I have this equation $f(x) = \frac{1}{4}x^4-2x^3+6x^2-13x+4$
I am asked to calculate by hand the one real root out of it. When derivated $x^3-6x^2+12x-13=0$
This is where I am stuck at. I know the root is $\sqrt[3]{5}+2$.
I have been trying to calculate the answer in many different ways and still not getting the correct answer. Using common divisor etc. What am I doing wrong?
$x^3-6x^2+12x-13= (x - 2)^3 -5=0$
Therefore, the only real root is $\sqrt[3]{5}+2$
Basically, you had to complete the "cube" which is a pretty standard technique.