The question I'm trying to answer is for what values of $p$ does $f(x)=x^3 + px^2 + qx - 4$ have three real roots. It is given that $(x-1)$ is a factor of $f(x)$ and that $p + q = 3$.
Using algebraic division you can get it down to, assuming I've done it right: $$f(x)=(x-1)(x^2 + (p+1)x + 4)$$
If you use the discriminant on the quadratic you can work out its number of real roots, but does it have to have real roots for $f(x)$ to have real roots? Or is the number of roots of the quadratic irrelevant to the number of roots of the cubic?
Your polynomial division is correct, indeed we have $$ f(x)=x^3 + px^2 + qx - 4=(x-1)(x^2 + (p+1)x + 4)=(x-1)g(x) $$ if we assume $p+q=3$ and set $g(x):=x^2 + (p+1)x + 4$
Every root (at most two) of the quadratic polynomial $g(x)$ is automatically a root of $f(x)$, so the number of roots of $g(x)$ is essential for the number of roots of the cubic $f(x)$.
Now we can proceed in finding the remaining real roots which are with respect to $p$ $$ x_{2,3}=-\frac{p+1}{2}\pm\sqrt{\frac{(p+1)^2}{4}-4} $$ so $x_{2,3}$ are real iff $$ \frac{(p+1)^2}{4}-4\geq0\iff (p+1)^2\geq16\iff \pm(p+1)\geq4 $$ and this means $$ p\geq3 \text{ or }p\leq-5 $$ and then the cubic $f(x)$ has three real roots $$ \left\{1,-\frac{p+1}{2}-\sqrt{\frac{(p+1)^2}{4}-4}=:x_2,-\frac{p+1}{2}+\sqrt{\frac{(p+1)^2}{4}-4}=:x_3\right\} $$ so we would have in total the factorization of the cubic $$ f(x)=(x-1)(x-x_2)(x-x_3) $$ In general, a polynomial can have imaginary roots and also real roots at the same time, but the algebraic multiplicity always adds up to the order of the polynomial.
For example the cubic polynomial $s(x)$ $$ s(x)=(x+1)(x^2+1) $$ has one real root $x_1=-1$ and two complex roots $x_2=i,x_3=-i$.