How to solve this 3rd degree polynomial? I looked up the factoring method, but I think this one is calculated using a calculator. With a basic calculator, how do I set this up?
$16x^3 - 18x^2 - 2x  - 1 = 0$
I factored it to become...  $2x(8x^2-9x-2)=1$
In my note book I had it as $x = 1.06272$. I would like to learn how I can set it up for the basic (non-graphing) calculator. 
 A: Use Newton-Rapshon.


*

*Calculate the derivative $f'(x)=48x^2-36x-2$.

*Choose an arbitrary point where to start the iteration (usually one you think is close to a root). I'll choose $x_0=1$.

*Calculate the next point of the iteration as $$ x_1=x_0-\frac{f(x_0)}{f'(x_0)},$$ which in this case is $x_1=1-\frac{-5}{10}=1.5$

*Repeat the process by defining $$x_{n+1} = x_n-\frac{f(x_n)}{f'(x_n)}.$$ In this case, the next term is $$x_2=1.5-\frac{f(1.5)}{f'(1.5)}=1.5-\frac{9.5}{52}=1.317307692.$$
And I'll do it a couple more times because the convergence is pretty cool. You can verify that:
$$x_3 = 1.266976947\text{ and } x_4=1.263154315.$$
Which are actually quite close to the root, which is approximately $x=1.26313285$.
A: Alpha finds the real root to be about $1.2631$, far from your notebook.  You can see that the root is between $1$ and $2$ by checking the signs (which you should have done as part of the rational root test).  If I had a non-programmable calculator, I would use bisection, testing $1.5, 1.25, 1.375$, etc.
