## Precalculus (6th Edition) Blitzer

The zeroes are, $x=2,3,-1$.
To calculate the first zero of the provided function, we will proceed by the hit and trial method. If for any rational number $x$, the function is evaluated to be zero, then that value of $x$ will be one of the roots of the given equation. Test 1- Take $x=1$ $\therefore f(1)={{1}^{3}}-4\cdot {{1}^{2}}+1+6=1-4+1+6=4\ne 0$ So, $x=1$ is not a zero. Test 2- Take $x=2$ $\therefore f(2)={{2}^{3}}-4\cdot {{2}^{2}}+2+6=8-16+2+6=4=0$ Therefore, $x=2$ is one of the zeros. So, take out the factor of $\left( x-2 \right)$ from the function when equated to zero \begin{align} & {{x}^{3}}-4{{x}^{2}}+x+6=0 \\ & {{x}^{3}}-2{{x}^{2}}-2{{x}^{2}}+4x-3x+6=0 \end{align} Make the factor of $(x-2)$ in each term \begin{align} & {{x}^{2}}\left( x-2 \right)-2x\left( x-2 \right)-3\left( x-2 \right)=0 \\ & \left( x-2 \right)\left( {{x}^{2}}-2x-3 \right)=0 \end{align} Either, $\left( x-2 \right)=0$ or $\left( {{x}^{2}}-2x-3 \right)=0$ When \begin{align} & \left( x-2 \right)=0 \\ & x=2 \end{align} When \begin{align} & \left( {{x}^{2}}-2x-3 \right)=0 \\ & \left( x-3 \right)\left( x+1 \right)=0 \\ \end{align} Either, $\left( x-3 \right)=0$ or $\left( x+1 \right)=0$ When \begin{align} & \left( x-3 \right)=0 \\ & x=3 \end{align} When \begin{align} & \left( x+1 \right)=0 \\ & x=-1 \end{align} Hence, the zeros of the function are $2,3,-1$.