Given $P(x)=x^4+a{x}^3+b{x}^2+cx+d$ such that $x=0$ is the only real root of $P^{\prime }(x)=0$. If $P(-1)<P(1)$, then in the interval $[-1,1]$:

$P^{\prime }(x)$ has only one root at $x=0$. This means $P(x)$ either has a global minimum or global maximum at $x=0$.Since the leading coefficient of $P(x)$ is $1>0$, $P(x)\to \infty$ as $x\to \pm \infty$. Thus, $x=0$ must be a global minimum.$P(x)$ is decreasing for $x<0$ and increasing for $x>0$.In $[-1,1]$, the minimum occurs at $x=0$.$P(-1)$ and $P(1)$ are candidates for the maximum.Since $P(-1)<P(1)$ and the function increases for $x\in [0,1]$, $P(1)$ is the maximum in $[-1,1]$.But $P(-1)$ is not the minimum (which is $P(0)$).