How do you FOIL #x^2+111x-104000# ?

1 Answer
George C.
Jan 20, 2018

#x^2+111x-104000#

#= (x+111/2-sqrt(428321)/2)(x+111/2+sqrt(428321)/2)#

Explanation:

I think you are wanting to factor #x^2+111x-104000#

We find:

#4(x^2+111x-104000)#

#= 4x^2+444x-416000#

#= (2x)^2+2(2x)(111)+111^2-428321#

#= (2x+111)^2-(sqrt(428321))^2#

#= ((2x+111)-sqrt(428321))((2x+111)+sqrt(428321))#

#= (2x+111-sqrt(428321))(2x+111+sqrt(428321))#

So:

#x^2+111x-104000#

#= 1/4(2x+111-sqrt(428321))(2x+111+sqrt(428321))#

#= (x+111/2-sqrt(428321)/2)(x+111/2+sqrt(428321)/2)#

Note that #428321 = 107 * 4003# has no square factors, so #sqrt(428321)# is in simplest form.

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