This week's problem was correctly answered by anemone, magneto, MarkFL, and mente oscura. You can find anemone's solution below.
[sp]If there is a straight line that will intersect the curve $y=x^4+cx^3+12x^2-5x+2$ at four distinct points, that means the curve of $y$ must have three critical points or more precisely, two inflexion points.
From the definition of inflexion point, we know we can find its two x-coordinates of the inflexion point by setting the second derivative of the original function as zero. In this instance, the second derivative of the quartic function yields a quadratic function and if we solve for $x$ of the quadratic function by letting it equals zero, we must obtain two distinct $x$ values. This is achievable iff the discriminant is greater than zero.
$y'=4x^3+3cx^2+24x-5$
$y''=12x^2+6cx+24$
The discriminant of the second derivative must be greater than zero, i.e. $(6c)^2-4(12)(24)>0$. Solving this inequality for $c$, we will find the range of value of $c$ that will allow a straight line to cut the curve $y=x^4+cx^3+12x^2-5x+2$ at four distinct points.
$(6c)^2-4(12)(24)>0$
$2^23^2c^2-2^73^2>0$
$c^2-2^5>0$
$c^2-4^2(2)>0$
$(c-4\sqrt{2})(c+4\sqrt{2})>0$
This gives the solution set of $c$ as:
${c: c<-4\sqrt{2}\; \text{or}\; c>4\sqrt{2}}$[/sp]