This question has been asked but I am stuck with my method.
I have shown that the roots can not be on the real and imaginary axis. Since the coefficients are real and thus the roots must be in conjugate pairs. So If there is no root in first quadrant (or right half plane), then we have 2 roots in second quadrant and third quadrant respectively. (this is the answer to this question)
So I am going to show that there is no root in right half plane. Consider the integral along the semi-circle in right half plane with radius $R$ $$\int_{C}\dfrac{f'}{f}dz=\int_{-Ri}^{Ri}+\int_{-\pi/2}^{\pi/2}\dfrac{4z^3+3z^2+8z+2}{z^4+z^3+4z^2+2z+3}dz$$
For the second part, $$\int_{-\pi/2}^{\pi/2}\dfrac{4z^3+3z^2+8z+2}{z^4+z^3+4z^2+2z+3}dz=i\int_{-\pi/2}^{\pi/2}\dfrac{4z^4+3z^3+8z^2+2z}{z^4+z^3+4z^2+2z+3}d\theta=4\pi i$$ as $R\rightarrow \infty$.
For the first part, \begin{align*} &\int_{-Ri}^{Ri}\dfrac{4z^3+3z^2+8z+2}{z^4+z^3+4z^2+2z+3}dz\\ &=\int^R_{-R}\dfrac{3y^2-2+i(4y^3-8y)}{y^4-4y^2+3-i(y^3+2y)}idy\\ \\ &=\int^R_{-R}\dfrac{(3y^2-3)(y^4-4y^2+3)-(4y^3-8y)(y^3-2y)+i[(4y^3-8y)(y^4-4y^2+3)+(y^3-2y)(3y^2-2)]}{(y^4-4y^2+3)^2+(y^3+2y)^2}idy\\ &=\int^R_{-R} i(even \,\,part)-(odd \,\,part) dy\\ &=\int^R_{-R} i(even \,\,part) \,\,dy \end{align*} It is expected that $\int^R_{-R} i(even \,\,part) \,\,dy=-4\pi i$, then we are done, but how?
Also is there any method using Rouche theorem?
The coefficients are positive, so the polynomial has no positive real roots.
We check that $f$ has no purely imaginary roots. Let $z=yi$, then $$f=y^4-4y^2+3+i(2y-y^3).$$ If $f=0$, we have $y=0$ or $\pm\sqrt{2}$ from the imaginary part, but $f\neq 0$ by substituting these values of $y$ into the real part of $f$.
We check that there is no negative real roots. On the real axis, we have $$f'=4z^3+3z^2+8z+2 \text{ and } f''=12z^2+6z+8>0.$$ From these, we see that $f$ is convex and have local minima. Let $x_1$ such that $f'(x_1)=4x_1^3+3x_1^2+8x_1+2=0$ and $f(x_1)$ is the global minimum. Then on the real axis, \begin{align*} f(z)&\geq f(x_1)=x_1^4+x_1^3+4x_1^2+2x_1+3\\ &=x_1^4+\dfrac{1}{4}(4x_1^3+16x_1^2+8x_1+12)\\ &=x_1^4+\dfrac{1}{4}(13x_1^2+10)>0. \end{align*}
It is enough to consider the number of roots in the right half plane. We take a contour $C$ to be a right half circle with radius $R$, $$\int_C \dfrac{f'}{f}dz=\int_{-\pi/2}^{\pi/2}+\int_{Ri}^{-Ri} \dfrac{f'}{f}dz.$$
For the first integral, $z=Re^{i\theta}$ \begin{align*} \int_{-\pi/2}^{\pi/2} \dfrac{f'}{f}dz&=\int_{-\pi/2}^{\pi/2} \dfrac{4z^3+3z^2+8z+2}{z^4+z^3+4z^2+2z+3}Re^{i\theta}id\theta\\ &\to4\pi i \text{ as }R \to \infty \end{align*}
For the second integral, $$\int_{Ri}^{-Ri} \dfrac{f'}{f}dz=\int_{Ri}^{-Ri} \dfrac{df}{f-0}.$$ It is the winding number of $f$ around 0. If we set $z=yi$, we have $f=y^4-4y^2+3+i(2y-y^3)$. We find all the zero of the real and imaginary parts, that is $y=\pm 1,\pm \sqrt{3}$ and $y=0,\pm\sqrt{2}$.
\begin{array}{|c|c|c|c|c|c|c|c|c|} \hline y & -\infty & -\sqrt{3} & -\sqrt{2} & -1 & 0 & 1 & \sqrt{2} & \sqrt{3} & \infty \\ \hline Re(f) & + & 0 & - & 0 & + & 0 & - & 0 & + \\ \hline Im(f) & 0 & + & 0 & - & 0 & + & 0 & - & 0 \\ \end{array}
Thus, $\dfrac{1}{2\pi i}\displaystyle\int_{Ri}^{-Ri} \dfrac{f'}{f}dz \to -2$ as $R \to \infty$. Combining our results, $\displaystyle\int_C \dfrac{f'}{f}dz \to 0$ as $R \to \infty$, which implies that there is no root in the right half plane. Thus there are 0, 2, 2, 0 roots in quadrant I, II, III, IV respectively.