If we put $a = c = 0$, polynomial $x^4 + b x^2 + d$ can be written as product of four linear terms if and only if quadratic polynomial $y^2 + b y + d$ has two nonnegative roots. Therefore $P(0, b, 0, d) \ge 0$ if and only if $b \le 0$, $d \ge 0$ and $b^2 - 4d \ge 0$. For a fixed $b \le 0$ let $Q_b(d) = P(0, b, 0, d)$. Now, $Q_b(d) \ge 0$ if and only if $0 \le d \le \frac{b^2}{4}$ and hence by continuity of $Q_b$, $Q_b(b^2/4)$ must be zero. This implies that for all $b \le 0$, one variable polynomial $P(0, b, 0, \frac{b^2}{4}) = 0$, and hence this polynomial is always zero. This means that polynomial $x^4 + b x^2 + \frac{b^2}{4}$ has four real roots for all values of $b$. Contradiction! $\square$