1. Differentiate the polynomial

Given $P(x)=x^4+ax^3+bx^2+cx+d$ so $P'(x)=4x^3+3ax^2+2bx+c.$

We are told that $x=0$ is the only real root of $P'(x)=0$.

Since $P'(0)=c=0$, we get $c=0.$ Thus $P'(x)=4x^3+3ax^2+2bx=x\bigl(4x^2+3ax+2b\bigr).$

Because $x=0$ is the only real root of $P'(x)=0$, the quadratic $4x^2+3ax+2b$ must have no real roots.

Also, since its leading coefficient is positive, we have $4x^2+3ax+2b>0 \quad \text{for all real }x.$ Hence for $x\neq 0$, $P'(x)=x\bigl(4x^2+3ax+2b\bigr)$ has the same sign as $x$.

So:

• for $x<0$, $P'(x)<0$
• for $x>0$, $P'(x)>0$

Therefore, $P(x)$ is:

• decreasing on $(-\infty,0)$
• increasing on $(0,\infty)$

Thus $x=0$ is a minimum point of $P$.

2. Compare values on the interval $[-1,1]$

Since $P$ decreases on $[-1,0]$, we get $P(-1)>P(0).$ Since $P$ increases on $[0,1]$, we get $P(1)>P(0).$ So the minimum on $[-1,1]$ occurs at $x=0$, not at $x=-1$. Hence: $P(-1) \text{ is not the minimum.}$

Now we are additionally given $P(-1)<P(1).$ Among the endpoints, $P(1)$ is larger. Since the only interior critical point $x=0$ is a minimum, the maximum on the closed interval $[-1,1]$ must occur at one of the endpoints. Because $P(1)>P(-1),$ we conclude that $P(1) \text{ is the maximum on }[-1,1].$

3. Check options

• A: $P(-1)$ is not minimum but $P(1)$ is the maximum of $P$ on $[-1,1]$ ✅
• B: false
• C: false
• D: false

Therefore, the correct option is $\boxed{A}.$

4. Comparison with stored answer

Stored correct answer: A

My derived answer is also A, so they agree.