Solving a Nice Quartic Equation | #Polynomials

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[jpolowin]

Substitute y = x-3, giving us (y + 1)^5 - y^5 = 1. Expand and cancel the quintic terms.

5y^4 + 10y^3 + 10y^2 + 5y + 1 = 1
5y(y^3 + 2y^2 + 2y + 1) = 0 giving us y = 0 => x = 3
The cubic expression has the obvious root of y = -1 => x = 2
Factor to give the quadratic y^2 + y + 1 = 0 which gives the two complex values for y, from which the two complex values for x are easily found.

[XJWill]

Very contrived equation. It is easier to just substitute x = t + 5/2, and then expanding we get
t^4 + t^2 / 2 - 3/16 = 0
Solving the quadratic in t^2 give
t^2 = 1/4 OR t^2 = -3/4
So the roots for x are
x1 = 1/2 + 5/2 = 3
x2 = -1/2 + 5/2 = 2
x3 = 5/2 + i * sqrt(3) / 2
x4 = 5/2 - i * sqrt(3) / 2
Incidentally, if the equation did not come already "gift wrapped" to suggest the x = t + 5/2 substitution, once
the given equation is fully expanded to a quartic, the highest two terms are
x^4 - 10*x^3
which means to depress the quartic by removal of the x^3 term, we need the substitution
x = t - (-10)/4 = t + 5/2
What a coincidence! <- sarcasm, since the problem was so contrived

[Entroper]

p can also = -1, which yields the complex solutions:

uv = -1, u - v = 1
v = -1/u, u - (-1/u) = 1
u^2 + 1 = u, u^2 - u + 1 = 0
u = (1 +/-sqrt(-3))/2 = 1/2 + sqrt(3)i/2, 1/2 - sqrt(3)i/2
x = 5/2 + sqrt(3)i/2, 5/2 - sqrt(3)i/2

[tontonbeber]

Of course x=2 and x=3 are obvious solutions, then divide polynomial to obtain a new quadratic whoses solutions are probably complex, as I can visualise mentally the curves of (x-2)^5 and (x-3)^5 which are in fact the same curve x^5 translated on x axis ...
But ok lets go brute force ...
x5 - 10 x4 + 40 x3 - 80 x2 + 80 x - 32
-(x5 - 15 x4 + 90 x3 -270 x2 +405 x -243) = 1
5 x4 - 50 x3 +190 x2 -325 x +211-1 = 0
x4 - 10x3 + 38x2 - 65x + 42 divide by x2 - 5x + 6
- (x4 - 5x3 + 6x2) => x2
- 5x3 + 32x2 - 65x + 42
-(- 5x3 + 25x2 - 30x) => -5x
7x2 - 35x + 42 => +7
So factorise into (x2-5x+6)(x2-5x+7)
First quadratic gives obvious real solutions 2 and 3
Second quadratic gives complex solutions 5/2 +/- i V(3)/2

[MrGeorge]

x = 2 and x = 3 are two trivial solutions. (just by observation) If we expand the two brackets the x⁵ terms will cancel out so we will get a quartic polynomial with two known solutions. We just need a long division and the quadratic formula to obtain the other two solutions. But I am just too lazy to do this right now 😋

[pyrite]

rewrite (x-3)^5 as ((x-2)-1))^5, and expand using binomial theorem, (a-b)^5 = a^5-b^5-5ab(a-b)(a^2-ab+b^2)

after substitution,



and then factor and simplify for,

5(x-2)(x-3)(x^2-5x+7)=0

which gives you your four solutions

x=2
x=3
x=(5+sqrt(3)i)/2
x=(5-sqrt(3)i)/2

[danielwhidden]

wouldn't this be a quintic equation? lol