Pls solve in English not French

Given that when $P(x)$ is divided by $(x-1)$, the remainder is $-6$, we have $P(1) = -6$.

Given that $(x+2)$ is a factor of $P(x)$, we have $P(-2) = 0$.

Dividing by $-2$:

Subtract equation $(*)$ from equation $(**)$:

Substitute $a=2$ into equation $(*)$:

So, the polynomial is $P(x) = 2x^3 - 3x^2 - 11x + 6$.
The values are $\boxed{a=2, b=-11}$.

Since $(x+2)$ is a factor, $x=-2$ is a root. We can use synthetic division to find the quadratic factor.

-2 | 2   -3   -11   6
   |     -4    14  -6
   ------------------
     2   -7     3   0

The quotient is $2x^2 - 7x + 3$.
So, $P(x) = (x+2)(2x^2 - 7x + 3) = 0$.
Now, we solve the quadratic equation $2x^2 - 7x + 3 = 0$.
We can factor this quadratic:

This gives us two more roots:

The solutions to $P(x)=0$ are $x = -2$, $x = \frac{1}{2}$, and $x = 3$.
The solutions are $\boxed{x = -2, x = \frac{1}{2}, x = 3}$.