Algebraic Operations and Polynomials

Dividing one polynomial by another

Dividing $f(x)$ by $(x - a)$ works exactly like long division with numbers. You get a quotient and a remainder, and the remainder is always of lower degree than the divisor — so dividing by a linear expression always leaves a constant.

$$f(x) = (x-a)\,Q(x) + R$$

That identity is the whole of this topic. Everything below is a consequence of it.

The remainder theorem

Put $x = a$ into the identity above. The $(x-a)$ term becomes zero and takes $Q(x)$ with it, whatever $Q(x)$ happens to be, leaving:

So you never have to carry out the division to find a remainder. Substitute and you have it.

For a divisor of the form $(ax - b)$, the value that makes it zero is $x = \dfrac{b}{a}$, so the remainder is $f\!\left(\dfrac{b}{a}\right)$.

The factor theorem

If the remainder happens to be zero, the division was exact — which is what "factor" means.

This is the remainder theorem with $R = 0$; it is not a separate idea. It gives you a way to test a candidate factor in one line, and a way to hunt for one.

To find a factor of a cubic, try small values: $x = \pm 1, \pm 2, \pm 3$, and so on. You only need to test factors of the constant term, because if $(x - a)$ divides $f(x)$ exactly then $a$ must divide the constant term.

Factorising a cubic

1. Find one root by trial, using the factor theorem. That gives you one linear factor.
2. Divide the cubic by that factor to get a quadratic.
3. Factorise the quadratic in the usual way.

For $f(x) = x^3 - 4x^2 + x + 6$: testing $x = -1$ gives $-1 - 4 - 1 + 6 = 0$, so $(x+1)$ is a factor. Dividing leaves $x^2 - 5x + 6$, which factorises to $(x-2)(x-3)$. So

$$x^3 - 4x^2 + x + 6 = (x+1)(x-2)(x-3)$$

and the equation $f(x) = 0$ has roots $x = -1$, $x = 2$ and $x = 3$.

Finding unknown coefficients

If a polynomial contains unknown letters and you are told something about a factor or a remainder, each piece of information gives you one equation.

• "$(x - a)$ is a factor" gives $f(a) = 0$.
• "the remainder on division by $(x - a)$ is $R$" gives $f(a) = R$.

Two unknowns need two such statements, and you solve the pair simultaneously.

Algebraic fractions

Factorise the top, factorise the bottom, cancel any bracket that appears in both.

$$\frac{2x^2 + 5x - 3}{x^2 - 9} = \frac{(2x-1)(x+3)}{(x-3)(x+3)} = \frac{2x-1}{x-3}$$

Strictly, the cancelled expression is equal to the original everywhere except where the cancelled bracket is zero, since the original is undefined there. CSEC rarely asks you to state it, but if a question says "state the values of $x$ for which the expression is undefined", it is asking about exactly those points.