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

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

Factorize 2(a+b)+(a+b)/6+(a+b)^2-22(a+b)+(a+b)/6+(a+b)^2-2?