We are required to simplify
[tex]\begin{gathered} \frac{2x^2+3x-35}{x+5} \\ \end{gathered}[/tex]To solve the problem, we need to factorize the numerator first
[tex]\begin{gathered} \frac{2x^2+3x-35}{x+5}\text{ =}\frac{2x^2+10x-7x-35}{x+5} \\ \\ =\frac{2x(x+5)-7(x+5)}{x+5} \\ \\ =\frac{(x+5)(2x-7)}{(x+5)} \\ \\ (x+5)\text{ is a common factor},\text{ divide the numerator and denominator by (x+5)} \\ =2x-7 \end{gathered}[/tex]The answer is 2x - 7
