1. Rewrite the equation in a simpler form

   Given $e^{4x}+e^{3x}-4e^{2x}+e^x+1=0$

   Let $t=e^x$ Since $e^x>0$ for all real $x$, we have $t>0$

   Then the equation becomes $t^4+t^3-4t^2+t+1=0$

2. Factor the polynomial

   Observe that the polynomial is symmetric: $t^4+t^3-4t^2+t+1$

   Try factorization of the form $t^4+t^3-4t^2+t+1=(t^2+at+1)(t^2+bt+1)$

   Expanding: $(t^2+at+1)(t^2+bt+1)=t^4+(a+b)t^3+(ab+2)t^2+(a+b)t+1$

   Comparing coefficients:

   \qquad ab+2=-4$$ so $$ab=-6$$ Numbers satisfying $a+b=1$ and $ab=-6$ are $a=3$, $b=-2$. Hence $$t^4+t^3-4t^2+t+1=(t^2+3t+1)(t^2-2t+1)$$ Therefore, $$ (t^2+3t+1)(t-1)^2=0 $$

3. Solve for positive $t$

   So either $t-1=0 \implies t=1$ or $t^2+3t+1=0$

   For the quadratic, $t=\frac{-3\pm\sqrt{9-4}}{2}=\frac{-3\pm\sqrt5}{2}$

   Both values are negative, since $\frac{-3+\sqrt5}{2}<0, \qquad \frac{-3-\sqrt5}{2}<0$

   But $t=e^x>0$, so these are not allowed.

   Thus the only valid solution is $t=1$

4. Convert back to $x$

   Since $e^x=1$ we get $x=0$

5. Count the number of real roots

   There is only one real root.

6. Check options

   • A: $1$ ✅
   • B: $2$ ❌
   • C: $3$ ❌
   • D: $4$ ❌

Therefore, the correct answer is A.