Question

How do you factor the trinomial `n^2+4n-21`?

Answer 1

We need to play with factors of `21` (`+-1, +-3, +-7, +-21`) which add to `+4`:

`n^2 + 4n - 21 => (n + 7)(n - 3)`

Answer 2

`n^2+4n-21 = (n-3)(n+7)`

Explanation:

Note that `(n+2)^2 = n^2+4n+4`, which matches the given expression in the first two terms.

So we find:

`n^2+4n-21 = n^2+4n+4-25`

`color(white)(n^2+4n-21) = (n+2)^2-5^2`

`color(white)(n^2+4n-21) = ((n+2)-5)((n+2)+5)`

`color(white)(n^2+4n-21) = (n-3)(n+7)`