I am to find the zeros and multiplicities of $f(x)=x^3−4x^2+x−4$.
The solution provided in the answers section of my book is 4 with multiplicity 1. I arrived at $2\pm\sqrt(8)$.
My working:
$$x^3-4x^2+x-4$$
$$x(x^2-4x+1)-4$$
Then, focusing on the quadratic in the middle, I used 3BlueOneBrownVideo (start from ~23 minutes) to find the zeros:
$$m = \frac{-b}{2}=\frac{4}{2}=2$$
$$d^2=m^2-p=2^2+4=8$$
$$r,s=m\pm\sqrt{d^2}$$
$$r,s=2\pm\sqrt{8}$$
The zeros I arrive at are therefore $2\pm\sqrt{8}$
How can I arrive at 4 per the solution? Granular baby steps much appreciated.
Best Answer
I think the following at least is right: $$x^3-4x^2+x-4=x^2(x-4)+1(x-4)=(x-4)(x^2+1),$$ which gives $$\{4,\pm i\}$$