Quick questions on Remainder and factor theorem explained: O-Level A-Maths

Dividing a polynomial $f(x)$ by a divisor $g(x)$ gives a quotient $q(x)$ and a remainder $r(x)$ whose degree is below that of $g(x)$:

The factor theorem is the special case where the remainder is zero. If $f(a) = 0$, then $(x - a)$ divides $f(x)$ exactly, so $(x - a)$ is a factor:

To factorise a cubic, find one root by trial (test small integer factors of the constant term), use the factor theorem to confirm a linear factor, then divide to get a quadratic, which you factorise by the usual methods.

For a divisor such as $(2x - 1)$, the value that makes it zero is $x = \tfrac{1}{2}$, so the remainder is $f\!\left(\tfrac{1}{2}\right)$ and $(2x - 1)$ is a factor exactly when $f\!\left(\tfrac{1}{2}\right) = 0$. The principle is unchanged; only the test value differs.

A common Paper 2 question gives two facts, such as two remainders, or one factor and one remainder, for a polynomial with two unknown coefficients. Each fact gives an equation in the unknowns through $f$ evaluated at a value; solving the pair of simultaneous equations pins down both coefficients.

Evaluate $f(a)$ carefully: $2(2)^3 = 16$, a frequent source of error.

Find the remainder when $x^3 + 2x^2 - x + 5$ is divided by $(x - 1)$. [2 marks]

Show that $(x - 3)$ is a factor of $x^3 - 4x^2 + x + 6$. [2 marks]

Given that $(x + 2)$ is a factor of $x^3 + kx - 2$, find $k$. [3 marks]