Answer :
(i) P(x) =
Now for zeroes, putting the given values in x.
P(1/2) = 2(1/2)3 + (1/2)2 - 5(1/2) + 2
= (1/4) + (1/4) - (5/2) + 2
= (1 + 1 - 10 + 8)/2
= 0/2 = 0
P(1) =
P(-2) =
Thus, 1/2, 1 and -2 are zeroes of given polynomial.
Comparing given polynomial with ax3 + bx2 + cx + d and Taking zeroes as α, β, and γ, we have
Now, We know the relation between zeroes and the coefficient of a standard cubic polynomial as
Substituting value, we have
Since LHS = RHS, Relation verified.
Thus, all three relationships between zeroes and the coefficient is verified.
(ii) p(x) = x3 – 4x2 + 5x – 2
Now for zeroes , put the given value in x.
P(2) = =
P(1) =
P(1) =
Thus, 2, 1 , 1 are the zeroes of the given polynomial.
Now,
Comparing the given polynomial with ax3 + bx2 + cx + d, we get
4 = 4
5 = 5
αβγ =
2 × 1 × 1 = 2
2 = 2
Thus, all three relationships between zeroes and the coefficient is verified.
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