Factorize #2(a+b)+(a+b)/6+(a+b)^2-2#?

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Answer 1 · 2017-03-13T07:07:40

#2(a+b)+(a+b)/6+(a+b)^2-2=(a+b+13/12+sqrt457/12)(a+b+13/12-sqrt457/12)#

#2(a+b)+(a+b)/6+(a+b)^2-2#

= #(2+1/6)(a+b)+(a+b)^2-2#

= #13/6(a+b)+(a+b)^2-2#

= #(a+b)^2+13/6(a+b)-2#

Let #a+b=x#, then quadratic polynomial inside square bracket becomes #x^2+13/6x-2# and as

#x^2+13/6x-2#

= #x^2+2xx13/12xx x+(13/12)^2-2-(13/12)^2#

= #(x+13/12)^2-2-169/144#

= #(x+13/12)^2-457/144#

= #(x+13/12)^2-(sqrt457/12)^2#

= #(x+13/12+sqrt457/12)(x+13/12-sqrt457/12)#

and substituting #x# with #a+b#, we get

#2(a+b)+(a+b)/6+(a+b)^2-2=(a+b+13/12+sqrt457/12)(a+b+13/12-sqrt457/12)#