# Using the remainder theorem, determine whether (x-4) and (x-1) are factors of the expression x^{3}+3x^{2}-22x-24. Hence, by use of long division, find all remaining factors of the expression.

a) Using the remainder theorem, determine whether $$\displaystyle{\left({x}-{4}\right)}$$ and $$\displaystyle{\left({x}-{1}\right)}$$ are factors of the expression $$\displaystyle{x}^{{{3}}}+{3}{x}^{{{2}}}-{22}{x}-{24}$$.
b) Hence, by use of long division, find all remaining factors of the expression.

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Step 1
Given: $$\displaystyle{x}^{{{3}}}+{3}{x}^{{{2}}}-{22}{x}-{24}$$
a) $$\displaystyle{\left({x}-{4}\right)}$$ Factor $$\displaystyle{x}-{4}={0}$$
$$\displaystyle{x}-{4}$$
$$\displaystyle{\left({4}\right)}^{{{3}}}+{3}{\left({4}\right)}^{{{2}}}-{22}{\left({4}\right)}-{24}$$
$$=64+48-88-24$$
$$\displaystyle={112}-{112}={0}$$
$$\displaystyle{\left({x}-{1}\right)}$$ Factor $$\displaystyle{x}-{1}={0}$$
$$\displaystyle{x}={1}$$
$$\displaystyle{\left({1}\right)}^{{{3}}}+{3}{\left({1}\right)}^{{{2}}}-{22}{\left({1}\right)}-{24}$$
$$\displaystyle={1}+{3}-{22}-{24}$$
$$\displaystyle={4}-{46}$$
$$\displaystyle=-{40}$$(not a factor)
Step 2
b) $$\displaystyle{\left({x}-{4}\right)}$$ Factor of $$\displaystyle{x}^{{{3}}}+{3}{x}^{{{2}}}-{22}{n}-{24}$$
$$\displaystyle{x}={4}$$
$$\begin{array}{|c|c|} \hline 4 & 1 & 3 & -22 & -24 \\ \hline & 0 & 4 & 28 & 24 \\ \hline & 1 & 7 & 6 & 0\\ \hline \end{array}$$
$$\displaystyle{x}^{{{2}}}+{7}{x}+{6}$$
$$\displaystyle={x}^{{{2}}}+{6}{x}+{x}+{6}$$
$$\displaystyle={x}{\left({x}+{6}\right)}+{1}{\left({x}+{6}\right)}$$
$$\displaystyle={\left({x}+{6}\right)}{\left({x}+{1}\right)}$$
$$\displaystyle{\left({x}+{6}\right)}{\left({x}+{1}\right)}$$ are the remaining factors.