Question

How do you factor `4n^2-15n-25`?

Answer 1

`4n^2-15n-25 = (4n+5)(n-5)`

To find this use the rational roots theorem to deduce that if `p/q` is a root of `4n^2-15n-25 = 0` in lowest terms, then `p` must be a divisor of `25` and `q` a divisor of `4`.

This gives a limited number of possible rational roots and corresponding factors.

In addition, if the factors were of the form `(2n+a)(2n+b)` (to get the `4n^2` leading term) then the coefficient of the middle term would be even - which it isn't.

So the factorisation must be something like `(4n+a)(n+b)`.

It does not take long to find the appropriate choice of factorisation of `-25` into `a` and `b` to give the required factorisation of the quadratic.