Roots of polynomial $-4x^3+3x^2-2x+1=0$

Question

I've calculated the derivative of the function $f(x)=-4x^3+3x^2-2x+1$ and came up with $f'(x)=-12x^2+6x-2$ which is always negative and since $\lim_{x\to\infty} f(x)=-\infty$ and $\lim_{x\to-\infty} f(x)=\infty$ I assumed that $f(x)=0$ has only one real solution and two other complex ones. But I'm confused since the polynomial has integer coefficients, shouldn't there be a way to determine the real solution?

No one says there isn't a way to determine the real solution... — JSCB, Mar 17 '13 at 08:05
http://wolframalpha.com/input/?i=solve+4x%5E3-3x%5E2%2B2x-1&x=9&y=13 It's ugly, but it's there. — Arthur, Mar 17 '13 at 08:12

score 1 · Accepted Answer · answered Mar 17 '13 at 08:14

1

You are right about the polynomial $f$ having only one real root. Since $f'$ is always negative $f$ is strictly decreasing ...

To find the root you can use this formula.

answered Mar 17 '13 at 08:14

P..

14,929

Roots of polynomial $-4x^3+3x^2-2x+1=0$

1 Answers1