## When should I use factorising?

Factorising is a great tool if the roots of the equation are simple, usually integers. It is the quickest method to use (assuming it is applicable) and can often be done in your head once you get practiced. For example: \[ 0 = x^2+7x+12 = (x+3)(x+4) \\[6pt] x = -3\hspace{6 pt}\text{or }-4 \] This is a great candidate for factorising and we can get the roots easily. See more useful examples on the factorising page.

## When not to use factorising?

If the roots are fractions or surds, it's not going to be so easy to spot the factorisation. It's likely you wouldn't want to use factorising in cases such as: \[0 = x^2 - 2x - 2 \Rightarrow x = (1+\sqrt{3})\hspace{6 pt}\text{or }(1-\sqrt{3})\] \[0 = x^2 - \frac{29}{10}x + \frac{21}{10} \Rightarrow x = \frac{3}{2}\hspace{6 pt}\text{or }\frac{7}{5}\] These examples would be better suited to the Quadratic Formula or Completing the Square.

#### If the \(x^2\) coefficient isn't \(1\)

Usually there's no need to try and factorise in these cases. Although it sometimes factorises with 'nice' integer values
and you may spot it straight away, the other methods aren't particularly slow once you are well practiced. Here I show
factorising vs the Quadratic Formula to solve \(0 = 2x^2-2x-12\):

**Factorising:**
\[
0 = 2x^2-2x-12 = (2x+4)(x-3)
\\[6pt] x = -2\hspace{6 pt}\text{or }3
\]
**Quadratic Formula:**
\[\begin{align}
x &= \frac{-(-2)\pm\sqrt{(-2)^2-4\times2\times(-12)}}{2\times2}
\\[6pt] &= \frac{2\pm10}{4}
\\[6pt] &= -2\hspace{6 pt}\text{or }3
\end{align}\]
Bearing in mind the extra thinking you'll have to do to find the combination which works for factorising vs the routine approach to
the Quadratic Formula, you can weigh up pros and cons yourself. Also remember the fact that you may waste time trying to factorise
something which can't be factorised!

## When should I use completing the square?

Completing the square is easiest when the \(x\) coefficient is **even**. This is simply because the method requires that you half this value
and if it's even then you avoid more fractions! However, this method can be very felxible and when well practiced can be applied to most situations.

## When not to use completing the square

As with factorising, completing the square doesn't lend itself to equations where the \(x^2\) coefficient isn't \(1\). There is a way around this,
if you divide through every term by the \(x^2\) coefficient to *make it* \(1\). For example;
\[3x^2 + 4x + 9 = 0 \Rightarrow x^2 + \frac{4}{3}x + 3 = 0\]
But you can see that this may give fractions which can give you more hassle than it's worth and you might be able to find the solution using the
Quadratic Formula more quickly.

## When should I use the Quadratic Formula?

The Quadratic Formula should be in your arsenal to use whenever one of the other methods doesn't seem easier or quicker. It will work **every time** if
there are roots, and if there aren't then you will be able to see by the discriminant early on in
your workings! If you are going to have a "go to" method for solving quadratic equations then this should be it... Not to say you shouldn't practice the
other methods!