Question

Find all possible rational roots using the rational root theorem.

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

plus/minus 1/3 and plus/minus 1?

Answer 1

Look at the coefficient of your leading term and your last term

3x^3 <-- leading term

1 <-- last term

finding numbers that can divide into each term evenly

Let q correspond to the leading term:

q = +-1, +-3 (only numbers that can divide evenly with 3)

let p correspond to the last term

p = +1 (only numbers that can divide evenly into 1)

now possible rational roots is just:

q/p

+-1/+-1, +-3/+-1

Answer 2

oops ignore the last part:

It's p/q

so roots are

+-1/+-1 and +-1/+-3

http://www.mathwords.com/r/rational_root_theorem.htm

Answer 3

let f(x) = 3x^3 + 2x^2-1
f(1) = 3 + 2 - 1 ≠ 0
f(-1) = -3 + 2 - 1 ≠ 0
f(1/3) = 1/9 + 2/9 - 1 ≠ 0
f(-1/3) ≠ 0

So there are no "nice" roots
looking at the graph, there appears to be a real solution at appr x = 0.5
and two complex roots (the graph crosses the x-axis only once)