\(\left(x+a\right)\left(x+5\right)=x^2+3x+b\)
\(\Leftrightarrow x^2+5x+ax+5a=x^2+3x+b\)
\(\Leftrightarrow x^2+\left(5+a\right)x+5a=x^2+3x+b\)
Đồng nhất ta có : \(\hept{\begin{cases}5+a=3\\5a=b\end{cases}\Rightarrow\hept{\begin{cases}a=-2\\b=-10\end{cases}}}\)