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Find the value of the polynomial at (i) x = 0 (ii) x = –1 (iii) x = 2
(i) 
(ii)
(iii)
Find p(0), p(1) and p(2) for each of the following polynomials:
(i) p(y) =
(ii) p(x) =
(iii) p(t) =
(iv) p(x) = (x – 1) (x + 1)
(i) p(y) = y2 − 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 + 2t2 − t3
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) = x3
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
Verify whether the following are zeroes of the polynomial, indicated against them.
(i) p(x) = ,
(ii) p(x) =
(iii) p(x) =
(iv) p(x) = (x + 1) (x – 2), x = – 1, 2
(v) p(x)
(vi) p(x)
(vii) p(x) =
(viii) p(x)
(i) If 
is a zero of given polynomial p(x) = 3x + 1, then 
should be 0.
Therefore, is a zero of the given polynomial.
(ii) If 
is a zero of polynomial p(x) = 5x − π , then
should be 0.
Therefore, is not a zero of the given polynomial.
(iii) If x = 1 and x = −1 are zeroes of polynomial p(x) = x2 − 1, then p(1) and p(−1) should be 0.
Here, p(1) = (1)2 − 1 = 0, and
p(− 1) = (− 1)2 − 1 = 0
Hence, x = 1 and −1 are zeroes of the given polynomial.
(iv) If x = −1 and x = 2 are zeroes of polynomial p(x) = (x +1) (x − 2), then p(−1) and p(2)should be 0.
Here, p(−1) = (− 1 + 1) (− 1 − 2) = 0 (−3) = 0, and
p(2) = (2 + 1) (2 − 2 ) = 3 (0) = 0
Therefore, x = −1 and x = 2 are zeroes of the given polynomial.
(v) If x = 0 is a zero of polynomial p(x) = x2, then p(0) should be zero.
Here, p(0) = (0)2 = 0
Hence, x = 0 is a zero of the given polynomial.
(vi) If 
is a zero of polynomial p(x) = lx + m, then 
should be 0.
Here, 
Therefore, 
is a zero of the given polynomial.
(vii) If 
and 
are zeroes of polynomial p(x) = 3x2 − 1, then
Hence, is a zero of the given polynomial. However, is not a zero of the given polynomial.
(viii) If 
is a zero of polynomial p(x) = 2x + 1, then 
should be 0.
Therefore, is not a zero of the given polynomial.
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 (vii) p(x) = cx + d,
, c, d are real numbers.
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
Therefore, for, the value of the polynomial is 0 and hence, 
is a zero of the given polynomial.
(iv) p(x) = 3x − 2
p(x) = 0
3x − 2 = 0
Therefore, for, the value of the polynomial is 0 and hence, is a zero of the given polynomial.
(v) p(x) = 3x
p(x) = 0
3x = 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.
(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
Therefore, for, the value of the polynomial is 0 and hence, is a zero of the given polynomial.
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