Solve x^6 + 25x^5 + 192x^4 - 7394x^3 + 48936x^2 - 113304x + 79488=0

[mente oscura]

Hello.:)

Find the 6 reals roots:

P(x)=x^6-25x^5-192x^4+7394x^3-48936x^2+113304x-79488

Regards.

 

[lfdahl]

Spoiler

I have factorized $P(x)$ ”the hard way”:
\[P(x)=(x+18)(x-23)(x^2-6x+6)(x^2-14x+32)
\\\\
P(x) = 0 \Rightarrow
x \in\left \{ -18,23,3\pm \sqrt{3},7\pm \sqrt{17} \right \}\]

There must be a much more elegant way. I do hope someone appears with a better reply(Whew)

lfdahl

 

[mente oscura]

Spoiler

I have factorized $P(x)$ ”the hard way”:
\[P(x)=(x+18)(x-23)(x^2-6x+6)(x^2-14x+32)
\\\\
P(x) = 0 \Rightarrow
x \in\left \{ -18,23,3\pm \sqrt{3},7\pm \sqrt{17} \right \}\]

There must be a much more elegant way. I do hope someone appears with a better reply(Whew)

lfdahl

Hello, Idahl.
Thank you, for taking part in the challenge.

But, what calculations have you realized?

(Muscle) ?

Regards. (Med venlig hilsen)

 

[lfdahl]

Spoiler

Hello mente oscura

I have checked the roots numerically and used polynomial division knowing that I was looking for the multiplum of two quadratic polynomials:

$P(x)=(x+18)(x-23)(x^2+ax+b)(x^2+cx+d)$

where:

$(x^2+ax+b)(x^2+cx+d) = x^4-20x^3+122x^2-276x+192$Con vistas mejores

lfdahl