Question 1000987: Form a polynomial f(x) with real coefficients having the degree and zeros. Degree 5; zeros: -6;-i; 5+i. Answer by Boreal(15235) (Show Source):
You can put this solution on YOUR website! The roots are 5, +/- i and 5+i, 5-i, since complex roots are conjugate
The polynomial has the form (x+6)(x^2+1)and another quadratic, whose roots are 5+i and 5-i
ax^2+bx+c=0
let a=1
b must be some multiple of -5 (quadratic formula, -b...)
sqrt(b^2-4ac)=-1
using -10 for b, and 1 for a,
x=(1/2) (-10+/-sqrt(100-(4)(1)26))
x=(1/2) (-10 +/- sqrt (-4))
x=(1/2) (-10 +/- 2i)
x=-5 +/- i
the last quadratic is x^2-10x+26
The polynomial is
(x+6)(x^2+1)((x^2-10x+26)
