Algebra

Modulus

What $|x|$ means

$|x|$ is the size of $x$, ignoring its sign: $|3| = 3$ and $|-3| = 3$. It is never negative.

The graph of $y = |ax + b|$ is the straight line $y = ax + b$ with any part below the $x$-axis reflected upwards, giving a V shape with its vertex on the axis at $x = -\dfrac{b}{a}$.

Solving modulus equations

Squaring both sides removes the modulus. It is safe when both sides are moduli.

$(x - 1)^2 = (2x + 3)^2 \quad\Rightarrow\quad 3x^2 + 14x + 8 = 0 \quad\Rightarrow\quad (3x + 2)(x + 4) = 0$

So $x = -\dfrac{2}{3}$ or $x = -4$.

When one side has no modulus, sketch both graphs, or consider each case and check the answers.

$2x - 3 = x$ gives $x = 3$. $2x - 3 = -x$ gives $x = 1$. Both check, since the right side is positive in each case.

Solving modulus inequalities

Find the critical values as for the equation, then use a sketch to pick the region.

The critical values are $x = 1$ and $x = 3$. The V is below the line between them, so $1 < x < 3$.

$(x - 2)^2 > (x + 4)^2 \quad\Rightarrow\quad -4x + 4 > 8x + 16 \quad\Rightarrow\quad x < -1$

Polynomials

Algebraic division

A polynomial $\mathrm{P}(x)$ divided by a divisor gives a quotient and a remainder:

$\mathrm{P}(x) = \text{divisor} \times \text{quotient} + \text{remainder}$

The remainder has a lower degree than the divisor: a number for a linear divisor, and a linear expression for a quadratic divisor.

Divide each leading term by $x$, multiply back, subtract and bring down the next term:

The quotient is $2x^2 - x - 3$ and the remainder is $10$.

The remainder and factor theorems

When a question has two unknowns, two conditions give simultaneous equations.

$\mathrm{P}(2) = 8 + 4a + 2b + 6 = 0$, so $2a + b = -7$.

$\mathrm{P}(-1) = -1 + a - b + 6 = 12$, so $a - b = 7$.

Adding gives $3a = 0$, so $a = 0$ and $b = -7$.

Factorising a cubic

Find one factor by trying small values, divide, then factorise the quadratic.

$\mathrm{P}(1) = 1 - 2 - 5 + 6 = 0$, so $(x - 1)$ is a factor. Dividing gives $(x - 1)(x^2 - x - 6) = (x - 1)(x - 3)(x + 2)$.

The solutions are $x = 1$, $x = 3$ and $x = -2$.

Partial fractions (Paper 3)

The forms to use

A fraction with factorised denominator can be split into simpler fractions, which are easier to integrate and to expand.

Write $5x + 1 = A(x - 1) + B(x + 1)$.

Put $x = 1$: $6 = 2B$, so $B = 3$. Put $x = -1$: $-4 = -2A$, so $A = 2$.

$\frac{2}{x + 1} + \frac{3}{x - 1}$

Write $x + 5 = A(x + 1)^2 + B(x - 1)(x + 1) + C(x - 1)$.

Put $x = 1$: $6 = 4A$, so $A = \dfrac{3}{2}$. Put $x = -1$: $4 = -2C$, so $C = -2$. Compare $x^2$ terms: $0 = A + B$, so $B = -\dfrac{3}{2}$.

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

Binomial expansion for any rational power (Paper 3)

The expansion

When $n$ is not a positive integer, the expansion never ends and only works for small $x$.

Replace $x$ with $-2x$ and put $n = -3$:

$1 + (-3)(-2x) + \frac{(-3)(-4)}{2}(-2x)^2 = 1 + 6x + 24x^2$

It is valid for $|2x| < 1$, so $|x| < \dfrac{1}{2}$.

When the first term is not 1

Take the first term out as a factor, including its power.

$\sqrt{4 + x} = 4^{\frac{1}{2}}\left(1 + \frac{x}{4}\right)^{\frac{1}{2}} = 2\left[1 + \frac{1}{2}\cdot\frac{x}{4} + \frac{\frac{1}{2}\left(-\frac{1}{2}\right)}{2}\left(\frac{x}{4}\right)^2\right] = 2 + \frac{x}{4} - \frac{x^2}{64}$

It is valid for $|x| < 4$.

Partial fractions are often used first, so that each part can be expanded separately and the results added.