- #1

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any way to solve 3x^8-20x^6-12x^4+12x^2+1 by hand

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- Thread starter nameVoid
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- #1

- 241

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any way to solve 3x^8-20x^6-12x^4+12x^2+1 by hand

- #2

- 290

- 2

After some factoring, you get

[tex]

( x^2+1) (3 x^6-2 x^4 +11x^2+1)

[/tex]

The second factor is a cubic in [tex]x^2[/tex] so you can use the cubic formula, but it's messy.

- #3

- 241

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how did you factor that what is this cubic formula

- #4

- 290

- 2

To give you an idea of what the solutions looks like, I used mathematica to solve it and the six solutions from the cubic formula all look like this

[tex]

x = -\sqrt{\frac{23}{9}-\frac{43 5^{2/3}}{9 \left(2 \left(863+27 i \sqrt{69}\right)\right)^{1/3}}+\frac{43 i 5^{2/3}}{3 \sqrt{3} \left(2 \left(863+27 i \sqrt{69}\right)\right)^{1/3}}-\frac{\left(5 \left(863+27 i \sqrt{69}\right)\right)^{1/3}}{9 2^{2/3}}-\frac{i \left(5 \left(863+27 i \sqrt{69}\right)\right)^{1/3}}{3 2^{2/3} \sqrt{3}}}

[/tex]

- #5

- 290

- 2

how did you factor that what is this cubic formula

You can check to see if it might be factorable over Z with eisenstien's criterion, and I guessed one of the factors (which wasn't difficult because the constant is 1) then used polynomial division. If you had to solve something like that factoring wouldn't really be that practical (unless the coefficients were symmetric or something) but there was an easy factor in this. In fact Eisenstein's can't even be used to check if the second factor can be factored, so this isn't something you would do by hand.

The cubic formula is like the quadratic formula but for cubics. It's very long and not worth memorizing because the answers almost always end up messy.

Last edited:

- #6

- 241

- 0

well this equation comes up on the second derivative when trying to sketch the the curve y=(2-x^2)/(1+x^4)

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