The polynomial of degree $3, P(x)$, has a root of multiplicity 2 at $x=3$ and a root of multiplicity 1 at $x=-1$. The $y$-intercept is $y=-2.7$; Find a formula for $P(x)$.
\[
P(x)=
\]
Answer
\(\boxed{P(x) = -0.3*(x - 3)^2*(x + 1)}\) is the final answer.
Step 1 ：The polynomial of degree 3 with a root of multiplicity 2 at x=3 and a root of multiplicity 1 at x=-1 can be written in the form of \(P(x) = a*(x-3)^2*(x+1)\).
Step 2 ：The y-intercept is the value of the function at x=0, so we can substitute x=0 into the function and set it equal to -2.7 to solve for a.
Step 3 ：After solving, we find that \(a = -0.3\).
Step 4 ：Substituting the value of a back into the polynomial, we get \(P(x) = -0.3*(x - 3)^2*(x + 1)\).
Step 5 ：\(\boxed{P(x) = -0.3*(x - 3)^2*(x + 1)}\) is the final answer.
