Question

it is given that the polynomial P(x) =x^3 ax^2 bx c and P(x) =0 has three distint positive integral roots and p(14)=231. let Q(x)=x^2-2x 232. It is also given that the polynomial equation P(Q(x))=0 has no real roots .Then (a) =

Answer 1

Given: It is given that the polynomial P(x) =x^3 ax^2 by c and P(x) =0 has three distinct positive integral roots and p(14)=231. let Q(x)=x^2-2x 232. It is also given that the polynomial equation P(Q(x))=0.

To Find: Value for (a)

Solution:

−2b<0, then the equation has one real and two imaginary roots

Let y=f(x)= x

Let n be the root of ()=0P(x)=0

Then by rational root theorem also

(6)=3⇒216−36+6+=3

P(6)=3⇒216−36a+6b+c=3

⇒3|

Therefore, the value for a= 3/c.

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