How do I find the points at which the following functions intersect? And how many times do they intersect? I have a function $f(x) = 4x^3 − 2x^2 − 4x − 2 $
And a line $y = 4x -1$
And I need to know where they intersect, I know that I have to do $4x^3 − 2x^2 − 4x − 2 = 4x -1$
But I can't find the x-values and I'm not allowed to use graphs.
As for how many times they intersect, I think it's three because the polynomial is third degree

The 
discriminant
of the cubic $ax^3+bx^2+cx+d$ is
\begin{align} 
\Delta(a,b,c,d) &=18abcd-4b^3d+b^2c^2-4ac^3-27a^2d^2
,
\end{align}  
in particular for the cubic 
\begin{align} 
4x^3-2x^2-8x-1
\tag{1}\label{1}
\end{align}
the discriminant is
\begin{align} 
\Delta(4,-2,-8,-1)&=6832>0
,
\end{align}
hence, \eqref{1} has three distinct real roots.
Dividing \eqref{1} by $4$
and substituting $t+\tfrac16$ for $x$, we get
a reduced cubic equation
\begin{align} 
t^3+pt+q=&0
\tag{2}\label{2}
,
\end{align}
\begin{align} 
\text{where}\quad
p&=-\tfrac{25}{12}
,\quad
q=-\tfrac{16}{27}
.
\end{align} 
Applying 
Trigonometric solution for three real roots
to \eqref{2},
we can find the roots as
\begin{align}
t_k&=
2\sqrt{-\frac{p}3}
\cos\left(\frac1{3}\arccos 
\left({\frac{3q}{2p}}\sqrt{-\frac3p}\right)
-\frac{2\pi k}3\right)
\quad \text{for}\quad k=0,1,2
,\\
\end{align}
\begin{align} 
t_0&=\phantom{-}\tfrac53\cos(\tfrac13\arccos(\tfrac{64}{125}))
\approx  1.56878
,\\
t_1&=-\tfrac53\cos(\tfrac13\arccos(\tfrac{64}{125})+\tfrac\pi3)
\approx -0.29702
,\\
t_2&=-\tfrac53\sin(\tfrac13\arccos(\tfrac{64}{125})+\tfrac\pi6)
\approx -1.27176
.
\end{align}  
Then the corresponding roots of \eqref{1} are
\begin{align} 
x_0&=t_0+\tfrac16\approx 1.73545
,\\
x_1&=t_1+\tfrac16\approx -0.13036
,\\
x_2&=t_2+\tfrac16\approx -1.10509
.
\end{align}
