How do you factor #4n^2- 26n- 42#?

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Answer 1 · 2015-05-23T20:55:55

#4n^2-26n-42 = (2n-13/2)^2-(13/2)^2-42#

#= (2n-13/2)^2-169/4-168/4#

#= (2n-13/2)^2-337/4#

#= (2n-13/2)^2-(sqrt(337)/2)^2#

#= (2n-13/2-sqrt(337)/2)(2n-13/2+sqrt(337)/2)#

Are you sure about the #-# sign on #42#?

#4n^2-26n+42 = (2n-13/2)^2-(13/2)^2+42#

#= (2n-13/2)^2-169/4+168/4#

#= (2n-13/2)^2-1/4#

#= (2n-13/2)^2-(1/2)^2#

#= (2n-13/2-1/2)(2n-13/2+1/2)#

#= (2n-7)(2n-6)#