Algebra

📚The mathematics

The modulus function

\( |x| \) is the distance of \(x\) from zero, so it is never negative: \( |x| = x \) when \( x \ge 0 \) and \( -x \) when \( x < 0 \). The graph of \( y = |\mathrm{f}(x)| \) is the graph of \( \mathrm{f} \) with everything below the \(x\)-axis reflected upwards.

To solve \( |ax + b| = c \), split into two cases and solve both:

Then check each answer in the original equation, because case-splitting and squaring can both manufacture roots that do not work.

For inequalities \( |ax+b| < c \), the useful reading is “the distance from \( -\tfrac{b}{a} \) is less than…”, which gives a single interval \( -c < ax + b < c \). For \( |ax+b| > c \) you get two separate regions. Where both sides carry a modulus, squaring both sides is legitimate — since both are non-negative — and usually the fastest route: \( |A| = |B| \iff A^{2} = B^{2} \).

Polynomial division

Dividing \( \mathrm{P}(x) \) by a linear or quadratic divisor gives a quotient and a remainder:

The remainder always has degree lower than the divisor — a constant when dividing by a linear expression, and linear (\( ax + b \)) when dividing by a quadratic. Long division works; so does comparing coefficients, which is often faster when the quotient's shape is obvious.

Insert a zero placeholder for any missing power. Dividing \( x^{3} - 7 \) means treating it as \( x^{3} + 0x^{2} + 0x - 7 \), and skipping that step misaligns the whole calculation.

The factor and remainder theorems

For a divisor \( (ax - b) \), substitute \( x = \dfrac{b}{a} \): the root of the divisor is what you put in. So the remainder on dividing by \( (2x - 1) \) is \( \mathrm{P}\!\left(\tfrac{1}{2}\right) \).

To factorise a cubic: find one root by trial, testing the factors of the constant term (\( \pm 1, \pm 2, \dots \)); that gives one linear factor by the factor theorem; divide to get a quadratic; then factorise or use the formula on that. Two unknown constants in the cubic means the question will give you two conditions — set up two simultaneous equations.

P3 Partial fractions

Three standard forms, and recognising which one applies is most of the skill:

Note the patterns: a repeated linear factor needs a term for each power, and an irreducible quadratic factor needs a linear numerator \( Bx + C \), not just a constant.

Method: multiply through by the denominator, then substitute the values of \(x\) that make each bracket zero — those kill most of the unknowns at once. For the leftovers, compare coefficients or substitute any convenient value.

If the numerator's degree is greater than or equal to the denominator's, divide first to get a polynomial plus a proper fraction, then decompose the fraction.

P3 The binomial expansion for rational \(n\)

Unlike the P1 version, this series is infinite and only valid for \( |x| < 1 \). Two consequences follow.

First, the bracket must start with a 1. To expand \( (4 + x)^{1/2} \), factor out: \( \left[4\left(1 + \tfrac{x}{4}\right)\right]^{1/2} = 2\left(1 + \tfrac{x}{4}\right)^{1/2} \). The factor comes outside and the expansion applies to what is left.

Second, state the validity condition, and adjust it for what actually sits in the bracket. For \( \left(1 + \tfrac{x}{4}\right)^{n} \) the condition is \( \left|\tfrac{x}{4}\right| < 1 \), that is \( |x| < 4 \). Quoting \( |x| < 1 \) regardless is a standard lost mark.

Partial fractions and this expansion combine constantly: decompose first, expand each simple fraction with a negative index, then add. The validity condition for the combined result is the most restrictive of the individual ones.