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?
3x^3 + 2x^2 - 1 = 0
plus/minus 1/3 and plus/minus 1?
Answers
DonHo
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
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
DonHo
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
It's p/q
so roots are
+-1/+-1 and +-1/+-3
http://www.mathwords.com/r/rational_root_theorem.htm
Reiny
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)
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)