A polynomial P is positive for `x lt 0`, and the area of the region bounded by `P(x)`, then x-axis, and the vertical lines `x=0 and x= K` is `K^(2) (K+3)//3`. The polynomial `P(x)` is

Options

A. `x^(2) + 2x`
B. `x^(2) + x+1`
C. `x^(2) + 2x+1`
D. `x^(3) +1`

Answer

To solve the problem, we need to find the polynomial \( P(x) \) given that it is positive for \( x < 0 \) and the area bounded by \( P(x) \), the x-axis, and the vertical lines \( x = 0 \) and \( x = K \) is given by

\[ \frac{K^2 (K + 3)}{3}. \]

### Step 1: Set up the integral for the area
The area \( A \) under the curve \( P(x) \) from \( x = 0 \) to \( x = K \) can be expressed as:

\[ A = \int_0^K P(x) \, dx. \]

According to the problem, this area is equal to

\[ \frac{K^2 (K + 3)}{3}. \]

### Step 2: Differentiate the area with respect to \( K \)
To find \( P(x) \), we can differentiate both sides of the equation with respect to \( K \):

\[ \frac{d}{dK} \left( \int_0^K P(x) \, dx \right) = \frac{d}{dK} \left( \frac{K^2 (K + 3)}{3} \right). \]

Using the Fundamental Theorem of Calculus, the left-hand side becomes:

\[ P(K). \]

### Step 3: Differentiate the right-hand side
Now we differentiate the right-hand side:

\[ \frac{d}{dK} \left( \frac{K^2 (K + 3)}{3} \right) = \frac{1}{3} \frac{d}{dK} (K^3 + 3K^2). \]

Using the power rule for differentiation:

\[ \frac{d}{dK} (K^3) = 3K^2 \quad \text{and} \quad \frac{d}{dK} (3K^2) = 6K. \]

So, we have:

\[ \frac{d}{dK} (K^3 + 3K^2) = 3K^2 + 6K. \]

Thus,

\[ P(K) = \frac{1}{3} (3K^2 + 6K) = K^2 + 2K. \]

### Step 4: Substitute \( K \) with \( x \)
Since \( P(K) \) is a polynomial in \( K \), we can replace \( K \) with \( x \) to express \( P(x) \):

\[ P(x) = x^2 + 2x. \]

### Conclusion
The polynomial \( P(x) \) is:

\[ \boxed{x^2 + 2x}. \]