Let $f(x)=x^4-4x+1.$ We need the number of distinct real roots of $x^4-4x+1=0.$

1. Use calculus to study the graph

Differentiate: $f'(x)=4x^3-4=4(x^3-1)=4(x-1)(x^2+x+1).$

Since $x^2+x+1>0$ for all real $x$, the only real critical point is $x=1.$

Now check monotonicity:

• For $x<1$, $x^3-1<0 \Rightarrow f'(x)<0$, so $f(x)$ is decreasing.
• For $x>1$, $x^3-1>0 \Rightarrow f'(x)>0$, so $f(x)$ is increasing.

Hence, $x=1$ is the only minimum point of $f(x)$.

2. Value at the minimum

$f(1)=1^4-4(1)+1=1-4+1=-2.$ So the minimum value is negative.

Also, $\lim_{x\to \pm\infty} f(x)=+\infty$ because the leading term is $x^4$.

3. Conclude the number of real roots

Since:

• $f(x)\to +\infty$ as $x\to -\infty$,
• $f$ decreases until $x=1$,
• $f(1)=-2<0$,
• then $f$ increases to $+\infty$ as $x\to +\infty$,

the graph must cross the $x$-axis exactly:

• once for some $x<1$,
• once for some $x>1$.

So there are exactly 2 distinct real roots.

4. Quick check by sign change

• $f(0)=1>0$

• $f(1)=-2<0$
  So one root lies in $(0,1)$.

• $f(1)=-2<0$

• $f(2)=16-8+1=9>0$
  So another root lies in $(1,2)$.

Also, because there is only one turning point, there cannot be more than $2$ real roots.

5. Option check

• A: $4$ ❌
• B: $2$ ✅
• C: $1$ ❌
• D: $0$ ❌

Therefore, the number of distinct real roots is $\boxed{2}.$