Polynomials

Division of polynomials

When $f(x)$ is divided by $d(x)$:

$$f(x) = d(x) \cdot q(x) + r(x)$$

where $q(x)$ is the quotient and the remainder $r(x)$ has a lower degree than $d(x)$.

Remainder theorem

When $f(x)$ is divided by $(x - a)$, the remainder is $f(a)$.

When $f(x)$ is divided by $(ax - b)$, the remainder is $f\left(\dfrac{b}{a}\right)$.

Common mistake: dividing by $(x + 1)$ means $a = -1$, so the remainder is $f(-1)$, not $f(1)$.

Factor theorem

$(x - a)$ is a factor of $f(x)$ if and only if $f(a) = 0$.

Factorising a cubic

1. Try small factors of the constant term ($\pm 1, \pm 2, \ldots$) until $f(a) = 0$.
2. Divide $f(x)$ by $(x - a)$ to get a quadratic.
3. Factorise the quadratic.

Example: $f(x) = x^3 - 7x + 6$. $f(1) = 1 - 7 + 6 = 0$, so $(x - 1)$ is a factor.

$$x^3 - 7x + 6 = (x - 1)(x^2 + x - 6) = (x - 1)(x - 2)(x + 3)$$

Finding unknown coefficients

Each piece of information gives one equation. "Remainder 5 when divided by $(x - 2)$" gives $f(2) = 5$; "$(x + 3)$ is a factor" gives $f(-3) = 0$. Solve the equations simultaneously.

HCF, LCM and rational functions

Factorise first. The HCF takes common factors with the lowest powers; the LCM takes all factors with the highest powers. To add or subtract rational functions, use the LCM of the denominators:

$$\frac{1}{x - 1} - \frac{2}{x^2 - 1} = \frac{(x + 1) - 2}{(x - 1)(x + 1)} = \frac{1}{x + 1}$$

Exam tip: in a two-part question, part (b) usually says "hence". Use the factor you found in part (a) to start the division.