Polynomial Question (1 Viewer)

[Jase]

Consider the polynomial P(z) = z^3 + az^2 + bz + c where a,b,c are real.

If P(xi) = 0 where x is real and non zero:

explain why P(-xi) = 0;
show that P(z) has one real zero;
and hence show that c=ab, where b>0

I ended up screwing around with everything until i got c=-ab, and i'm not quite sure how i even arrived at that.

 

[Rorix]

-c = product of roots = x^2y therefore c= - x^2 y
-a= sum of roots = xi - xi + y = y therefore a= -y
b= product of roots two at a time = xiy - xiy + (-xi)(xi) = x^2

i.e. c=ab

 

[Jase]

ah.. cool thanks.