Question

Explain this: I read that to factorize you need to find numbers that adds or minus to the middle value and multiply to give the last value. But how did this come up? The middle quantities when multiplied don't make up 12. Please explain this.

`12^2-7y-12`=0

Then:
`12^2-16y+9y-12`=0

Answer 1

What you start describing is appropriate for monic quadratics, but if the leading coefficient is not `1` then you need something more...

Explanation:

If you have a monic quadratic (i.e. leading coefficient is `1`) with zeros `alpha` and `beta` then note that:

`(x-alpha)(x-beta) = x^2-(alpha+beta)x+alpha beta`

If the leading coefficient is not `1` then it gets more complicated.

One way of simplifying things a bit is an "AC" method.

In the example:

`12x^2-7x-12 = 0`

we multiply the first coefficient `A=12` by the last coefficient `C=12` ignoring the sign to get `AC=144`.

Then noting that the sign of the constant term is `-`, we look for a pair of factors of `AC=144` which differ by `B=7`.

(If the sign on the constant term was `+` then we would instead look for a pair of factors whose sum is `B=7`)

The pair `16, 9` works in that `16 * 9 = 144` and `16 - 9 = 7`

In order to find this pair you can reason as follows:

The prime factorisation of `144` is:

`144 = 2^4 * 3^2`

Note that `7` is divisible by neither `2` nor `3`.

So if we split `144` into a product of two factors with difference `7`, then all of the powers of `2` must be in one of the factors and all of the powers of `3` must be in one of the factors. That gives two possibilities to try:

`144 * 1" "` giving difference `144-1 = 143`

`16 * 9" "` giving difference `16-9 = 7` as required.

We then use this pair to split the middle term and factor by grouping:

`0 = 12x^2-7x-12`

`color(white)(0) = (12x^2-16x)+(9x-12)`

`color(white)(0) = 4x(3x-4)+3(3x-4)`

`color(white)(0) = (4x+3)(3x-4)`

Hence roots `x=-3/4` and `x=4/3`

Another way we might have approached this quadratic would be to divide it by `12` first to give a monic quadratic, then look for sum and product as first described. However, then you would be looking for two rational numbers with sum `7/12` and product `-1`, which hardly makes the problem easier.

Footnotes

To understand why the AC method works, consider a quadratic equation:

`ax^2+bx+c = 0`

Multiplying by `a` we get:

`0 = a(ax^2+bx+c)`

`color(white)(0) = a^2x^2+abx+ac`

`color(white)(0) = (ax)^2+b(ax)+ac`

`color(white)(0) = x_1^2+bx_1+ac`

where `x_1 = ax`

Having arrived at a monic quadratic in `x_1`, we can then look for a pair of factors of the constant term `ac` with sum `-b`. Ignoring signs, that means looking for a pair of factors of `AC` with sum or difference `B`, according to whatever signs are going on.

This will give us a factorisation of `x_1^2+bx_1+ac` which we then translate into a factorisation of `ax^2+bx+c`