Degree 3; zeros; -8 , 5- i

Question

Information is given about a polynomial f(x) whoae coeffiecients are real numbers

Answer 1

there is one given real root, -8, so x+8 is a factor.

There are two complex zeros, so (x-(5-i)) and (x-(5+i)) are also factors [complex conjugate pair of zeros].

 

p(x) = (x+8)(x-5-i)(x-5+i) = 0

p(x) = (x+8)(x² -10x +26) = 0.

 

I will let you get the final result, a cubic polynomial.