Does $2x^5 - x^4 - 22x^3 - 23x^2 + 22x +24 = 0$ have exact solutions in radicals?
A mysterious commenter said on Youtube this was the "easiest quintic equation of my life," and I'm suspecting a troll is afoot. In fact, I believe the opposite is true, this is unsolvable in radicals and you need to prove it.
Symbolab gave up.
Update: Aahaan now says "bro i just took a bunch of linear equations and multiplied them to get this. Now we can factor it into them." The similar equation
$$x^5 - 2x^4 - 22x^3 - 23x^2 + 22x +24 = 0$$
has roots $-2, 1$ and reduces to a straightforward after polynomial division, so there might've been a typo.
Your polynomial $2x^5-x^4-22x^3-23x^2+22x+24$ has no solution in radical it's Galois group is the symmetric group if $5$ objects, on the other hard, $x^5-2x^4-22x^3-23x^2+22x+24$ factors into $(x-1)(x+2)(x^3-3x^2-17x-12)$
Since the polynomials are related in shape $$2x^5-x^4-22x^3-23x^2+22x+24 = x^5-2x^4-22x^3-23x^2+22x+24 +(x^5+x^4)$$ $$2x^5-x^4-22x^3-23x^2+22x+24 = (x-1)(x+2)(x^3-3x^2-17x-12)+x^4(x+1)$$ There is no common linear factor, I believe it's a typo