Question

How do you determine the factors of `x^2-12x-20`?

Answer 1

`x^2-12x+20 = (x-10)(x-2)`

Explanation:

Given:

`x^2-12x+20`

`color(white)()`
Method 1 - Fishing for factors

Note that:

`(x-a)(x-b) = x^2-(a+b)x+ab`

So if we can find `a, b` such that their sum is `12` and product is `20` then we can deduce the factorisation.

It's fairly quick to spot the answer, but here's an example of the sort of reasoning you can apply to help find such factors:

`20 = 2*2*5`

So splitting `20` into a pair of factors amounts to sharing `2`, `2` and `5` between two "buckets", the value of each bucket being the product of the prime factors it contains. The only way this can result in an even sum is if the value of the buckets are both odd or both even. Since we have have a factor `2`, we need both buckets to be even. So there has to be at least one `2` in each bucket.

That leaves one other factor `5`, which put in one or other of the buckets results in the split: `{2, 5 | 2}` or `{2 | 2, 5}`

So basically the pair of factors we are looking for is `10, 2` and hence:

`x^2-12x+20 = (x-10)(x-2)`

`color(white)()`
Method 2 - Completing the square

Rather than fishing for factors, which in some cases is somewhat more involved than the current example, and in some cases will not work at all, we can complete the square then use the difference of squares identity:

`a^2-b^2 = (a-b)(a+b)`

like this:

`x^2-12x+20 = x^2-12x+36-16`

`color(white)(x^2-12x+20) = (x-6)^2-4^2`

`color(white)(x^2-12x+20) = ((x-6)-4)((x-6)+4)`

`color(white)(x^2-12x+20) = (x-10)(x-2)`

In the case of any monic quadratic, you can complete the square like this:

`x^2+2bx+c = x^2+2bx+b^2-b^2+c`

`color(white)(x^2+2bx+c) = (x^2+2bx+b^2)-(b^2-c)`

`color(white)(x^2+2bx+c) = (x+b)^2-(sqrt(b^2-c))^2`

`color(white)(x^2+2bx+c) = ((x+b)-sqrt(b^2-c))((x+b)+sqrt(b^2-c))`

`color(white)(x^2+2bx+c) = (x+b-sqrt(b^2-c))(x+b+sqrt(b^2-c))`