# 4n squared + 68n + 250 factor completely

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Factor `4n^2+68n+250` :

`2(2n^2+34n+125)` factor out greatest common factor

This will not factor further in the rationals. The discriminant is 156 which is not a perfect square.

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Somewhat beyond the typical Algebra I answer, but if you want to have an irreducible factorization then:

`4n^2+68n+250=(2n-sqrt(39)+17)(2n+sqrt(39)+17)`

`2n^2+34n+125=0` (divided by 2)

`Delta= 1156- 4xx2xx125=156 >0` has two solution:

`x=(-34+-2sqrt(39))/2` `=-17+-sqrt(39)`

So: `4n^2+68n+250=` `2(n+17+sqrt(39))(n+17-sqrt(39))`