The shit blog of Paul Chris Jones

Hi Paul. I love your blog, and I think you're very handsome. But first, tell me why for the love of God there are three ways to solve quadratic equations when I only need one.

You only need the quadratic formula to solve quadratic equations. (The quadratic formula gives you the vertex co-ordinates, too.)

But, there is another method called factorisation. Personally I would rather ignore this method as using the quadratic formula works so well. But exam boards like to you to learn how to do factorisation too, so we'll have to learn it because of that.

The third method is called completing the square and is really just a type of factorisation. We'll get on to that in another post.

What is factorisation? No wait, what is a factor?

A factor is a number where if you multiply it by another number, you get a third number.

For example, for 4 x 2 = 8, 4 and 2 are factors of 8.

But for 1/2 x 16 = 8, 1/2 isn't a factor. That's because non-integers can't be factors (because we're racist to non-integers). And that means 16 isn't a factor of 8 either.

So what is factorisation?

Factorisation is a way of solving quadratic equations. Or any other type of polynominal, come to think of it. Actually, this is where factorisation has an advantage over the quadratic formula. Factorisation can be used to solve other polynomials such as cubic or even quartic equations. There are formulas for other polynomials though - e.g. the cubic formula and the quartic formula. But these are a lot larger than than the relatively simple quadratic formula.

Sorry, back to the topic. Factorisation changes a quadratic equation into two factors, so it looks like this:

y = (ax+b)(cx+d)

where a, b, c and d are numbers.

Here's an example factorised equation:

y = x² -3x -4 = (x+1)(x-4)

Do you know how to expand a factorised equation? If not, read this.

Factorisation is the opposite of expansion. You want to put the numbers into brackets.

Why do I want to put the numbers into brackets?

Good question, padawan. The answer is: you can then easily solve a quadratic equation, i.e. you can find the x values at a given y value.

To see how, let's use the above example again:

y = (x+1)(x-4)

Let's assume we want to find the values for x when y = 0.

We know that the only way to get to zero from two factors is if at least one of the factors is zero. For exampe, 6 x 0 = 0, 0 x 29 = 0, 0 x 0 = 0

So we take each factor in turn and treat it as though it's zero.

So, for y = 0 = (x+1)(x-4):

• For (x+1) to equal zero, x must equal -1. And what do you know? That is one of the x co-ordinates when y = 0.
• For (x-4) to equal zero, x must equal 4, which is the other x co-ordinate when y = 0.

Can I find the vertex using factorisation?

Yes, but using logic instead of a formula. If the graph has two roots then the line of symmetry must be halfway between these two roots. And the vertex lies on the line of symmetry.

For example, if we know that x = 2 and 4 when y = 0, then the line of symmetry here is x = 3, the midpoint between x= 2 and x = 4. Therefore the vertex x co-ordinate = 3.

One last thing - how do I actually do factorisation??

Sorry, I almost forgot to tell you!

Look at the following example equation. I have substituted all numbers with letters:

ax² + bx + c = (dx+e)(fx+g)

Hopefully you know how to expand a factorised equation, because that knowledge is necessary here.

The first step of expansion is to multiply dx and fx:

(dx)(fx) = ax²

The part we're interested in is this:

df = a

the value of a = d multiplied by f.

• usually a = 1. If so, then both usually d = 1 and f = 1 too. In theory they could both also = -1.
• if a = 2, then usually d = 2 and f = 1 (or the other way around, but it doesn't matter at this stage.)
• if for example a = 16, then you have lots of possibilities. Perhaps d = 4 and f = 4, or d = 2 and f = 8.

Next, notice that:

eg = c

As with the value of a, if c is a large number then you could have lots of possibilities for the values of e and g. The next step helps you narrow these possibilities down to the actual answer. Notice that:

ef + dg = b

If you have an idea what f and d may be, you can substitute them here. Usually f and d will both equal 1, which will simplify matters greatly.

The work of factorisation is in comparing the possibilities for each equation until you find a set of values for a, b, and c that satisfies them all.

You now have all the tools to factorise an equation. Factorisation requires logic and some guesswork but there are only one set of values for a, b and c that will work, so you'll know when you've got the right answer.

I'm still confused. Can you show me an example?

Since I'm in a generous mood, let's go through an example question then:

Solve x² -3x -4 = 0 with factorisation

So the question is asking us: when y = 0, what does x equal? First, here's our equation with all the numbers substituted for letters:

ax² + bx + c = (dx+e)(fx+g)

We already know the values of a, b, and c:

a = 1, b = -3, c = -4

Let's start with a = df:

a = 1

a = df

Therefore df = 1

So perhaps d = 1 and f = 1?

Moving on to c = eg:

c = -4

c = eg

Therefore eg = -4

Perhaps e = 1, g = -4

Or maybe e = -1. g = 4

Or maybe even e = 2, g = -2

Moving on to ef + dg = b:

b = -3

b = ef + dg

Therefore ef + dg = -3

Assuming d =1 and f =1, we can simplifyy to e + g = -3

Now for a bit of detective work:

I have noticed that when e =1 and g =-4, all requirements are satisfied.

Yay! Now for the final steps:

Therefore x² -3x -4 = (x+1)(x-4)

Which means x = -1 and 4 when y = 0

In the final post of the 'solving quadratic equations' series, we'll look at the method of completing the square.