How to factorize $a^2-b^2-a+b+(a+b-1)^2$?

Question

The answer is $(a+b-1)(2a-1)$ but I have no idea how to get this answer.

Answer 1

$$(a+b-1)^2=a^2+b^2+1-2a-2b+2ab$$

$$\implies a^2-b^2-a+b+(a+b-1)^2=2a(a+b)-\{2a+(a+b)\}+1$$ $$ =2a\{a+b-1\}-\{a+b-1\}=?$$

Answer 2

Using $a^2-b^2=(a+b)(a-b)$ we find $a^2-b^2-a+b=(a+b-1)(a-b)$. The first factor also occurs in the remaining summand.

Answer 3

$$a^2-b^2-a+b+(a+b-1)^2\Leftrightarrow (a-b)(a+b)-(a-b)+(a+b-1)^2\\\implies(a-b)(a+b-1)+(a+b-1)^2=\boxed{(a+b-1)(2a-1)}$$

Answer 4

Expand it and treat it as a quadratic expression in $a.$ That is if $C\ne 0$ then $Ca^2+Da+E=\;[a-(D- F)/2C]\;[a-(D+F)/2C)]\;$ where $F=\sqrt {D^2-4CE}.$

In this Q, we get $D^2-4CE=4b^2-4b+1=(2b-1)^2$ so it will simplify.

That is $$a^2-b^2-a+b+(a+b-1)^2=$$ $$= a^2-b^2-a+b +(a^2+2ab+b^2)-2(a+b)+1=$$ $$=2a^2+2ab -3a+b-1=$$ $$=2a^2+a(2b-3)+(b-1).$$ Let $C=2, D=2b-3,$ and $E=b-1.$