91Ó°ÊÓ

The given function is$f\left(x\right)=4x^3+4x^2-7x+2$

The given function has a degree 3. So, there will be a maximum of 3 complex zeroes.

We can see from the function that it has two sign variations. So, there will be two or no positive real zero.

$$\begin{gathered} f(-x)=4{(-x)}^3+4{(-x)}^2-7(-x)+2 \\ =-4x^3+4x^2+7x+2 \end{gathered}$$

Since there is one sign change in $f(-x)$, there is one negative real zero.

We will list the integers which are factors of the constant term 2. $p=Â\pm 1,Â\pm 2$

Now, we will list the integers which are factors of the leading coefficient 4.

$q=Â\pm 1,Â\pm 2,Â\pm 4$

Next, list the possible potential rational zeroes which is the ratio $\frac{p}{q}$

$\frac{p}{q}:Â\pm 1,Â\pm 2,Â\pm 4,Â\pm \frac{1}{2},Â\pm \frac{1}{4}$

Thus, the potential rational zero of function are$Â\pm 1,Â\pm 2,Â\pm 4,Â\pm \frac{1}{2}andÂ\pm \frac{1}{4}$

Using the synthetic division check whether 1 is a zero of function or not.

Since, $f\left(1\right)≠0$, 1 is not a zero of the function.

Using the synthetic division check whether $-2$is a zero of function or not.

Thus, the depressed equation of function is

$4x^2-4x+1=0$

Now, factorize the above equation by grouping.

$$\begin{gathered} 4(x^2-x+\frac{1}{4})=0 \\ 4{(x-\frac{1}{2})}^2=0 \end{gathered}$$

Thus, ${(x-\frac{1}{2})}^2$is a polynomial of degree 2, the zero $\frac{1}{2}$has a multiplicity of 2.

Hence, the three complex zeroes of function are$\left\{-2,\frac{1}{2}(multiplicity2)\right\}$