Everything from the 4 lessons of this topic on a few pages: the key idea of each lesson, the diagram to remember, the points to learn and the traps examiners set. Read it, cover it, test yourself.
\(|x|\) is the distance of \(x\) from 0. Two facts do most of the work: \(|a| = |b| \iff a^2 = b^2\), and \(|x - a| < b \iff a - b < x < a + b\). When one side is NOT a modulus, any answer found by splitting into cases must be checked in the original equation.
The vertex of \(y = |2x - 3|\) is at \(x = 1.5\), not \(x = 3\) or \(x = -1.5\). Solve \(ax + b = 0\) properly; the most common slip is forgetting to divide by \(a\).
In Case 2, \(-(2x - 1)\) is \(-2x + 1\), not \(-2x - 1\). Put brackets round the whole side before you change its sign.
Never square when one side might be negative: \(2x + 5 < |x + 1|\) squared gives a wrong answer, because squaring a negative left side flips which side is bigger. Square only when both sides are moduli (or are known to be non-negative).
\(p(x) = (\text{divisor}) \times (\text{quotient}) + \text{remainder}\), where the remainder has lower degree than the divisor. The remainder when \(p(x)\) is divided by \((ax + b)\) is \(p\left(-\dfrac{b}{a}\right)\), and \((ax + b)\) is a factor exactly when \(p\left(-\dfrac{b}{a}\right) = 0\).
Subtracting a negative term is where marks go: \((2x^2 + 5x) - (2x^2 - 4x) = 9x\), not \(x\). Put the line you subtract in brackets and change every sign.
For the divisor \((x + 3)\) you substitute \(x = -3\), not \(x = 3\); for \((2x - 1)\) you substitute \(x = \dfrac{1}{2}\), not \(x = 1\) or \(x = 2\).
A remainder condition is \(p(c) = r\), not \(p(c) = 0\). Only a factor gives zero.
Choose the form from the denominator, multiply through to get an identity in \(x\), then find the constants by substituting the root of each linear factor and comparing coefficients for whatever is left. A repeated factor \((cx + d)^2\) needs BOTH \(\dfrac{B}{cx + d}\) and \(\dfrac{C}{(cx + d)^2}\); a quadratic factor \(cx^2 + d\) needs a numerator \(Bx + C\). If the numerator has the same degree as the denominator, include a constant term as well.
Factorise the denominator before choosing the form. \(x^2 - 2x - 3\) is not a 'quadratic factor' needing \(Bx + C\): it is \((x - 3)(x + 1)\). Only a quadratic that does not factorise, such as \(x^2 + 3\), gets the \(Bx + C\) numerator.
When substituting \(x = -\dfrac{1}{2}\), every bracket must be evaluated at \(-\dfrac{1}{2}\): \((x - 1) = -\dfrac{3}{2}\), not \(-1\). A single slip here makes \(C\) wrong, so always check with a coefficient comparison.
Writing only \(\dfrac{A}{x + 1} + \dfrac{C}{(2x + 1)^2}\) loses the middle term. With two unknowns the identity cannot hold for every \(x\), and you will find the 'answers' disagree when you check.
\(\dfrac{B}{x^2 + 3}\) with a constant numerator is the wrong form: it gives too few unknowns, and the \(x\) coefficients will not match. Over an irreducible quadratic always write \(Bx + C\).
Leaving out the constant term is the classic error here. Without it the identity \(2x^2 + x + 3 \equiv B(2x + 1) + C(x - 1)\) is impossible: the right side has no \(x^2\) term. Compare the degrees of the numerator and denominator before you write down the form.
\((1 + x)^n = 1 + nx + \dfrac{n(n - 1)}{2!}x^2 + \dfrac{n(n - 1)(n - 2)}{3!}x^3 + \dots\), valid for \(|x| < 1\) when \(n\) is not a positive integer. For \((a + bx)^n\), first take out \(a^n\): \(a^n\left(1 + \dfrac{b}{a}x\right)^n\), valid for \(\left|\dfrac{bx}{a}\right| < 1\).
\((-4x)^2 = 16x^2\), not \(-4x^2\) or \(4x^2\). Write the bracket in every term, then square or cube both the number and the sign.
\((2 - x)^{-3}\) is NOT \(2\left(1 - \dfrac{x}{2}\right)^{-3}\). The factor taken out is \(2^{-3} = \dfrac{1}{8}\): it carries the same power as the bracket.
Write \(\dfrac{2}{2x - 1}\) as \(-2(1 - 2x)^{-1}\) before expanding: the standard series needs the constant term in the bracket to be \(+1\). Expanding \((2x - 1)^{-1}\) 'as if' it were \((1 + 2x)^{-1}\) gives the wrong signs.