The quadratic polynomial $P(x),$ with real coefficients, satisfies
\[P(x^3 + x) \ge P(x^2 + 1)\]for all real numbers $x.$  Find the sum of the roots of $P(x).$
Let $P(x) = ax^2 + bx + c.$  Then
\[a(x^3 + x)^2 + b(x^3 + x) + c \ge a(x^2 + 1)^2 + b(x^2 + 1) + c\]for all real numbers $x.$  This simplifies to
\[ax^6 + ax^4 + bx^3 - (a + b)x^2 + bx - a - b \ge 0.\]This factors as
\[(x - 1)(x^2 + 1)(ax^3 + ax^2 + ax + a + b) \ge 0.\]For this inequality to hold for all real numbers $x,$ $ax^3 + ax^2 + ax + a + b$ must have a factor of $x - 1.$  (Otherwise, as $x$ increases from just below 1 to just above 1, $x - 1$ changes sign, but $(x^2 + 1)(ax^3 + ax^2 + ax + a + b)$ does not, meaning that it cannot be nonnegative for all real numbers $x.$)  Hence, setting $x = 1,$ we get $a + a + a + a + b = 0,$ so $4a + b = 0.$

Then by Vieta's formulas, the sum of the roots of $ax^2 + bx + c = 0$ is $-\frac{b}{a} = \boxed{4}.$
4
Level 5
Intermediate Algebra
Problem: The quadratic polynomial $P(x),$ with real coefficients, satisfies
\[P(x^3 + x) \ge P(x^2 + 1)\]for all real numbers $x.$  Find the sum of the roots of $P(x).$
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To solve the problem, we start by letting the quadratic polynomial be expressed in the standard form: \( P(x) = ax^2 + bx + c \), where \( a, b, c \) are real coefficients and \( a \neq 0 \).

Next, we analyze the inequality \( P(x^3 + x) \ge P(x^2 + 1) \). We will substitute \( x^3 + x \) and \( x^2 + 1 \) into the polynomial:

1. Calculate \( P(x^3 + x) \):
   \[
   P(x^3 + x) = a(x^3 + x)^2 + b(x^3 + x) + c
   \]
   Expanding this gives:
   \[
   P(x^3 + x) = a(x^6 + 2x^4 + x^2) + b(x^3 + x) + c
   \]

2. Calculate \( P(x^2 + 1) \):
   \[
   P(x^2 + 1) = a(x^2 + 1)^2 + b(x^2 + 1) + c
   \]
   Expanding this gives:
   \[
   P(x^2 + 1) = a(x^4 + 2x^2 + 1) + b(x^2 + 1) + c
   \]

3. Set up the inequality:
   \[
   a(x^6 + 2x^4 + x^2) + b(x^3 + x) + c \ge a(x^4 + 2x^2 + 1) + b(x^2 + 1) + c
   \]
   Simplifying this leads to:
   \[
   ax^6 + 2ax^4 + bx^3 + bx + c \ge ax^4 + 2ax^2 + a + bx^2 + b + c
   \]
   Rearranging terms results in:
   \[
   ax^6 + (2a - a)x^4 + (b - 2a - b)x^2 + (b - a)x + (c - c - b) \ge 0
   \]
   This simplifies to:
   \[
   ax^6 + ax^4 - ax^2 + (b - a)x \ge 0
   \]

4. For the inequality to hold for all \( x \), the leading coefficient \( a \) must be non-negative, and the polynomial must not change sign. This suggests that \( P(x) \) could be a perfect square.

5. Assume \( P(x) = a(x - r)^2 \) for some root \( r \). The sum of the roots of \( P(x) \) is given by \( 2r \).

6. To satisfy the inequality, we can test specific values of \( x \). For instance, if we set \( x = 0 \), we find:
   \[
   P(0) \ge P(1)
   \]
   This leads to conditions on \( c \) and \( a \).

7. Testing \( P(x) = x^2 - 4x + 4 \) (which has roots at \( x = 2 \)), we find:
   \[
   P(x) = (x - 2)^2
   \]
   This satisfies the inequality since it is non-negative for all \( x \).

8. The sum of the roots of \( P(x) = 0 \) is \( 2 + 2 = 4 \).

Thus, the final answer is \( 4 \).<|END_OF_LATENT|> The final answer is $4$.