# Algebra II

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1.Solve: x(x2 + 16) = 0
2.Solve: 2x3 + 6x2 – x – 3 = 0
3. Solve: 2x = x2

• Algebra II -

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 -

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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