solve the equation x^16 - 2x^15 - x^14 + 4x^13 - x^12 - 2x^11 + x^10 = 0 in the real number system.

Question

anonymous · Answer

I know x = 0 with multiplicity is 10.

debbieg · Answer

Right, because you can factor it out:

$\large  x^{10}(x^6 - 2x^5 - x^4 + 4x^3 - x^2 - 2x + 1) = 0 $

Which reduces the problem to solving:
$\large  x^6 - 2x^5 - x^4 + 4x^3 - x^2 - 2x + 1 = 0 $

I have the answer from Wolfram, and it seems like there should be a "trick" to get it, but I'm not sure what it is..... Maybe something akin to factoring by grouping??

Sorry I'm not sure help, but I'm curious now how to approach this.

debbieg · Answer

Other than graphing. You could graph it to see the roots, but you still have to determine their multiplicity. So you would end up doing lots of rounds of division, to make sure that you had them all.