Posted by **Christian** on Friday, January 11, 2008 at 9:11pm.

1.Solve: x(x2 + 16) = 0

2.Solve: 2x3 + 6x2 – x – 3 = 0

3. Solve: 2x = x2

- Algebra II -
**drwls**, Friday, January 11, 2008 at 9:41pm
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^2-1. That can also be factored, giving you the other two solutions.

3. Rewrite it as x(x-2) = 0

That's an easy one. What are the two x values that satisfy that equation?

- Algebra II -
**Damon**, Friday, January 11, 2008 at 9:45pm
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 x-3 would be factors because of that lonely 3 at the end)

(x+3)(2x^2-1)) = 0

x = -3 is a solution

two others come from 2x^2-1 = 0

x^2 = 1/2

x = + or - (1/2) sqrt( 2)

3. Solve: 2x = x2

that is x^2 - 2x = 0

which is

x (x-2) = 0

x = 0 or x = 2

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