I think you're factoring this...? You can do that by grouping here. Put the 4 terms in groups of 2 and factor out the greatest common factor. What's left behind is a common binomial factor that can also then be factored out. Like this:
[tex](10 n^{3} -25 n^{2})-(16n+40) [/tex]
Factoring what we can out of both sets gives us this:
[tex]5 n^{2} (2n-5)-8(2n-5)[/tex]
The 2n-5 is a common factor between both the sets of terms, which in turn can also be factored out:
[tex](2n-5)(5 n^{2}-8) [/tex]
I'm not exactly sure what you are doing with this, but we can keep factoring for the sake of completeness. The 5 n-squared term can also be factored:
[tex]5 n^{2}-8=0 [/tex] and [tex]5 n^{2} =8[/tex] and [tex] n^{2} = \frac{8}{5} [/tex] so [tex]n= \frac{ \sqrt{8} }{ \sqrt{5} } [/tex]. Simplifying that down its simplest is this:
[tex]n=+/- \frac{2 \sqrt{10} }{5} [/tex]
So your three factors for that polynomial are
[tex](n- \frac{5}{2} ),(n- \frac{2 \sqrt{10} }{5} ),(n+ \frac{2 \sqrt{10} }{5} )[/tex],
