# 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!

It's $$x^2(x-4)-4(x-4)=0$$ or $$(x-4)(x^2-4)=0.$$ Can you end it now?

We have that

$$x^3-4x^2-4x+16=x(x^2-4x+4)-8x+16=x(x-2)^2-8(x-2)=$$

$$=(x-2)(x^2-2x-8)=(x-2)(x+2)(x-4)=0$$

$$\frac{x^3-4x^2-4x+16}{\sqrt{x^2-5x+4}}=\frac{(x-4)(x-2)(x+2)}{\sqrt{(x-4)(x-1)}}==\frac{\sqrt{x-4}(x-2)(x+2)}{\sqrt{x-1}}$$ The fraction will be $$=0$$ if the numerator is $$0$$ and the denominator is not $$0$$. That is true for $$x=2,-2,4$$.

• In this case it is still zero even at $x=4$ though, the denominator is not a power one but rather one half ! Oct 9, 2020 at 19:22

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}

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.

• Thank you for the response! I don't understand how from $x(x^2-4x-4)=-16$ we see that any integer solution must divide $16$. Can you clarify that for me? Thank you in advance! Oct 10, 2020 at 10:43
• @LYI Assuming integer solutions, both $x$ and $x^2-4x-4$ are integers. If their product is $-16$, both those factors must divide 16. Oct 10, 2020 at 17:09

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:

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

2. 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.

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 : $$S=(2,-2)$$