Algebra II
posted by Christian .
1.Solve: x(x2 + 16) = 0
2.Solve: 2x3 + 6x2 – x – 3 = 0
3. Solve: 2x = x2

1. x = 0, 4i or 4i
(i is the square root of 1)
2. One of then roots is 3. I got that by trying small integers. That means (x+3) is a factor. Divide x+3 into
2x^3 + 6x2 – x – 3 and you get the other factor, 2x^21. That can also be factored, giving you the other two solutions.
3. Rewrite it as x(x2) = 0
That's an easy one. What are the two x values that satisfy that equation? 
I assume x2 means x^2  to the second power
1.Solve: x(x2 + 16) = 0
If x is zero, the equation is true, so x = 0 is a solution.
The other solutions are from x^2+16 = 0
x^2 = 16
x = + or  sqrt (16)
sqrt (16) = 4 sqrt (1)
sqrt (1) is usually called i
so x = + or  4i
2.Solve: 2x3 + 6x2 – x – 3 = 0
factor out (x+3)
( I guessed x + 3 or x3 would be factors because of that lonely 3 at the end)
(x+3)(2x^21)) = 0
x = 3 is a solution
two others come from 2x^21 = 0
x^2 = 1/2
x = + or  (1/2) sqrt( 2)
3. Solve: 2x = x2
that is x^2  2x = 0
which is
x (x2) = 0
x = 0 or x = 2