Question

How do you factor `15n^(2)-n-28`?

Answer 1

Playing with factors of `15` and `28` always us to factor there expression:

`15n^2 - n - 28 => (5n - 7)(3n + 4)`

Answer 2

The factors are `(3n+4)(5n-7)`

Explanation:

All the information is given in the expression:

Find factors of `15 and 28` whose products differ (because of `-28`)

by `1` (because of the `-1n` term.)

The signs will be different `(-28)` and there should be more negatives, (because of the `-1n` term.)

NOte that `" "-1n" "` is odd, which must come from the difference between an odd and an even number.

Therefore `2 xx 14` are not possible as factors of `28` because they will give even products.

`15 and -28`
`darr" "darr`
`3" "+4" "rarr 5 xx +4 = +20" "larr` even
`5" "-7" "rarr 3 xx -7 = -ul21" "larr` odd
`color(white)(xxxxxwwwwwwwwwwwwxxxx)-1`

The factors are `(3n+4)(5n-7)`