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# How do you simplify and divide $\left( 6{{\omega }^{5}}-18{{\omega }^{2}}-120 \right)\div \left( \omega -2 \right)$?

Last updated date: 21st Jun 2024
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Hint: In this question we have to divide the $\left( 6{{\omega }^{5}}-18{{\omega }^{2}}-120 \right)$ by $\left( \omega -2 \right)$. To solve the question we will use the long division method. For this first we will write the all terms in decreasing power of the variable order including the missing terms. Then we start dividing the terms by following the step by step procedure.

We have been given that $\left( 6{{\omega }^{5}}-18{{\omega }^{2}}-120 \right)\div \left( \omega -2 \right)$.
Here we have dividend $=\left( 6{{\omega }^{5}}-18{{\omega }^{2}}-120 \right)$, divisor $=\left( \omega -2 \right)$.
$\Rightarrow \left( 6{{\omega }^{5}}+0{{\omega }^{4}}+0{{\omega }^{3}}-18{{\omega }^{2}}+0\omega -120 \right)$
\omega -2\overset{6{{\omega }^{4}}+12{{\omega }^{3}}+24{{\omega }^{2}}+30\omega +60}{\overline{\left){\begin{align} & 6{{\omega }^{5}}+0{{\omega }^{4}}+0{{\omega }^{3}}-18{{\omega }^{2}}+0\omega -120 \\ & \underline{6{{\omega }^{5}}-12{{\omega }^{4}}} \\ & 12{{\omega }^{4}} \\ & 12{{\omega }^{4}}-24{{\omega }^{3}} \\ & \underline{\overline{\begin{align} & 24{{\omega }^{3}}-18{{\omega }^{2}} \\ & 24{{\omega }^{3}}-48{{\omega }^{2}} \\ \end{align}}} \\ & 30{{\omega }^{2}} \\ & \underline{30{{\omega }^{2}}-60\omega } \\ & 60\omega -120 \\ & \underline{60\omega -120} \\ & 0 \\ \end{align}}\right.}}
So on dividing the given expression $\left( 6{{\omega }^{5}}-18{{\omega }^{2}}-120 \right)\div \left( \omega -2 \right)$ we get the quotient as $6{{\omega }^{4}}+12{{\omega }^{3}}+24{{\omega }^{2}}+30\omega +60$.
In algebraic long division methods we need to follow the same steps as we follow in the arithmetic. We need to perform division until no term is left in the divisor. We can also check our answer by using the formula that $\text{Dividend=quotient}\times \text{divisor+remainder}\text{.}$. So by substituting the values we can verify the answer.