${{x}^{4}}+{{x}^{3}}-{{x}^{2}}+ax+b=\left( {{x}^{2}}+x-2 \right)\left( {{x}^{2}}+cx+d \right)$
$\Leftrightarrow {{x}^{4}}+{{x}^{3}}-{{x}^{2}}+ax+b={{x}^{4}}+c{{x}^{3}}+d{{x}^{2}}+{{x}^{3}}+c{{x}^{2}}+dx-2{{x}^{2}}-2cx-2d$
$\Leftrightarrow {{x}^{4}}+{{x}^{3}}-{{x}^{2}}+ax+b={{x}^{4}}+\left( c+1 \right){{x}^{3}}+\left( d+c-2 \right){{x}^{2}}+\left( d-2c \right)x-2d$
Đồng nhất hệ số ta có:
$\Rightarrow \left\{ \begin{align} & 1=1 \\ & c+1=1 \\ & d+c-2=-1 \\ & d-2c=a \\ & b=-2d \\ \end{align} \right.$$\Leftrightarrow \left\{ \begin{align} & c=0 \\ & d=1 \\ & a=1 \\ & b=-2 \\ \end{align} \right.$