Question:

Step 1: Assume $P(x)$ is a polynomial of degree 2, i.e., $P(x) = ax^2 + bx + c$. Substitute this into the given equation $P(x) + P(2x) = 5x^2 - 18$: $$ (ax^2 + bx + c) + (a(2x)^2 + b(2x) + c) = 5x^2 - 18 $$ $$ ax^2 + bx + c + 4ax^2 + 2bx + c = 5x^2 - 18 $$ $$ 5ax^2 + 3bx + 2c = 5x^2 - 18 $$ Step 2: Equate the coefficients of the powers of $x$ on both sides of the equation: For $x^2$: $5a = 5 \Rightarrow a = 1$. For $x$: $3b = 0 \Rightarrow b = 0$. For the constant term: $2c = -18 \Rightarrow c = -9$. Thus, the polynomial is $P(x) = x^2 - 9$. Step 3: Determine the number of solutions for $P(x) = e^x$, which is $x^2 - 9 = e^x$. Let $\phi(x) = x^2 - 9 - e^x$. Evaluate $\phi(x)$ at several points: $\phi(0) = 0^2 - 9 - e^0 = -9 - 1 = -10$. $\phi(-3) = (-3)^2 - 9 - e^{-3} = 9 - 9 - e^{-3} = -e^{-3} < 0$. $\phi(-4) = (-4)^2 - 9 - e^{-4} = 16 - 9 - e^{-4} = 7 - e^{-4}$. Since $e^{-4}$ is a small positive number, $7 - e^{-4} > 0$. Since $\phi(-4) > 0$ and $\phi(-3) < 0$, by the Intermediate Value Theorem, there is at least one root for $x \in (-4, -3)$. Consider the derivatives: $\phi'(x) = 2x - e^x$ and $\phi''(x) = 2 - e^x$. $\phi''(x) = 0$ when $e^x = 2$, so $x = \ln 2 \approx 0.69$. For $x < \ln 2$, $\phi''(x) > 0$, so $\phi(x)$ is convex. For $x > \ln 2$, $\phi''(x) < 0$, so $\phi(x)$ is concave. The function $\phi(x)$ has a local maximum at $x = \ln 2$. $\phi(\ln 2) = (\ln 2)^2 - 9 - e^{\ln 2} = (\ln 2)^2 - 9 - 2 \approx (0.69)^2 - 11 \approx 0.47 - 11 = -10.53 < 0$. Since the local maximum is negative, there are no roots for $x > 0$. For $x < 0$, $\phi(x)$ is convex. As $x \to -\infty$, $x^2$ dominates $e^x$, so $\phi(x) \to \infty$. Since $\phi(x) \to \infty$ as $x \to -\infty$, $\phi(-4) > 0$, and $\phi(-3) < 0$, there is exactly one root for $x < -3$. Therefore, there is exactly 1 solution for $P(x) = e^x$. Step 4: Evaluate the definite integral $\displaystyle\int_0^{\infty} \dfrac{dx}{P(x)+25}$. Substitute $P(x) = x^2 - 9$: $$ \int_0^{\infty} \dfrac{dx}{(x^2 - 9) + 25} = \int_0^{\infty} \dfrac{dx}{x^2 + 16} $$ This is a standard integral of the form $\int \frac{1}{x^2+a^2} dx = \frac{1}{a}\arctan\left(\frac{x}{a}\right)$. Here $a=4$. $$ \int_0^{\infty} \dfrac{dx}{x^2 + 4^2} = \left[ \frac{1}{4}\arctan\left(\frac{x}{4}\right) \right]_0^{\infty} $$ $$ = \frac{1}{4}\left( \lim_{x \to \infty}\arctan\left(\frac{x}{4}\right) - \arctan\left(\frac{0}{4}\right) \right) $$ $$ = \frac{1}{4}\left( \frac{\pi}{2} - 0 \right) = \frac{\pi}{8} $$ Step 5: The number of solutions for $P(x) = e^x$ is 1. The value of the integral $\displaystyle\int_0^{\infty} \dfrac{dx}{P(x)+25}$ is $\dfrac{\pi}{8}$.