91Ó°ÊÓ

The given function is$f\left(x\right)=2x^4+2x^3-11x^2+x-6$

As the leading coefficient of the function, $a_4=2$and the constant term is $-6$, the potential rational zeroes are $Â\pm 1,Â\pm 2,Â\pm 3,Â\pm 6,Â\pm \frac{1}{2}andÂ\pm \frac{3}{2}$

Test the potential zero 1, using the synthetic division.

Since, the remainder is $-12$, 1 is not a zero of the function.

Next, take the value 2 and perform the synthetic division.

The remainder of this is 0. So, 2 is a zero and $x-2$is a factor of the function.

The remainder of $f(-3)=0$. So, $-3$is a zero and $x+3$is a factor of the function.

Using the above entries in the bottom row of the synthetic division, the function can be factored.

$$\begin{gathered} f\left(x\right)=(x-2)(2x^3+6x^2+x+3) \\ =(x-2)(x+3)(2x^2+1) \end{gathered}$$

Use the zero product property to check the remaining zeroes satisfy the new equation.

$$\begin{gathered} 2x^2+1=0 \\ 2x^2=-1 \\ x=\sqrt{-\frac{1}{2}} \end{gathered}$$

Since, the obtained number is an imaginary number, it is not a real zero of the function.

Thus, the real zeroes of the function are$\left\{-3,2\right\}$