# A polynomial and one of its factors is given. Factor the polynomial COMPLETELY given that one of its factors is x+4. Your final answer should be in factored form. f(x)=x^{7}+4x^{6}+7x^{4}+28x^{3}-8x-32

A polynomial and one of its factors is given. Factor the polynomial COMPLETELY given that one of its factors is $$\displaystyle{x}+{4}$$. Your final answer should be in factored form.
$$\displaystyle{f{{\left({x}\right)}}}={x}^{{{7}}}+{4}{x}^{{{6}}}+{7}{x}^{{{4}}}+{28}{x}^{{{3}}}-{8}{x}-{32}$$

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