A-Level Maths & Further Maths/Algebraic Expressions

Fractions

Fractions are something that you should be comfortable using at this level, and should be second nature. Recall that ${\frac {a+b+c}{x}}={\frac {a}{x}}+{\frac {b}{x}}+{\frac {c}{x}}$ is true, but in general ${\frac {x}{a+b}}\neq {\frac {x}{a}}+{\frac {x}{b}}$. It should also be reminded that the denominator cannot be zero.

Factorising Expressions

A factor is a number or variable that appears in multiple parts of an expression; it can be 'taken out' to help simplify an expression. For example, $ax+ay$ has a factor of $a$, so the expression can be simplified to $a(x+y)$.

Expanding Expressions

Expanding expressions basically reverses factorising. It is easy to make a mistake when expanding brackets, so care should be taken. A common example is $(x+y)^{2}=x^{2}+y^{2}+2xy$, where many students make the mistake of suggesting it is $x^{2}+y^{2}$. Another important one to remember is the difference of two squares: $x^{2}-y^{2}=(x+y)(x-y)$.

Indices

There are some basic rules to remember when working with indices:

$a^{1}=a$

$a^{0}=1$

$a^{m}\cdot a^{n}=a^{m+n}$

$a^{m}/a^{n}=a^{m-n}$

$(a^{m})^{n}=a^{m\cdot n}$

$a^{1/n}={\sqrt[{n}]{a}}$

$a^{-m}={\frac {1}{a^{m}}}$

These will be revisited when looking at logarithms.

Surds

Surds are expressions involving square roots. Like indices, the basic rules are:

${\sqrt {ab}}={\sqrt {a}}{\sqrt {b}}$

${\sqrt {\frac {a}{b}}}={\frac {\sqrt {a}}{\sqrt {b}}}$

$a={\sqrt {a^{2}}}={\sqrt {a}}{\sqrt {a}}$

$a-b=({\sqrt {a}}+{\sqrt {b}})({\sqrt {a}}-{\sqrt {b}})$

Rationalising the Denominator

By definition, square roots are irrational. This makes them difficult to work with, especially when they are in the denominator of a fraction, when trying to simplify expressions. To help with this, we use the process of 'rationalising the denominator'. To do so, we multiply the numerator and denominator by the same number, as ${\frac {a}{a}}=1$, so we avoid changing the value of the fraction, but still change the numbers we are working with.

Example

To rationalise ${\frac {2+{\sqrt {3}}}{\sqrt {6}}}$ the full working out would be

${\frac {2+{\sqrt {3}}}{\sqrt {6}}}={\frac {2+{\sqrt {3}}}{\sqrt {6}}}\times {\frac {\sqrt {6}}{\sqrt {6}}}={\frac {2{\sqrt {6}}+{\sqrt {3}}{\sqrt {6}}}{{\sqrt {6}}{\sqrt {6}}}}={\frac {2{\sqrt {6}}+3{\sqrt {2}}}{6}}$.