SOLUTION: Solve algebraically for all values of x: x^4 +4x^3 +4x^2 = -16x * I have already got an answer from my classmates that the values were x=0, x=-4 , x=2i and x=-2i, I got the f

Algebra ->  Expressions-with-variables -> SOLUTION: Solve algebraically for all values of x: x^4 +4x^3 +4x^2 = -16x * I have already got an answer from my classmates that the values were x=0, x=-4 , x=2i and x=-2i, I got the f      Log On


   

Question 1013015: Solve algebraically for all values of x:
x^4 +4x^3 +4x^2 = -16x
* I have already got an answer from my classmates that the values were x=0, x=-4 , x=2i and x=-2i, I got the first two but wondering how to get the imaginary parts. Thank you ^^
Found 2 solutions by MathLover1, ikleyn:
Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!

x%5E4+%2B4x%5E3+%2B4x%5E2+=+-16x
x%5E4+%2B4x%5E3+%2B4x%5E2+%2B16x=0
%28x%5E4+%2B4x%5E3%29+%2B%284x%5E2+%2B16x%29=0
x%5E3%28x+%2B4%29+%2B4x%28x%2B4%29=0
%28x%5E3+%2B4x%29%28x%2B4%29=0
x%28x%5E2+%2B4%29%28x%2B4%29=0
solutions:
x=0 -> real solution
if x%5E2+%2B4=0 =>x%5E2=-4 => x=sqrt%28-4%29=> x=sqrt%28%28-1%292%5E2%29=> x=2i:there are two complex solutions and they are x=2i or x=-2i
if x%2B4=0 => x=-4 -> real solution


Answer by ikleyn(52788) About Me  (Show Source):
You can put this solution on YOUR website!
.
Solve algebraically for all values of x:
x^4 +4x^3 +4x^2 = -16x
-------------------------------------------- 

x%5E4+%2B4x%5E3+%2B4x%5E2+%2B+16x = 0.

Factor taking x out of parentheses

x%2A%28x%5E3+%2B+4x%5E2+%2B+4x+%2B+16%29 = 0.

Factor further the third degree polynomial by grouping

x%2A%28%28x%5E3+%2B+4x%5E2%29+%2B+%284x+%2B+16%29%29 = 0,

x%2A%28x%5E2%28x%2B4%29+%2B+4%2A%28x%2B4%29%29 = 0,

x%2A%28x%5E2+%2B+4%29%2A%28x%2B4%29 = 0.

Everything is clear now, isn't it?