I Get It Now :)

If $f(x)-\ell(x)$ has four zeroes, where $\ell(x)$ is linear and $f(x) = x^4+9x^3+cx^2+9x+4,$ then the second derivative has at least two zeroes (by two applications of Rolle's theorem ). But the second derivative is just $f''(x) = 12x^2+54x+2c.$ This has two zeroes if and only if the discriminant $54^2-96c$ is positive, which happens if and only if $c < 243/8.$

Conversely, suppose $c < 243/8.$ Then $f''(x) = 12x^2+54x+2c$ has two zeroes, say $m$ and $n$ with $m < n.$ Then $f''(x)$ is negative between $m$ and $n,$ so $f'(x)$ is decreasing on $[m,n].$ So we can find an $a$ such that $f'(m) - a$ is positive but $f'(n)-a$ is negative. This implies that $f'(x)-a$ has three zeroes: it goes to $-\infty$ on the left and $+\infty$ on the right, but in between it is positive at $m$ but negative at $n.$ So (by the Intermediate Value Theorem ) there is one zero in $(-\infty,m),$ one in $(m,n),$ and one in $(n,+\infty).$

Now since $f'(x)-a$ has three zeroes, let's say $c_1 < c_2 < c_3,$ and its antiderivative $f(x)-ax$ is a monic quartic, the shape of the graph of $f(x)-ax$ looks like a (smoothed) W; that is, $f(c_1)-ac_1$ and $f(c_3)-ac_3$ are both less than $f(c_2)-ac_2.$ Let $b$ be a number less than $f(c_2)-ac_2$ but greater than $f(c_1)-ac_1$ and $f(c_3)-ac_3.$ Then $f(x)-ax-b$ goes to $+\infty$ on the left, is negative at $c_1,$ positive at $c_2,$ negative at $c_3,$ and goes to $+\infty$ on the right. This implies (again by the IVT) that $f(x)-ax-b$ has four zeroes, as desired.