Revise Algebra for Cambridge 9709 (2.1)

The modulus of a number is its 'absolute value' or magnitude, irrespective of its sign. For a real number $x$, its modulus is denoted by $|x|$. For example, $|5| = 5$ and $|-5| = 5$. This simple concept leads to interesting graphs, equations, and inequalities.

$|x| = \begin{cases} x & \text{if } x \ge 0 \\ -x & \text{if } x < 0 \end{cases}$

When solving equations involving modulus, such as $|f(x)| = |g(x)|$, a reliable method is to square both sides. Since the square of a number and the square of its negative are identical, squaring removes the modulus signs. For example, $(|A|)^2 = A^2$. However, be cautious as this can sometimes introduce extraneous solutions, so it's good practice to check your answers in the original equation. An alternative is to consider the two cases: $f(x) = g(x)$ and $f(x) = -g(x)$.

Polynomials are expressions involving variables raised to positive integer powers, such as $3x^3 - 2x^2 + 5x - 7$. The Factor and Remainder Theorems are powerful shortcuts for investigating their properties without resorting to full polynomial division every time.

In exams, a common task is to find the roots of a cubic polynomial. You'll typically use the Factor Theorem to find the first root by testing simple integer values for $x$ (like 1, -1, 2, -2) that are factors of the constant term. Once you find a value 'a' for which $P(a)=0$, you know $(x-a)$ is a factor. You can then use polynomial long division or synthetic division to divide $P(x)$ by $(x-a)$ to find the remaining quadratic factor, which can then be factorised or solved.

Partial fractions provide a way to decompose a complicated algebraic fraction into a sum of simpler fractions. This is an essential skill for integration in P3, but in P2 we focus on the algebraic decomposition itself. The process depends on the form of the factors in the denominator.

For distinct linear factors: $\frac{px+q}{(x-a)(x-b)} \equiv \frac{A}{x-a} + \frac{B}{x-b}$

For repeated linear factors: $\frac{px^2+qx+r}{(x-a)^2(x-b)} \equiv \frac{A}{x-a} + \frac{B}{(x-a)^2} + \frac{C}{x-b}$

You will be familiar with the binomial expansion for positive integer powers from P1. Here, we extend the principle to cases where the power, $n$, is a rational number (e.g., a fraction or a negative number). This results in an infinite series expansion, which is only valid for a specific range of $x$ values.

For any rational number $n$, if $|x| < 1$, then: $(1+x)^n = 1 + nx + \frac{n(n-1)}{2!}x^2 + \frac{n(n-1)(n-2)}{3!}x^3 + \dots$

If the expression is not in the form $(1+x)^n$, you must manipulate it first. For an expression like $(a+bx)^n$, you must factor out the 'a' term: $(a+bx)^n = a^n(1 + \frac{b}{a}x)^n$. You can then apply the formula to the bracket, remembering to multiply the entire resulting series by $a^n$. The validity condition will be $|\frac{b}{a}x| < 1$.