How do you factor #5a^2 - ab + 6b^2#?
1 Answer
Jul 5, 2016
#5a^2-ab+6b^2=1/20(10a-(1+sqrt(119)i)b)(10a-(1-sqrt(119)i)b)#
Explanation:
The discriminant
#Delta = (-1)^2-4(5)(6) = 1-120 = -119#
So this quadratic can only be factored using Complex coefficients:
#5a^2-ab+6b^2#
#=1/20(100a^2-20ab+120b^2)#
#=1/20((10a-b)^2+119b^2)#
#=1/20((10a-b)^2-(sqrt(119)i b)^2)#
#=1/20((10a-b)-sqrt(119)i b)((10a-b)+sqrt(119)i b)#
#=1/20(10a-(1+sqrt(119)i)b)(10a-(1-sqrt(119)i)b)#
Footnote
If the last sign in the quadratic was a minus then this would be much simpler:
#5a^2-ab-6b^2=(5a-6b)(a+b)#
Was there a typo in the question?