How do you find irrational zeros?

Question

I know how to find possible rational zeros p/q and also how to find out if they work as zeros.

But how do you find irrational zeros?

Ex. x^3 + 2x^2 - 4x + 1
rational zeros @ 1

Answer 1

Ok, you have already solved the fist part.

x^3 + 2x^2 - 4x + 1 = 0
(x - 1)(x^2 +3x - 1) = 0

rational zero at x = 1

good job! Now you have:

x^2 + 3x - 1 = 0

I'm going to solve this by competing the square.
(you could use the quadratic formula)

(x + 1.5)^2 = x^2 + 3x + 1.5^2
x^2 + 3x - 1 = (x + 1.5)^2 - 1 -1.5^2 = 0
(x + 1.5)^2 = 1 +1.5^2
x = -1.5 + sqr(1 + 1.5^2) and -1.5 - sqr(1 + 1.5^2)

Answer 2

If you know there is a zero at 1, then you know you can factor (x-1) out of the equation. So
x^3 +2x^2 -4x +1 = (x-1)(x^2 +3x - 1). So to find the other "zeroes" or solutions, you need to solve that quadratic equation x^2 + 3x -1 = 0. If you apply the quadratic formula to this, you will get the solutions. Doing it in my head (if I remember the formula correctly), you get (-3 +/- sqrt(13))/2.