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 $.

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 is 1 it is not normally written so $ 1x^2 $ is normally written $x^2$.

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. | ||

$2x+5x$ | $=$ | $7x$ |

$2x+5xy$ | $=$ | $2x+5xy$ |

$4ax-3ax$ | $=$ | $ax$ |

$4ax^2-3ax$ | $=$ | $4ax^2-3ax$ |

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. | ||

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

$4(2x-4y)$ | $=$ | $8x-16y$ |

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

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 **F**irst, **O**utside, **I**nside and **L**ast.

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

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 $x^2+x-6$

$x^2+x-6$ | = | $x^2-2x+3x-6$ |

= | $x(x-2)+3(x-2)$ | |

= | $(x+3)(x-2)$ |

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.

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 $(x+4)^2-4^2+15$ | = | $0$ |

$(x+4)^2-16+15$ | = | $0$ |

$(x+4)^2-1$ | = | $0$ |

$(x+4)^2$ | = | $1$ |

$x+4$ | = | $\pm 1$ |

so $x$ | = | $-3$ or $-5$ |

If $x=-3$ then $x+3=0$ and if $x=-5$ then $x+5=0$ | ||

So $x^2 + 8x +15$ | = | $(x+3)(x+5)$ |

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