## Precalculus (6th Edition) Blitzer

The required solution is $\left\{ \left( -3,0 \right),\left( 2,20 \right),\left( -2,4 \right) \right\}$
The given equations are, \begin{align} & -4x+y=12\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left( \text{I} \right) \\ & y={{x}^{3}}+3{{x}^{2}}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \left( \text{II} \right) \end{align} Solve the equation (I): Put the value of $y={{x}^{3}}+3{{x}^{2}}$ in equation (I) \begin{align} & -4x+\left( {{x}^{3}}+3{{x}^{2}} \right)=12 \\ & {{x}^{3}}+3{{x}^{2}}-4x-12=0 \\ & {{x}^{2}}\left( x+3 \right)-4\left( x+4 \right)=0 \\ & \left( {{x}^{2}}-4 \right)\left( x+3 \right)=0 \end{align} On solving further: \begin{align} & \left( x+3 \right)\left( x-2 \right)\left( x+2 \right)=0 \\ & x=-3,x=2,x=-2 \\ \end{align} Substitute the value of $x=-3,x=2,x=-2$ into equation (II): For, \begin{align} & x=-3 \\ & y={{\left( -3 \right)}^{3}}+3\left( -{{3}^{2}} \right) \\ & =-27+27 \\ & =0 \end{align} For \begin{align} & x=2 \\ & y={{\left( 2 \right)}^{3}}+3{{\left( 2 \right)}^{2}} \\ & =8+12 \\ & =20 \end{align} For \begin{align} & x=-2 \\ & y={{\left( -2 \right)}^{3}}+3{{\left( -2 \right)}^{2}} \\ & =-8+12 \\ & =0 \end{align} Thus, the solution set of system is $\left\{ \left( -3,0 \right),\left( 2,20 \right),\left( -2,4 \right) \right\}$.