SOLUTION: find the roots of the polynomial p(x) = x^4+4x^3+6x^2+4x+5=0 given that one of the roots is x= -i

Algebra ->  Square-cubic-other-roots -> SOLUTION: find the roots of the polynomial p(x) = x^4+4x^3+6x^2+4x+5=0 given that one of the roots is x= -i      Log On


   

Question 976056: find the roots of the polynomial
p(x) = x^4+4x^3+6x^2+4x+5=0 given that one of the roots is x= -i
Answer by josgarithmetic(39621) About Me  (Show Source):
You can put this solution on YOUR website!
Another root is x=i, because complex roots for polynomial functions come as conjugate pairs. Those two roots are as %28x-%28-i%29%29%28x-i%29=%28x-i%29%28x%2Bi%29
x%5E2-i%5E2
x%5E2-%28-1%29
highlight_green%28x%5E2%2B1%29, a factor of the polynomial p(x).

Use polynomial division for x%5E4%2B4x%5E3%2B6x%5E2%2B4x%2B5 as dividend
and x%5E2%2B1 as divisor. The quotient represents the rest of the factors, as quadratic, degree two.
-
The division process not shown here; but result is x%5E2%2B4x%2B5 as the quotient.

Use general solution method for a quadratic equation to find the zeros of this factor:
roots are
x=%28-4%2B-+sqrt%284%5E2-4%2A5%29%29%2F2
x=%28-4%2B-+sqrt%28-4%29%29%2F2
x=%28-4%2B-+2i%29%2F2
highlight%28x=-2%2B-+i%29 or -2-i and -2+i.

(along with -i and i ).