Let us take your concrete example:-
$$p(x) = a_0 + a_2x^2 + a_4x^4 + a_6x^6$$
$$q(x) = b_0 + b_4x^4 + b_6x^6 + b_8x^8$$
We can express the above two polynomials in vector format:-
$$\mathbf{p}=[a_0,0,a_2,0,a_4,0,a_6]$$
$$\mathbf{q}=[b_0,0,0,0,b_4,0,b_6,0,b_8]$$
where the $k$th entry (for $k\in\{0,1,2,..\}$) represents the coefficient of $x^k$.
Once we have the vectors, we need to pad $\mathbf{p}$ with $8$ zeros (the length of vector $\mathbf{q}$ minus $1$), and we need to pad $\mathbf{q}$ with $6$ zeros (the length of vector $\mathbf{p}$ minus $1$), so we have:-
$$\mathbf{p'}=[a_0,0,a_2,0,a_4,0,a_6,\color{red}{0,0,0,0,0,0,0,0}]$$
$$\mathbf{q'}=[b_0,0,0,0,b_4,0,b_6,0,b_8,\color{red}{0,0,0,0,0,0}]$$
The reason why we need zero padding is to ensure that when we carry out the FFT on the vectors, multiply the FFTs element wise and carry out an Inverse FFT (IFFT), the result will correspond to a linear convolution.
It is worth noting that the length of $\mathbf{p'}$ and $\mathbf{q'}$ is $15$. As the FFT is designed to optimally process an integer power of $2$ number of samples, we can pad $\mathbf{p'}$ and $\mathbf{q'}$ by an extra $0$ element, so that we have $16$ samples to process.
$$\mathbf{p''}=[a_0,0,a_2,0,a_4,0,a_6,\color{red}{0,0,0,0,0,0,0,0,0}]$$
$$\mathbf{q''}=[b_0,0,0,0,b_4,0,b_6,0,b_8,\color{red}{0,0,0,0,0,0,0}]$$
Next we carry out the FFT on vectors $\mathbf{p''}$ and $\mathbf{q''}$, and we denote the resulting vectors by $\mathbf{P}$ and $\mathbf{Q}$ respectively (each element will be a complex number):-
$$\mathbf{P}=[A_0,A_1,A_2,A_3,A_4,A_5,A_6,A_7,A_8,A_9,A_{10},A_{11},A_{12},A_{13},A_{14},A_{15}]$$
$$\mathbf{Q}=[B_0,B_1,B_2,B_3,B_4,B_5,B_6,B_7,B_8,B_9,B_{10},B_{11},B_{12},B_{13},B_{14},B_{15}]$$
We then multiply each element of $\mathbf{P}$ with the corresponding element of $\mathbf{Q}$, resulting in vector
$$\mathbf{R}=[C_0,C_1,C_2,...,C_{15}]$$
where $C_k=A_k\times B_k$.
The final step is to calculate the IFFT of vector $\mathbf{R}$, resulting in
$$\mathbf{r}=[r_0,r_1,r_2,...,r_{15}]$$
where element $r_k$ is the coefficient of $x^k$, for $\in{0,1,..,15}$. Note that $r_{15}$ should be $0$, as we zero padded both vectors by an extra $0$ term.
Assuming that the coefficients of the polynomials $p(x)$ and $q(x)$ were real, then the elements of $\mathbf{r}$ should be real (any deviations, in terms of a small imaginary component, could arise from the finite precision nature of how the FFT is implemented on a processor).
Given the nature of the polynomials $p(x)$ and $q(x)$, the elements $r_m$ for odd $m$ will be zero (or a very small number, also due to the finite precision effects).