If you're so good at math answer this problem........?

Question

The degree of a polynomial is 3 and it has the zeros: 6, -5+2i, and
f(2)= -363

Find the polynomial f(x)

Whoever answer this first and their answer agrees with the concensus will win 10 points and i'll forever be indebt of gratitude to you.

Answer 1

Any 3rd degree polynomial has at least one real root, here it is 6.
So the polynomila would be the result of the product

(x-6)*(ax^2+bx+c)

with a, b and c to be determined.
The parabolic (ax^2+bx+c) must have the -5+2i root, but a parabolic must have conjugate complex roots, that means its other root will be -5-2i, and it must thus be the product of
(x+5-2i)(x+5+2i), which is (x^2+10x+29).
Multiply this by the (x-6) we already have, and we have a polynomial that is x^3+4x^2-31x-174.
This polynomial will return a value of -212 if x=2. But you want it to be 363, so one simply needs to multiply the whole equation by the factor -363/212.

That is your polynomial:

-363/212 * (x^3+4x^2-31x-174).

Answer 2

Complex roots come in conjugate pairs. So the third root is -5-2i.
Thus we get A(x-6)(x-(-5+2i))(x-(-5-2i)).

Expanding the last pair gives A(x-6)(x^2 + 10x + 29).
Which gives A(x^3 + 4x^2 - 31x - 174).
Put in x = 2 and we get -212A = -363, so A = 363/212.

Thus, we get 363/212 (x^3 + 4x^2 - 31x - 174).
Or 363/212 x^3 + 363/53 x^2 - 11253/212 x - 31581/106, if you want to be messy.