What is $a+b$ if $ x^2-y^2+11x+7y+18$ can be factorized as $(x+y+a)(x-y+b)$? If the polynomial $x^2-y^2+11x+7y+18$ can be factorized as $(x+y+a)(x-y+b)$ what is the value of $a+b$?

Observe that the expression can be rearranged with respect to  (x) and factorized as
\begin{align} x^2-y^2+11x+7y+18 &amp;= x^2+11x-(y^2-7y-18) \\ &amp;= x^2+11x-(y+2)(y-9) \\ &amp;= (x+(y+2))(x-(y-9)) \\ &amp;= (x+y+2)(x-y+9). \end{align}
This implies  (a=2) and  (b=9), hence  (a+b=11).

I don't understand how they got from step 2, 
$ \\ x^2+11x-(y+2)(y-9)\\$, to step 3,  $\\ (x+(y+2))(x-(y-9)) \\ $

https://brilliant.org/practice/algebraic-manipulation-rearranging/?problem=algebra-problem-56641&amp;subtopic=advanced-polynomials&amp;chapter=algebraic-manipulation
Hint: &nbsp;consider $x^2+11x-y^2+7y+18$ as a quadratic in $\,x\,$, then its discriminant is:

$$
\Delta = 11^2 - 4(-y^2+7y+18) = 4y^2-28y+49=(2y-7)^2
$$

The roots in $\,x\,$ are $x_{1,2}=\big(-11 \pm (2y-7)\big)/2\,$, then the quadratic factors as $(x-x_1)(x-x_2).$
The solution is on this image Simplify like this