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
 A: 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}
A: Hint:
Method 1: You essentially need the root of:
$$4x^3-2x^2-8x-1=0$$
If you use Descartes' Rule, you can see that the above equation has a positive real root and two negative real roots.
So this tells you that you have three real roots and hence three intersections.
Now finding them is difficult without a graph so you may use Newton Raphson's method for each root or you may get an accurate root and you could factor it out to get the other roots but that would be pretty lengthy.
You should notice that the product of the roots: $$x_1*x_2*x_3=\frac{-(-1)}{4}=\frac{1}{4}$$
Also sum of the roots: $$x_1+x_2+x_3=\frac{-(-2)}{4}=\frac{1}{2}$$
In case you are using Newton Raphson's method you could use the above equations for the initial three good guesses required for the three roots.
A: Hint:
Method 2: you could take: $$y=x-\frac{1}{6}$$
This will eliminate the quadratic term and reduce the equation to:
$$y^3 - \frac{25 y}{12} - \frac {16}{27} = 0$$
For this equation we have a wicked formula known as the Cubic Formula. You can find the formula and similar methods here at Wiki.
