Find all possible rational roots using the rational root theorem.
3x^3 + 2x^2 - 1 = 0
plus/minus 1/3 and plus/minus 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
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
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)
To find all possible rational roots of a polynomial equation, you can use the Rational Root Theorem. The Rational Root Theorem states that if a polynomial equation has a rational root r = p/q, where p is the factors of the constant term and q is the factors of the leading coefficient.
In your equation 3x^3 + 2x^2 - 1 = 0, the constant term is -1 (the number without an x term), and the leading coefficient is 3 (the coefficient of the highest degree term).
1. Find the factors of the constant term (-1): The factors of -1 are 1 and -1.
2. Find the factors of the leading coefficient (3): The factors of 3 are 1 and 3.
3. Now, form all possible combinations of p/q, where p is a factor of the constant term and q is a factor of the leading coefficient. In this case, the possible rational roots are:
- p/q: ±1/1 and ±1/3
So, the possible rational roots of the equation 3x^3 + 2x^2 - 1 = 0 are ±1, and ±1/3.