Remember that the constant term $r$ equals $f(0)$. Try plugging $x = 0$ into your factored form $(x - 1)(x + 2)(x - a)$ and set that equal to 14 to solve for the unknown root.

Once you know all three roots, think about how $(x - r_1)(x - r_2)(x - r_3)$ expands. Which combination of the roots gives the coefficient of $x$?

In Desmos, enter $g(x) = (x - 1)(x + 2)(x - a)$. Desmos will create a slider for $a$, representing the unknown third root.

On a new line, enter $g(0)$. Then adjust the slider for $a$ until the value shown for $g(0)$ is 14. The corresponding value of $a$ is the third root of the polynomial.

Replace $a$ in $g(x)$ with the value you found, so $g(x)$ matches the actual polynomial. On a new line, type expand(g(x)) (or just retype the expression expanded) and look at the resulting cubic. The coefficient of $x$ in that expanded form is the value of $q$.

We are told that $(x - 1)$ and $(x + 2)$ are factors of the monic cubic

That means $x = 1$ and $x = -2$ are roots.

Since $f(x)$ is monic (leading coefficient 1) and all coefficients are integers, we can write it as

for some integer root $a$ that we need to determine.

The constant term $r$ is the value of the polynomial at $x = 0$.

Compute $f(0)$ using the factored form:

But we are told that $r = 14$, so

Solve for $a$:

So the three roots of $f(x)$ are $1$, $-2$, and $7$, and

For a monic cubic with roots $r_1, r_2, r_3$,

So the coefficient of $x$ is the sum of the pairwise products of the roots.

Here $r_1 = 1$, $r_2 = -2$, and $r_3 = 7$, so

Compute this sum:

Therefore, the value of $q$ is $-9$.