2 Algebra

Algebra is a way of generalising arithmetic by replacing numbers with letters. There are no rules about which letters to use but you will often find numbers replaced with $ a, b $ and $ c $ and variables represented by $ x, y $ and $ z $.

2.1 Terms

A term is a mathematical quantity that consists of a coefficient, a variable and an exponent. $ 3x^2 $ is a term. The coefficient is $ 3 $ the variable is $ x $ and the exponent is $ 2 $.

$ ax^n $ is a more generalised term. The coefficient is $ a $, the variable is $x$ and the exponent is $ n $.

The coefficient can have any numeric or algebraic value. If the value of the coefficient is 0 then the value of the term is zero. If the value of the coefficient is 1 then the coefficient is not normally written. $1x^2$ is usually written as $x^2$.

Two terms are like terms if each one has the same variable or variables and each variable is raised to the same power. $2x$ and $5x$ are like terms, $2x$ and $5xy$ are not. $4ax$ and $-3ax$ are like terms, $4ax^2$ and $-3ax$ are not.

Example 2.2: Collecting like terms

In the following expressions collect and simplify like terms.

2.3 Expanding Brackets

Expanding brackets means multiplying each term inside the brackets by the term outside the bracket. Take $ 5(2 + 4) = 5 \times 2 + 5 \times 4 = 10+20=30 $.

This can also be seen as $ 5(2 + 4) = 5 \times 6 = 30 $

In the same way the expression $ 5(x + y) $ implies $x+y$ multiplied by $5$ to produce $5x+5y$.

Example 2.3: Expanding brackets

Expand the brackets in the following expressions.

You will often find brackets multiplied by other brackets like this $(x+3)(x-2)$. In cases like these every term in the second bracket is multiplied every term in the first bracket. If there are $m$ terms in the first bracket and $n$ terms in the second there will be $m \times n$ terns when you expand the brackets. In this example we will get $2 \times 2 = 4$ terms like this

$(x+3)(x-2)= x \times x + x \times (-2) + 3 \times x + 3 \times (-2)$.

Note the brackets round the $-2$. They are there because $\times$ is an operator and so is $-$. Operators act on terms not on other operators so $x \times -2$ doesn't mean anything.

To complete our example we can collect like terms giving $(x+3)(x-2)= x^2 + x -6$. Notice the expression is written in decending powers of $x$. This is the correct way to write polynomials.

$(x+3)(x-2) = x^2-2x+3x-6 = x^2+x-6$.

Some people find the acronym FOIL helpful when expanding brackets. The letters stand for First, Outside, Inside and Last.

Expanding brackets

2.4 Factorisation

Factorisation is part of the process of simplification. Look at each term in an expression to see whether there are any common factors. If there are, collect them together.

Consider $2x-4y$. Both coefficients have the factor $2$ so we can take it outside a bracket giving:

$2x-4y = 2(x-2y)$

Now look at $8x-16y$. Both coefficients have the factor $8$ so we can take it outside a bracket giving:

$8x-16y = 8(x-2y)$

The final case from Example 2.3 is $-2ax+4ay$. Both coefficients have the factor $2a$ so we can take it outside a bracket giving:

$2ax-4ay = 2a(-x+2y)$ or $2ax-4ay = -2a(x-2y)$

2.4.1 Factorising Quadratic Expressions

The general form of a quadratic expression is $ax^2+bx+c$. Quadratic expressions can often be factorised in to two linear factors. Let's look at some examples. In section 2.3 we had the expression $x^2+x-6$

Example 2.4.1: Factorise $x^2+x-6$

 =$(x+3)(x-2)$ Expressions like $x^2+bx+c$

If you have an expression like $x^2+bx+c$ see if you can find two numbers whose product is $c$ and whose sum is $b$. In Example 2.4.1 $c=-6$. This can be factorised as $3$ and $-2$. $3 \times -2 = -6$ and $3-2 = 1$. This is called factorisation by inspection.

You can also factorise expressions like $ax^2+bx+c$ by inspection though the arithmetic is generally harder. Completing the Square

To complete the square for an expression like $x^2+bx+c$ we rearrange the expression to be $(x+b/2)^2 - (b/2)^2+c$. We set the expression equal to $0$ and subtract the numeric terms, $-(b/2)^2+c$, from both sides. Now we can take the square root of both sides and find the factors.

Imagine we have $x^2 + 8x +15$. $b=8$ so $b/2=4$.

$x^2 + 8x +15$=$(x+8/2)^2-(8/2)^2+15$
Let $(x+4)^2-4^2+15$=$0$
$x+4$=$\pm 1$
so $x$=$-3$ or $-5$
If $x=-3$ then $x+3=0$ and if $x=-5$ then $x+5=0$
So $x^2 + 8x +15$=$(x+3)(x+5)$ Quadratic Formula

If you cannot factorise a quadratic expression by inspection or by completing the square you can use the quadratic formula. If we have $ax^2+bx+c$ then $x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$

Taking the example from section 2.4.1 $x^2 + x -6$ so $a=1$, $b=1$ and $c=-6$

$x$=$\frac{-b \pm \sqrt{b^2-4ac}}{2a}$
 =$\frac{-1 \pm \sqrt{(-1)^2-4 \times 1 \times (-6)}}{2 \times 1}$
 =$\frac{-1 \pm \sqrt{1+24}}{2}$
 =$\frac{-1 \pm 5}{2}$
so $x$=$-3$ or $2$
If $x=-3$ then $x+3=0$ and if $x=2$ then $x-2=0$
So $x^2 + x - 6$=$(x+3)(x-2)$