How do I solve this guys?

How do I solve this guys?
[eqn]
\frac{x^3-5x+2}{x-10} = 4
[/eqn]

u put it on wolfram alpha

But it uses brute force (and it asks for monies).. is there no trick to solve this?

1) put equation into wolfram alpha
2) do your own fucking homework
3) cross-check against wolfram for correctness

I'm stuck at step 2 so I'm asking for pointers. What the fuck should I do. I can't find a factorisation to simplyify the denominator

change variables to y so that
y=x-10
x=y+10

wait

if you pirated mathematica would it show the step by step? I've only ever used wolfram alpha

mathematica isnt a better wolfram senpai,mathematica is more like matlab for symbolic computing .
if you want to see the whole solution on wolfram the best way to get it without paying is to pirate the mobile wolfram alpha app.
just torrent the apk and it will give you the premium features.

If I replace x with y+10 it just gets me somewhere like
[eqn]
y^3 + 30 y^2 + 291 y + 950 = 0
[/eqn]

but I still have a cubic equation that I don't know hot to solve?

use long division with polynomials. x-10 is the divisor and x^3-9x+42 is the dividend. This is fucking BASIC

x can't be 10 because the fraction would be indeterminate.

[math]
x \neq 10 \\ \\
x^3 - 5x + 2 = 4x - 40 \\ \\
x^3 - 9x = -42 \\ \\
x (x^2 - 9) = -42
[/math]

Since the right hand side of the equation is negative, exactly one of the factors must be negative. Assuming it's the left factor:

[math]
x < 0 \\ \\
\rightarrow x^2 - 9 > 0 \\ \\
x^2 > 9 \\ \\
|x| > 3 \\ \\
x ~ 0 \\ \\
\rightarrow x^2 - 9 < 0 \\ \\
x^2 < 9 \\ \\
|x| < 3 \\ \\
0 ~

...

Damn, math wizard beat the FUCK out of my babby maths. Good job

Is there a specific name for the method used?

Yeah it's called "high school maths"

x^2+(x/2)-(1/5)=4

x < -3 or x > 3
Since it's x2, it doesn't matter if x is positive or negative, only that x2 > 9 so that x2 - 9 > 0

This. Find one root and use long division. You'll end up with a quadratic polynomial and it will be easy to solve

Wrong.

[math]
x > 3 ~\rightarrow ~(x^2 - 9 > 0) \land (x > 0) ~\rightarrow~ x (x^2 - 9) > 0
[/math]

However, we established that [math]x (x^2 - 9) = -42 < 0[/math] which contradicts the above statement.

Use trigonometric form of Cardano's formula.

use symbolab, senpai

Got it to look like this:
[math]\frac{x^3-5x+2}{x-10} = 4\\ x^3-5x=[4(x-10)]-2 \\-x(x^2-9) = 42 [/math]