# 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 $\left(x-4\right)$ and $\left(x-1\right)$ are factors of the expression ${x}^{3}+3{x}^{2}-22x-24$.
b) Hence, by use of long division, find all remaining factors of the expression.
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Step 1
Given: ${x}^{3}+3{x}^{2}-22x-24$
a) $\left(x-4\right)$ Factor $x-4=0$
$x-4$
${\left(4\right)}^{3}+3{\left(4\right)}^{2}-22\left(4\right)-24$
$=64+48-88-24$
$=112-112=0$
$\left(x-1\right)$ Factor $x-1=0$
$x=1$
${\left(1\right)}^{3}+3{\left(1\right)}^{2}-22\left(1\right)-24$
$=1+3-22-24$
$=4-46$
$=-40$(not a factor)
Step 2
b) $\left(x-4\right)$ Factor of ${x}^{3}+3{x}^{2}-22n-24$
$x=4$
$\begin{array}{|ccccc|}\hline 4& 1& 3& -22& -24\\ & 0& 4& 28& 24\\ & 1& 7& 6& 0\\ \hline\end{array}$
${x}^{2}+7x+6$
$={x}^{2}+6x+x+6$
$=x\left(x+6\right)+1\left(x+6\right)$
$=\left(x+6\right)\left(x+1\right)$
$\left(x+6\right)\left(x+1\right)$ are the remaining factors.