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 ax^{3} + bx^{2} + 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)

Since LHS = RHS, Relation verified.

Since LHS = RHS, Relation verified.

Thus, all three relationships between zeroes and the coefficient is verified.

(ii) p(x) = x^{3} – 4x^{2} + 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 ax^{3} + bx^{2} + cx + d, we get

Now,

4 = 4

5 = 5

αβγ =

2 × 1 × 1 = 2

2 = 2

Thus, all three relationships between zeroes and the coefficient is verified.

Rate this question :

<span lang="EN-USKC Sinha - Mathematics

<span lang="EN-USKC Sinha - Mathematics

<span lang="EN-USKC Sinha - Mathematics

Find the quadratiKC Sinha - Mathematics

Find a quadratic KC Sinha - Mathematics

If the product ofKC Sinha - Mathematics