# How to find the roots of a higher than 4th degree polynomial? in this case, an 8th degree one

The equation I'm trying to find the roots of is: $3x^8-12x^4 +1 =0$

The roots I want to find are the real roots, not the complex ones. Using an online calculator , it says that the roots of this equation are:

$x = \dfrac{\sqrt{6-\sqrt{33}}}{\sqrt{3}} \approx 0.5401828449376001$

$x = -\dfrac{\sqrt{6-\sqrt{33}}}{\sqrt{3}} \approx −0.5401828449376001$

$x = \dfrac{\sqrt{\sqrt{33}+6}}{\sqrt{3}} \approx 1.406626835288273$

and

$x = -\dfrac{\sqrt{\sqrt{33}+6}}{\sqrt{3}} \approx -1.406626835288273$

How do I go about finding these roots? I don't know how to use integrals and the like yet as I have not gotten to that point yet in the book I'm studying on, and so far I only know how to use derivatives and limits, so if possible I'd like it if the solutions didn't include those. If the solution has to include those, I will check those out too to see if I understand them.

• Set $u=x^4$ therefore you have $3u^2 -12u+2 = 0$ to solve which seems to be straightforwards Aug 16, 2017 at 9:36
• oh! thanks. that was alot more simpler than I thought it'd be tbh Aug 16, 2017 at 9:41

Hint. Note that $$0=3x^8-12x^4 +1 =3(x^8-4x^4+4)-12 +1=3(x^4-2)^2-11.$$ Hence $$x^4=2\pm\sqrt{\frac{11}{3}}=\frac{6\pm\sqrt{33}}{3} \quad (\mbox{both positive real numbers})$$ Now you may find $x$ easily.

• thanks, but actually , I just noticed that there was a typo in the question I posted, it's actually $3x^4 - 12x^2 +1 =0$ instead of $+2$ , so I dont think this method will work with that too, right? Aug 16, 2017 at 9:44
• @user464154 The method works also in this case. Aug 16, 2017 at 9:46
• okay Thanks! I'll try it out now Aug 16, 2017 at 9:47

Oh, this is just quadratic equation. Take $t=x^4$.Then you get $3t^2-12t+1=0$ so $$t_{1,2} = \frac{6\pm\sqrt{33}}{3}$$

Then

$$x^4 =\frac{6\pm\sqrt{33}}{3} \Longrightarrow x= \pm \sqrt{\frac{6\pm\sqrt{33}}{3}}$$

• @user464154 Please think before accept the answer. Aug 17, 2017 at 12:47