Find the zeros of $f(x)=x^3−4x^2+x−4$

Question

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.

Answer 1

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 $$\{4,\pm i\}$$

Answer 2

Let $f(x) =x^3 - 4x^2 + x -4.$
Let $g(x) =x^3 - 4x^2 + x.$

Your initial approach was to find the zeroes of $g(x).$
This doesn't do you any good, because you are in effect identifying
where $f(x)$ crosses the horizontal line $y = -4$,
when what you want is to identify where $f(x)$
crosses the horizontal line $y = 0.$

Further, assuming that you have not studied the cubic equation, as far as non-creative pedestrian methods are concerned, in my opinion, you are absolutely stuck.

In my opinion, the whole key to the problem is to notice
that the coefficients in $f(x)$ look like $1,-4,1,-4.$
This suggests that $(x-4)$ is a factor of $f(x)$, which
turns out to be the case.

At this point, it is all downhill, assuming that you are comfortable with quadratic equations.

Addendum

It occurs to me that an alternative meta-cheating method is to assume that there is a rational root. Although I am not familiar with such tests, I suspect that such a test could be manually applied to discover the $(x-4)$ factor.

The idea behind my meta-cheating adjective is that it is reasonable to presume that the intent of the questioner is that the problem would yield to what you had recently been taught in class. I doubt whether you have ventured into raw cubic equations.