Question

How do you factor `n^ { 2} - 6n - 24= 0`?

Answer 1

`(n-3-sqrt(33))(n-3+sqrt(33)) = 0`

Explanation:

This can be factored by completing the square and using the difference of squares identity:

`A^2-B^2 = (A-B)(A+B)`

with `A=(n-3)` and `B=sqrt(33)` as follows:

`n^2-6n-24 = n^2-6n+9-33`

`color(white)(n^2-6n-24) = (n-3)^2-(sqrt(33))^2`

`color(white)(n^2-6n-24) = ((n-3)-sqrt(33))((n-3)+sqrt(33))`

`color(white)(n^2-6n-24) = (n-3-sqrt(33))(n-3+sqrt(33))`

So the factored form of the original equation can be written:

`(n-3-sqrt(33))(n-3+sqrt(33)) = 0`