Quick questions on Polynomial division and the remainder and factor theorems

If $P(x)$ is a polynomial of degree $n$ and $D(x)$ is a polynomial of degree $k \le n$, then there exist unique polynomials $Q(x)$ (the quotient) of degree $n - k$ and $R(x)$ (the remainder) of degree less than $k$ such that

If $P(x)$ is divided by $(x - a)$, the remainder is $P(a)$. In other words,

The factor theorem is the special case of the remainder theorem where the remainder is zero.

To divide $P(x)$ by $D(x)$, set up the long-division layout and at each step divide the leading term of the current dividend by the leading term of the divisor, multiply back, and subtract.

Long division is procedural, and the marks come from keeping the layout aligned and each subtraction correct. Below, $x^3 - 2x^2 - 5x + 6$ is divided by $(x - 3)$ one stage at a time. (Because $x = 3$ is a root, the remainder will turn out to be $0$, confirming $(x - 3)$ is a factor.)

To fully factorise a cubic $P(x)$ over the reals:

The rational root theorem is the engine that makes factorising higher polynomials tractable. For a polynomial with integer coefficients, any rational root $\frac{p}{q}$ in lowest terms must have $p$ dividing the constant term and $q$ dividing the leading coefficient. This produces a finite, often short, list of candidates to test. For $P(x) = 2x^3 - 3x^2 - 3x + 2$, the constant is $2$ (so $p \in \{\pm 1, \pm 2\}$) and the leading coefficient is $2$ (so $q \in \{1, 2\}$), giving candidates $\pm 1, \pm 2, \pm \frac{1}{2}$.

Two methods give the quotient after a factor is found. Long division is the general tool and works for any divisor, linear or quadratic, and is the only choice when the divisor has degree two or more. The equate-coefficients method (writing the quotient with unknown coefficients and comparing both sides) is often quicker for a linear divisor with a tidy quotient. Either is acceptable in the HSC, but show enough working that a marker can follow the steps; a bare quotient with no method risks lost marks if a sign is wrong.

For $P(x) = 2 x^3 + 3 x^2 - 8 x + 3$, candidates are $\pm 1, \pm 3, \pm \frac{1}{2}, \pm \frac{3}{2}$. Try in order and confirm with the factor theorem.