The problem asks for the polynomial $P(x)$ of degree 3, given the following conditions:

1. A root of multiplicity 2 at $x = 1$,
2. A root of multiplicity 1 at $x = -1$,
3. The y-intercept is $y = -0.3$, meaning $P(0) = -0.3$.

Step-by-step solution:

1. Express the polynomial with the given roots:
   • Since there is a root of multiplicity 2 at $x = 1$, one factor of $P(x)$ is $(x - 1)^2$.
   • Since there is a root of multiplicity 1 at $x = -1$, another factor is $(x + 1)$.

Thus, the polynomial can be written as: $P(x) = a(x - 1)^2(x + 1)$ where $a$ is a constant to be determined.

2. Use the y-intercept condition:

   • We are told that $P(0) = -0.3$. Plug $x = 0$ into the polynomial: $P(0) = a(0 - 1)^2(0 + 1) = a(1)(1) = a$ Therefore, $a = -0.3$.

3. Write the final polynomial: Substituting $a = -0.3$ into the polynomial, we get: $P(x) = -0.3(x - 1)^2(x + 1)$

This is the formula for $P(x)$.

Would you like more details or explanations?

Here are some related questions to deepen your understanding:

1. What does it mean for a root to have a "multiplicity" of 2 in a polynomial?
2. How would the graph of this polynomial behave at $x = 1$ and $x = -1$?
3. How would the polynomial change if the y-intercept were a different value?
4. What would happen if there were a third root with a different multiplicity?
5. How can we verify that the polynomial satisfies all given conditions?

Tip: Always check the degree of the polynomial to ensure it matches the number of roots and their multiplicities!