## Precalculus (6th Edition) Blitzer

The solutions of the equation $2{{x}^{3}}+11{{x}^{2}}-7x-6=0$ are $x=1,-6\text{ and}-\frac{1}{2}$.
Consider the provided equation, $2{{x}^{3}}+11{{x}^{2}}-7x-6=0$ …… (1) In equation (1), the constant term is $-6$ The factors of the constant term are $\pm 1,\pm 2,\pm 3\text{ and }\pm 6$. In equation (1), the leading term is 2 The factors of the coefficient of the leading term are $\pm 1\text{ and}\pm 2$. Now, the possible roots of the provided equation are \begin{align} & \frac{\text{Factors}\ \text{of}\ \text{the}\ \text{constant}\ \text{term}}{\text{Factors}\ \text{of}\ \text{the}\ \text{coefficient}\ \text{of}\ \text{the leading term}}=\frac{\pm 1,\pm 2,\pm 3,\pm 6}{\pm 1,\pm 2} \\ & =\pm 1,\pm 2,\pm 3,\pm 6,\pm \frac{1}{2},\pm \frac{3}{2} \end{align} Checking for $x=1$: We are checking whether $\left( x-1 \right)$ is a factor or not. Divide the polynomial $2{{x}^{3}}+11{{x}^{2}}-7x-6$ by $\left( x-1 \right)$. x-1\overset{2{{x}^{2}}+13x+6}{\overline{\left){\begin{align} & 2{{x}^{3}}+11{{x}^{2}}-7x-6 \\ & \underline{2{{x}^{3}}-2{{x}^{2}}} \\ & \ \text{ 13}{{x}^{2}}-7x-6 \\ & \ \text{ }\underline{13{{x}^{2}}-13x} \\ & \text{ 6}x-6 \\ & \ \text{ }\underline{6x-6} \\ & \text{ }\underline{\text{ 0}\ \text{ }} \\ \end{align}}\right.}} As, the remainder is zero . Thus, $\left( x-1 \right)$ is a factor of $2{{x}^{3}}+11{{x}^{2}}-7x-6=0$. Using the result from the division $2{{x}^{3}}+11{{x}^{2}}-7x-6=\left( x-1 \right)\left( 2{{x}^{2}}+13x+6 \right)+0$ \begin{align} & 2{{x}^{3}}+11{{x}^{2}}-7x-6=0 \\ & \left( x-1 \right)\left( 2{{x}^{2}}+13x+6 \right)+0=0 \\ & \left( x-1 \right)\left( 2{{x}^{2}}+12x+x+6 \right)=0 \\ & \left( x-1 \right)\left( 2x+1 \right)\left( x+6 \right)=0 \end{align} Equating all the factors to zero, we obtain the roots. $x=1,x=-6,x=-\frac{1}{2}$ Thus, the roots of the equation $2{{x}^{3}}+11{{x}^{2}}-7x-6=0$ are $x=1,-6\text{ and}-\frac{1}{2}$.