) = 0 which means that 0 is the zero of the polynomial $p(x) = x^2$.
(vi) $p(x) = lx + m$,x =-m/
p(-m/l) =l(-m/l) + m = -m + m = 0
p(-m/l) = 0 which means that(-m/l)is zero of the polynomial $p(x) = lx + m$
(vii) $p(x) = 3x^2 - 1$, x =-1/√3 and 2/√3
p(-1/√3) = 3(-1/√3)
^{2}-1=3(1/3)-1=1-1=0
p(2/√3) = 3(2/√3)
^{2}-1 = 3 ×(4/3)-1= 4-1 = 3
p(-1/√3) = 0 which means that-1/√3is zero of the polynomial p(x) = 3x
^{2}- 1.
p(2/√3) ≠ 0 which means that2/√3is not zero of the polynomial $p(x) = 3x^2 - 1$.
(viii) p(x) = 2x + 1 ,x =1/2
p(1/2)=2 ×(1/2)+ 1=1+1=2
p(1/2) ≠ 0. It means that 1/2 is not zero of the polynomial p(x)=2x+1.
Question 4.
Find the zero of the polynomial in each of the following cases:
(i) $p(x) = x + 5$
(ii) $p(x) = x - 5$
(iii) $p(x) = 2x + 5$
(iv) $p(x) = 3x - 2$
(v) $p(x) = 3x$
(vi) $p(x) = ax$, a ≠ 0
(vii) $p(x) = cx + d$, c ≠ 0, c, d are real numbers.
Solution:
Zero of a polynomial is that value of the variable at which the value of the polynomial is obtained as 0.
(i) $p(x) = x + 5$
p(x) = 0
x + 5 = 0
x = - 5
Therefore, for x = -5, the value of the polynomial is 0 and hence, x = -5 is a zero of the given polynomial.
(ii) $p(x) = x - 5$
p(x) = 0
x - 5 = 0
x = 5
Therefore, for x = 5, the value of the polynomial is0 and hence, x = 5 is a zero of the given polynomial.
(iii) $p(x) = 2x + 5$
p(x) = 0
2x + 5 = 0
2x = - 5
x = -5/2
Therefore, for x = -5/2 , the value of the polynomial is 0 and hence, x = -5/2is a zero of the given polynomial.
(iv) $p(x) = 3x - 2$
p(x) = 0
3x - 2 = 0
x=2/3, so 2/3 is the zero of the polynomial
(v) $p(x) = 3x$
p(x) = 0
3x = 0
x = 0
So x=0 is the zero of the polynomial
(vi) $p(x) = ax$
p(x) = 0
ax = 0
x = 0
Therefore, for x = 0, the value of the polynomial is 0 and hence, x = 0 is a zero of the given polynomial.
(vii) $p(x) = cx + d$
p(x) = 0
cx+ d = 0
x=-d/c
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