SOLUTION: The problem says to find all the solutions of the equation with the given value of x as a solution... I am so lost x^3+3x^2+16x-20=0 x=-2+4i

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Question 1004969: The problem says to find all the solutions of the equation with the given value of x as a solution... I am so lost
x^3+3x^2+16x-20=0
x=-2+4i
Found 2 solutions by josgarithmetic, MathLover1:
Answer by josgarithmetic(39616) About Me  (Show Source):
You can put this solution on YOUR website!
(1) Another solution is x=-2-4i.
(2) Form the quadratic factor according to the two complex solutions.
(3) Use polynomial division to find the last solution of the equation, using the quadratic expression as divisor and x%5E3%2B3x%5E2%2B16x-20 as the dividend.

Answer by MathLover1(20849) About Me  (Show Source):
You can put this solution on YOUR website!
x%5E3%2B3x%5E2%2B16x-20=0
using zero product theorem, we can write the 3rd degree function as
%28x-x%5B1%5D%29%28x-x%5B2%5D%29%28x-x%5B3%5D%29=0
given:
x%5B1%5D=-2%2B4i...if you have this root, you have to have x%5B2%5D=-2-4i too (complex roots always in pairs)
then we have

%28x-%28-2%2B4i%29%29%28x-%28-2-4i%29%29%28x-x%5B3%5D%29=0
%28x%2B2-4i%29%28x%2B2%2B4i%29%28x-x%5B3%5D%29=0
%28x%5E2%2B2x%2B4xi%2B2x%2B4%2B8i-4xi-8i-16i%5E2%29%28x-x%5B3%5D%29=0

%28x%5E2%2B2x%2B2x%2B4%2B16%29%28x-x%5B3%5D%29=0
%28x%5E2%2B4x%2B20%29%28x-x%5B3%5D%29=0
so,
x%5E3%2B3x%5E2%2B16x-20=%28x%5E2%2B4x%2B20%29%28x-x%5B3%5D%29
%28x%5E3%2B3x%5E2%2B16x-20%29%2F%28x%5E2%2B4x%2B20%29=%28x-x%5B3%5D%29
-----------------(x-1
%28x%5E2%2B4x%2B20%29|(x%5E3%2B3x%5E2%2B16x-20)
-------------------x%5E3%2B4x%5E2%2B20x........subtract
--------------------0-x%5E2-4x...........bring down -20
-------------------...-x%5E2-4x-20
-------------------...-x%5E2-4x-20...subtract
-------------------...........0......reminder
so, third product is %28x-x%5B3%5D%29=x-1=>x%5B3%5D=1
and your equation is factored as following:
%28x%2B2-4i%29%28x%2B2%2B4i%29%28x-1%29=0

and, roots are:
x%5B1%5D=-2%2B4i,
x%5B2%5D=-2-4i, and
x%5B3%5D=1