How do you factor the trinomial #x^2-11xy+60y^2 #?

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Answer 1 · 2016-01-04T10:46:14

#x^2-11xy+60y^2#

#= 1/4(2x-(11+sqrt(119) i)y)(2x-(11-sqrt(119) i)y)#

#x^2-11xy+60y^2# is in the form #ax^2+bxy+cy^2# with #a=1#, #b=-11# and #c=60#.

This has discriminant #Delta# given by the formula:

#Delta = b^2-4ac = (-11)^2-(4*1*60) = 121-240 = -119#

Since #Delta < 0# this trinomial has no linear factors with Real coefficients.

It does have factors of the form #(x-t_1y)(x-t_2y)# where #t_1# and #t_2# are the (Complex) roots of #t^2-11t+60 = 0# given by the quadratic formula:

#t = (-b+-sqrt(b^2-4ac))/(2a)#

#=(-b+-sqrt(Delta))/(2a)#

#=(11+-sqrt(-119))/2#

#=11/2+-sqrt(119)/2 i#

Hence:

#x^2-11xy+60y^2#

#= (x-(11/2+sqrt(119)/2 i)y)(x-(11/2-sqrt(119)/2 i)y)#

#= 1/4(2x-(11+sqrt(119) i)y)(2x-(11-sqrt(119) i)y)#