<s>[INST] Intersection of curve $ y = x^4 – 6x^3 + 12x^2 + cx + 1$ The number of integers in the range of 'c' such that there exists a line which intersects the curve   $ y = x^4 – 6x^3 + 12x^2 + cx + 1$ at four distinct points. 
My approach we need to intersect with line $y=mx+C$
Substituting we get  $x^4 – 6x^3 + 12x^2 + cx + 1-mx-C=0$ 
Now this is an equation of polynomial of degree 4
$x^4 – 6x^3 + 12x^2 + (c-m)x + 1-C=0$ 
All four roots needs to be real.
I don't have any idea how to check whether all the roots are real or not.
 [/INST] Consider equation 
$$
x^4 – 6x^3 + 12x^2 - 9x + 2 = 0.\tag{1}
$$
It is equivalent to
$$
(x - 2) (x - 1) (x^2 - 3 x + 1) = 0,\tag{2}
$$
and has $4$ real solutions:
$$
x_{1} = 1,\\
x_{2} = 2, \\
x_{3} = \dfrac{3-\sqrt{5}}{2}, \\
x_{4} = \dfrac{3-\sqrt{5}}{2}.\tag{3}
$$
Then for any $c\in\mathbb{R}$ we can rewrite $(1)$ in the form
$$
0 = x^4 – 6x^3 + 12x^2 -9x + 2 \\ 
= (x^4 – 6x^3 + 12x^2 +cx + 1) - ((9+c)x-1),
$$
which means that equation
$$
x^4 – 6x^3 + 12x^2 +cx + 1 = (9+c)x-1
$$
has those four real solutions: see $(3)$.
Other words, given curve and line $$y=(9+c)x-1$$ intersect in four distinct points with abscissas from list $(3)$.
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