Question

How do you factor `3n^2 - 8n + 4`?

Answer 1

`3n^2-8n+4 = (3n-2)(n-2)`

How did I find this?

We are looking for factors in the form `(an+b)(cn+d)`

`(an+b)(cn+d) = acn^2 + (bc+ad)n+bd`

Comparing this with our original quadratic we find:
`ac = 3`
`bc+ad = -8`
`bd = 4`

If `a`, `b`, `c` and `d` are integers then we are looking for integers that satisfy these equations.

`ac=3` has possible integer solutions:

`a=3`, `c=1`
`a=-3`, `c=-1`
`a=1`, `c=3`
`a=-1`, `c=-3`

We might as well pick the first of these, they will ultimately give very similar factorisations.

Let `a=3`, `c=1`

`bd` > 0. So `b` and `d` are either both positive or both negative. If they were both positive, then `bc+ad = b+3d` would be positive too, which it isn't. So both `b` and `d` are negative.

Notice that `b+3d = -8` is even. So either `b` and `d` are both odd or both even.

So we are looking for two negative whole numbers `b` and `d` such that `bd = 4` and `b` and `d` are both even.

That only leaves the possibility `b=d=-2`

Answer 2

`3n^2 - 8n +4`

We can Split the Middle Term of this expression to factorise it
In this technique, if we have to factorise an expression like `an^2 + bn + c`, we need to think of 2 numbers such that:

`N_1*N_2 = a*c = 3 xx 4 = 12`

and

`N_1 +N_2 = b = -8`
After trying out a few numbers we get `N_1 = -2` and `N_2 =-6`
`- 2 xx -6 = 12` and `(-2) +(-6)= -8`

`3n^2 - 8n +4 = 3n^2 - 6n - 2n +4`

`= 3n(n -2) - 2(n-2)`

`=color(green)( (3n -2)(n -2))`