Question 1174083
Absolutely, let's break down how to find such a polynomial.

**Understanding the Requirements**

* **Polynomial Form:** P(x) = ax² + bx + c
* **Divisibility:** P(b) must be divisible by (a - b).
* **Composite Quotient:** The result of P(b) / (a - b) must be a composite number (a number with factors other than 1 and itself).

**Steps to Find a Solution**

1.  **Calculate P(b):**
    * Substitute 'b' for 'x' in the polynomial: P(b) = ab² + b² + c

2.  **Calculate the Divisor (a - b):**
    * This is straightforward subtraction.

3.  **Calculate the Quotient:**
    * Divide P(b) by (a - b): (ab² + b² + c) / (a - b)

4.  **Check for Composite Quotient:**
    * The result of the division must be a composite number.

**Finding an Example**

Let's try some values:

* Let a = 6, b = 2, and c = 4.

* P(b) = (6 * 2²) + 2² + 4 = 24 + 4 + 4 = 32

* (a - b) = 6 - 2 = 4

* Quotient = 32 / 4 = 8

* 8 is a composite number (2 * 4).

**Therefore, the polynomial P(x) = 6x² + 2x + 4 satisfies the conditions.**

**Verification**

* P(2) = 6(2)² + 2(2) + 4 = 32
* a-b = 6-2 = 4
* 32/4 = 8.
* 8 is composite.