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.

Sinead Mcgee 2021-07-30 Answered
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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Expert Answer

Dora
Answered 2021-07-31 Author has 17734 answers

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.

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