$x^4 + 4x^3 - 2x^2 - 12x + k$ has 4 real roots. Find the condition on k. The question is: $f(x) = x^4 + 4x^3 - 2x^2 - 12x + k$ has 4 real roots. What values can k take? Please drop a hint!
 A: If this degree four polynomial has four real roots, then there is a local minimum between the first two and between the last two roots, and a local maximum between the second and hird root.
The location of the extrema does not change with $k$, only their $y$-values.
Thus we are looking for $k$ such that the two local minima have negative $y$-values and the local maximum has positive $y$-value.
Locate the local extrema as roots of the derivative
$$ 4x^3+12x^2-4x-12 = 4\cdot (x+3)(x+1)(x-1),$$
i.e. the minima are at $x=-3$ and $x=1$, the maximum is at $x=-1$.
The respective $y$-values are $y(-3)=-9+k$, $y(-1)=7+k$, and $y(1)=-9+k$. We conclude that the desired condition is
$$ -7<k<9.$$
(Here I assume that multiple roots are not allowed; otherwise the condition is $-7\le k\le 9$).
A: You can use this substitution $x=u-1$ to change your polynomial as
$$u^4-8u^2+k+7 $$
which has 4 real roots if and only if $\Delta=8^2-4(k+7)=4(9-k)\ge 0$ and $k+7\ge0$ or equivalently $-7\le k\le 9$.
A: Hint:
Lets call vieta for the rescue.
$f(x) = x^4 + 4x^3 - 2x^2 - 12x + k=0$, let $x_1, x_2,x_3$ and $x_4$are the roots.
$\sum x_i=-4$
$\sum x_ix_j=-2$
$\sum x_ix_jx_k=+12$
$x_1x_2x_3x_4=k$
$\sum (x_i)^2=(\sum x_i)^2-2\sum x_ix_j=16-(-4)=12$
$\dfrac{x_1^2+x_2^2+x_3^2+x_4^2}{4} \ge (x_1x_2x_3x_4)^\frac{1}{2}$
$3 \ge k^{\frac{1}{2}} \implies 9 \ge k$. Now check the lower bound using the discriminant formula.
HAIL VIETA!, like I always say.
