Solve $\frac{x^3-4x^2-4x+16}{\sqrt{x^2-5x+4}}=0$ Solve $$\dfrac{x^3-4x^2-4x+16}{\sqrt{x^2-5x+4}}=0.$$
We have $D_x:\begin{cases}x^2-5x+4\ge0\\x^2-5x+4\ne0\end{cases}\iff x^2-5x+4>0\iff x\in(-\infty;1)\cup(4;+\infty).$ Now I am trying to solve the equation $x^3-4x^2-4x+16=0.$ I have not studied how to solve cubic equations. Thank you in advance!
 A: It's $$x^2(x-4)-4(x-4)=0$$ or
$$(x-4)(x^2-4)=0.$$
Can you end it now?
A: $$\frac{x^3-4x^2-4x+16}{\sqrt{x^2-5x+4}}=\frac{(x-4)(x-2)(x+2)}{\sqrt{(x-4)(x-1)}}==\frac{\sqrt{x-4}(x-2)(x+2)}{\sqrt{x-1}}$$
The fraction will be $=0$ if the numerator is $0$ and the denominator is not $0$. That is true for $x=2,-2,4$.
A: We have that
$$x^3-4x^2-4x+16=x(x^2-4x+4)-8x+16=x(x-2)^2-8(x-2)=$$
$$=(x-2)(x^2-2x-8)=(x-2)(x+2)(x-4)=0$$
A: Use the fact that\begin{align}x^3-4x^2-4x+16=0&\iff x(x^2-4)-4(x^2-4)=0\\&\iff(x-4)(x^2-4)=0.\end{align}
A: The other answers pretty much covered all the aspects, I'm just adding an heuristic way of obtaining integer solutions of polynomial equations that can come handy sometimes. The equation $x^3-4x^2-4x+16=0$ can be rewritten as
$$
x(x^2-4x-4)= -16
$$
(it's just sending the constant term to the rhs and factoring $x$ on the lhs)
So you see that any integer solution must divide 16. Since the divisors of 16 are $\pm 2, \pm 4, \pm 8, \pm 16$ (and $\pm 1$, if you will), if there are any integer solutions, they must be in the set $\{-16,-8,\cdots, 8, 16\}$. In this case, trying solutions in this set will yield all three solutions to the equation.
Naturally, if there are no integer solutions, this gets you nowhere.
A: Alternate approach
My algebra abilities are limited.  I will show you how I would attack the
problem.
Given $$\frac{f(x)}{\sqrt{g(x)}} = 0$$
where (if I understand correctly) $x$ may be any real number
then my first step is to automatically
convert the equation to
$$\frac{[f(x)]^2}{g(x)} = 0$$
with the understanding that any values of $x$ that satisfy the second equation
must be manually examined to see if they also satisfy the first equation.
My next step, which I consider mandatory in this problem is to meta-cheat.
It can be presumed that you would not have been given this problem unless a
solution could be arrived at through the reasonable use of the tools that you
have been offered in your class.
Furthermore, attacking cubic equations (let alone $6^{\text{th}}$ degree equations) through
brute force is generally considered off limits, especially if you have not
been studying cubic equations in class.
At this point, there are only two possiblities:

*

*The teacher or book author is not of sound mind.


*There is some hidden elegance that you are expected to discover.
At this point, the only possible elegance that I can imagine (which allows for the possibility that my imagination is too narrow) consists of seeing if $f(x)$ and
$g(x)$ can be factored without much trouble.  If so, then you can remove
common factors, which will simplify the examination of
$$\frac{[f(x)]^2}{g(x)}.$$
There are two ways to handle this.  One way is to notice that
$g(x) = (x-1)(x-4)$ and then ask yourself whether either of those two factors
is also a factor of $f(x)$.
The other alternative, given that you are not supposed to use brute force
against a cubic, is to accidentally notice that the coefficients of $f(x)$ are
$$ 1, -4, -4, 16$$
This suggests in and of itself that $(x-4)$ might be a factor of $f(x)$.
However  you determine the common factor(s), and simplify the problem, at
this point the meta-cheating is concluded, and you can then attack the
simplified problem more easily.
By the way 
Once you factor $g(x) = (x-1)(x-4)$ 
you must then immediately presume that 
neither $x=1$ or $x=4$ can be considered as satisfactory
answers.
This is because either of those two values for $x$ would
cause the denominator in the original problem to $= 0$, which is forbidden.
A: Answer :
$\frac{x^3-4x^2-4x+16}{\sqrt{x^2 - 5x+4}}= \frac{x(x^2 - 4)-4(x^2 - 4)}{\sqrt{x^2 - 5x+4}}$=$\frac{(x^2 - 4)(x-4)}{\sqrt{x^2 - 5x+4}}$
$\sqrt{x^2 - 5x+4} = 0 $ if $  x =(1, 4) $
Suppose $x≠(1,4)$
$(x^2 - 4)(x-4)$ =0
$\Rightarrow$ $ (x - 2)(x+2)(x-4)=0$
$\Rightarrow $ the solution is ($x=2$ or $ x=- 2)$
Because if $x =4  $the denominator equal $0$
So $x=4$ not a solution
Finally :
$S=(2,-2) $
