How to solve an equation with $x^4$? Today, I had this question on a Maths test about Algebra. This was the equation I had to solve:
$$(1-x)(x-5)^3=x-1$$
I worked away the brackets and subtracted $x-1$ from both sides and was left with this:
$$-x^4+16x^3-90x^2+199x-124=0$$
Problem is, I haven't a clue how to solve this? First thing I tried was replacing $x^2$ with another variable like $u$ but that got me no further. Dividing the whole equation by $x^2$ (as is suggested by a lot of sites on this matter) also did not get me any further. I then tried something incredibly ludicrous;
$$(ax+b)(cx^3+dx^2+ex+f)=0$$
$$
\left\{
\begin{aligned} 
ac &= -1 \\ 
ad + bc &= 16 \\ 
ae + bd &= -90 \\
af + be &= 199 \\
bf &= -124
\end{aligned} 
\right. 
$$
which got even worse when there were 3 brackets;
$$(ax+b)(gx+h)(ix^2+jx+k)=0$$
$$
\left\{
\begin{aligned} 
ac &= agi &&= -1 \\
ad + bc &= agj + ahi + bgi &&= 16 \\
ae + bd &= agk + ahj + bgj + bhi &&= -90 \\ 
af + be &= ahk + bgk + bhj &&= 199 \\
bf &= bhk &&= -124
\end{aligned} 
\right. 
$$
only to be left with no result. 
When using Wolfram Alpha on this question, it performs a rather strange step I don't understand:
$$-x^4+16x^3-90x^2+199x-124=0$$
$$\downarrow$$
$$-((x-4)(x-1)(x^2-11x+31))=0$$
Could somebody explain how to properly tackle this problem? And if possible, also show me how to get the non-real answers for it. Thanks.
 A: Euh... I think you overcomplicated things here...
$(1-x)(x-5)^3=x-1$ is equivalent to $(1-x)[(x-5)^3+1]=0$
Either $x=1$ or $(x-5)^3=-1$...
A: There are already answers on how to get the real solutions, so I will only show you the non-real solutions.
You have obtained that $(x-5)^3=-1$. Expanding and simplifying we get:
$$x^3-15x^2+75x-124=0$$
However, we know that $x=4$ is a solution so we can say that $(x-4)(ax^2+bx+c)=0$. You can equate coefficients or use polynomial division, but as you have already found with Wolfram Alpha, $(ax^2+bx+c)=(x^2-11x+31)$.
To solve, complete the square:
$$x^2-11x+31=0$$
$$(x-\frac{11}{2})^2-\frac{121}{4}+31=0$$
$$(x-\frac{11}{2})^2=-\frac{3}{4}$$
$$x-\frac{11}{2}=\sqrt{-\frac{3}{4}}= \pm \frac{\sqrt{3}}{2}i$$
$$x=\frac {11 \pm i \sqrt{3}}{2}$$
A: From the beginning:
$$(1-x)(x-5)^3=x-1\\
(1-x)(x-5)^3+1-x=0\\
(1-x)(x-5)^3+(1-x)=0\\
(1-x)[(x-5)^3+1]=0\\$$
This implies $1-x=0$ or $(x-5)^3=-1$. I believe you can solve these.
A: I know that the problem has already been answered but I want to show you a more general method, let's suppose that you don't se how to rewrite the equation:
$-x^4+16x^3-90x^2+199x-124=0$
Or I prefer to write:
$x^4-16x^3+90x^2-199x+124=0$
You can use something called the Ruffini rule: search for integers divisor (both positive and negative) of the constant term and then set $x$ equals to the them and see if one of them is a solution. Starting from one we have:
$1-16+90-199+124=0$
So $x=1$ is a solution, now via Ruffini's rule, which can be seen here, we can rewrite the equation as:
$(x-1)(x^3-15x^2+75x-124)=0$
Now you have $3$ options to end this exercise:


$1.)$ Note that the second factor is a perfect cube;
$2.)$ Use Ruffini's rule again;
$3.)$ Use the general formula for third degree equation (which I'd not advise you to if your interested only in real solution).


This is a more general method so I hope this will help you in the future!
