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4n squared + 68n + 250   factor completely

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schooledmom | (Level 1) Salutatorian

Posted May 23, 2013 at 7:59 PM via web

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4n squared + 68n + 250

 

factor completely

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embizze | High School Teacher | (Level 1) Educator Emeritus

Posted May 23, 2013 at 8:28 PM (Answer #1)

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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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embizze | High School Teacher | (Level 1) Educator Emeritus

Posted May 23, 2013 at 8:33 PM (Reply #1)

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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)`

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oldnick | (Level 1) Valedictorian

Posted May 24, 2013 at 3:18 AM (Answer #2)

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`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))`

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