The polynomial of degree $3, P(x)$, has a root of multiplicity 2 at $x=1$ and a root of multiplicity 1 at $x=-1$. The $y$-intercept is $y=-0.6$. Find an equation for $P(x)$. \[ P(x)= \] Submit Question

Step 1 ：The polynomial of degree $3, P(x)$, has a root of multiplicity 2 at $x=1$ and a root of multiplicity 1 at $x=-1$. The $y$-intercept is $y=-0.6$.

Step 2 ：We can express the polynomial as $P(x) = a(x-1)^2(x+1)$, where $a$ is a constant.

Step 3 ：We can solve for $a$ by setting $P(0) = -0.6$ and solving for $a$.

Step 4 ：Substituting $x=0$ into the equation gives $P(0) = a(0-1)^2(0+1) = -a$.

Step 5 ：Setting this equal to $-0.6$ gives the equation $-a = -0.6$, which simplifies to $a = 0.6$.

Step 6 ：Substituting $a = 0.6$ back into the polynomial gives the final equation for $P(x)$.

Step 7 ：\(\boxed{P(x) = -0.6(x-1)^2(x+1)}\)