1. ## Find x

$\displaystyle 4x^3-4x+1=0$

Ok, here is how I tried to solve and where I stucked at

$\displaystyle 4x^3-4x+1=0$
$\displaystyle 4x(x^2-1)+1=0$
$\displaystyle 4x(x-1)(x+1)=-1$

I don't know what to do next

2. Alas, that won't work, but rewrite $\displaystyle -4x$ as $\displaystyle -2x-2x$ and factor by grouping.

3. ok, so

$\displaystyle 4x^3-4x+1=0$
$\displaystyle 4x^3-2x-2x+1=0$
$\displaystyle 2x(2x^2-1)-(2x-1)=0$

hmmm...

4. Ohhh, sorry, I thought it was a $\displaystyle x^2.$

I'll think in other method.

5. I tried to graph it out, x should have 3 values...

6. Originally Posted by jkami
$\displaystyle 4x^3-4x+1=0$

Ok, here is how I tried to solve and where I stucked at

$\displaystyle 4x^3-4x+1=0$
$\displaystyle 4x(x^2-1)+1=0$
$\displaystyle 4x(x-1)(x+1)=-1$

I don't know what to do next
Are you expected to get exact solutions by hand? In which case you have a difficult job ahead because I don't think there are rational roots. You will need to follow the process explained here: The "Cubic Formula"

Or are decimal approximations sufficient? In which case, use technology to get solutions to the required accuracy.

Posting the whole question, including the context it's taken from, will shed light on this if you cannot answer these questions.