Step 1

We use the synthetic division to get the other factor for the missing terms, we will use 0 as the coefficient.

\(\begin{array}{|c|c|} \hline -4 & 1 & 4 & 0 & 7 & 28 & 0 & -8 & -32 \\ \hline & & -4 & 0 & 0 & -28 & 0 & 0 & 32 \\ \hline & 1 & 0 & 0 & 7 & 0 & 0 & -8 & 0\\ \hline \end{array}\)

So the quotient is \(\displaystyle{x}^{{{6}}}+{7}{x}^{{{3}}}-{8}\)

Step 2

Then we try to factor \(\displaystyle{x}^{{{6}}}+{7}{x}^{{{3}}}-{8}\)

\(\displaystyle{x}^{{{6}}}+{7}{x}^{{{3}}}-{8}\)

\(\displaystyle={\left({x}^{{{3}}}+{8}\right)}{\left({x}^{{{3}}}-{1}\right)}\)

\(\displaystyle={\left({x}^{{{3}}}+{2}^{{{3}}}\right)}{\left({x}^{{{3}}}-{1}^{{{3}}}\right)}\)

\(\displaystyle={\left({x}+{2}\right)}{\left({x}^{{{2}}}-{2}{x}+{4}\right)}{\left({x}-{1}\right)}{\left({x}^{{{2}}}+{x}+{1}\right)}\)

Answer: \(\displaystyle{\left({x}+{4}\right)}{\left({x}+{2}\right)}{\left({x}^{{{2}}}-{2}{x}+{4}\right)}{\left({x}-{1}\right)}{\left({x}^{{{2}}}+{x}+{1}\right)}\)