Posted by Jack on Wednesday, July 1, 2009 at 2:38pm.
Hint: In school problems, most are conjured to work in whole numbers. Notice the coefficents 3,5,125. That begs me to try 3, 5 as roots (experience).
So inspection I see that -5 is a root.
So now I would divide the polynomial by (t+5) either long, or synthetic. I get
t^2-8t+25 . You can find those factors with the quadratic formula...I dont think they will be real numbers. (b^2-4ac is negative)
Let
f(t)= t^3 - 3t^2 - 15t + 125
Then factorize f(t) as
f(t)=(t+5)*(t^2-8*t+25)
from which t=-5 is the real root.
The other factor of (t^2-8*t+25) yields two complex roots, since the discriminant is less than zero.
The two complex roots are obtained by solving:
(t^2-8*t+25) = 0
or t=4+3i or t=4-3i.
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