[tex]\large \boldsymbol {} 1-st \ method \\\\ x^2+2x-35=0 \\\\ x^2+\underbrace{7x-5x}_{2x}-35=0 \\\\( x^2+7x)+(-5x-35) =0 \\\\ x\underline {(x+7)}-5\underline{(x+7)}=0 \\\\ \boxed{(x-5)(x+7)=0 } \\\\ ---------\\\\ 2-nd \ method \\\\ x^2+2x-35=0 \\\\ \begin {cases} x_1+x_2=-2 \\ x_1x_2=-35\end{cases} \Longrightarrow x_1=5 \ ; \ x_2=-7 \\\\\\ x^2+5x-35=(x-x_1)(x-x_2)=\boxed{(x-5)(x+7 )}[/tex]
[tex]\rm \large \boldsymbol {} Answer : \ \boxed{2) \ (x-5)(x+7 )}[/tex]
