How do you factor #36x^2 + 35x + 8#?

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Answer 1 · 2018-07-20T04:17:24

#36x^2+35x+8=(72x+35-sqrt73)(72+35+sqrt73)#

#x=(-b+-sqrt(b^2-4ac))/(2a)# where the equation is #ax^2+bx+c=0#

Therefore,
#a=36#, #b=35# and #c=8#

#x=(-35+-sqrt(35^2-4times36times8))/(2times36)#

#x=(-35+-sqrt73)/72#

#x=(-35+sqrt73)/72# or #x=(-35-sqrt73)/72#

#36x^2+35x+8#

#=(x-((-35+sqrt73)/72))(x-(-35-sqrt73)/72))#

#=(x+35/72-sqrt73/72)(x+35/72+sqrt73/72)#

#=(72x+35-sqrt73)(72+35+sqrt73)#