The polynomial f(x) = ax3 − x2 + bx + 6. (a) Find the values of the constants a and b. (b) Find the values of x for which f(x) = 0.

Since $(x-2)$ is a factor of $f(x)$, we know that $f(2) = 0$.
Substitute $x=2$ into the polynomial:

Divide the equation by 2:

When $f(x)$ is divided by $(x+1)$, the remainder is 10. This means $f(-1) = 10$.
Substitute $x=-1$ into the polynomial:

We have:

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

Subtract Equation 2 from Equation 1:

Substitute the value of $a$ into Equation 2:

The values of the constants are $\boxed{a = \frac{4}{3}, b = -\frac{19}{3}}$.

To make calculations easier, we can multiply the entire equation by 3 (since we are looking for roots, $f(x)=0$ implies $3f(x)=0$):

Since $(x-2)$ is a factor, $x=2$ is a root. We can use synthetic division with 2:

2 | 4   -3   -19   18
  |     8    10   -18
  --------------------
    4    5    -9    0

The quotient is $4x^2 + 5x - 9$.
So, $f(x) = (x-2)(4x^2 + 5x - 9) = 0$.

We can factor this quadratic equation. We look for two numbers that multiply to $4 \times (-9) = -36$ and add to 5. These numbers are 9 and $-4$.

From $(x-2)=0$, we get $x=2$.
From $(x-1)=0$, we get $x=1$.
From $(4x+9)=0$, we get $4x = -9 \implies x = -\frac{9}{4}$.

The values of $x$ for which $f(x)=0$ are $\boxed{x = 2, x = 1, x = -\frac{9}{4}}$.