Let f(x) = ax3 − x2 + bx + 6. If (x+2) is a factor of f(x) and the remainder when f(x) is divided by (x+1) is 10, find the values of a and b. Then, find the roots of f(x)=0.

Since $(x+2)$ is a factor of $f(x)$, we know that $f(-2) = 0$.

$$f(-2) = a(-2)^3 - (-2)^2 + b(-2) + 6 = 0$$ $$-8a - 4 - 2b + 6 = 0$$ $$-8a - 2b + 2 = 0$$

Dividing by $-2$, we get:

$$4a + b - 1 = 0 \implies 4a + b = 1 \quad (*)$$

When $f(x)$ is divided by $(x+1)$, the remainder is 10. So, $f(-1) = 10$.

$$f(-1) = a(-1)^3 - (-1)^2 + b(-1) + 6 = 10$$ $$-a - 1 - b + 6 = 10$$ $$-a - b + 5 = 10$$ $$-a - b = 5 \quad (**)$$

We have the system:

1. $4a + b = 1$
2. $-a - b = 5$

Add equation (1) and equation (2):

$$(4a + b) + (-a - b) = 1 + 5$$ $$3a = 6$$ $$a = 2$$

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

$$-(2) - b = 5$$ $$-2 - b = 5$$ $$-b = 7$$ $$b = -7$$

The values of the constants are $a=2$ and $b=-7$.

With $a=2$ and $b=-7$, the polynomial is $f(x) = 2x^3 - x^2 - 7x + 6$.
Since $(x+2)$ is a factor, $x=-2$ is a root. We can use synthetic division to find the other factors.

$$\begin{array}{c|cccc} -2 & 2 & -1 & -7 & 6 \\ & & -4 & 10 & -6 \\ \hline & 2 & -5 & 3 & 0 \end{array}$$

The quotient is $2x^2 - 5x + 3$.
So, $f(x) = (x+2)(2x^2 - 5x + 3)$.