Question

If \[\omega \] is the cube root of unity and \[x=\omega -{{\omega }^{2}}-2\] then the value of \[{{x}^{4}}+3{{x}^{3}}+2{{x}^{2}}-11x-6\] is equal to:

Answer 1

Hint: We solve this problem by forming an equation in \[x\] by using the given condition.
We have the standard results for the cube root of unity as
\[{{\omega }^{3}}=1\]
\[1+\omega +{{\omega }^{2}}=0\]
Then we divide the value of the required polynomial with the polynomial we get to find the remainder.
We have the formula of division that is
\[\text{Dividend}=\text{Divisor}\times \text{Quotient}+\text{Remainder}\]

Complete step by step answer:
We are given that \[\omega \] is the cube root of unity
We are also given that the value of \[x\] as
\[x=\omega -{{\omega }^{2}}-2\]
Now, let us try to form an equation of \[x\] by squaring the terms in the above equation by rearranging the terms then we get
\[\begin{align}
  & \Rightarrow x+2=\omega -{{\omega }^{2}} \\
 & \Rightarrow {{\left( x+2 \right)}^{2}}={{\left( \omega -{{\omega }^{2}} \right)}^{2}} \\
\end{align}\]
We know that the formula of algebra that is
\[\begin{align}
  & {{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}} \\
 & {{\left( a-b \right)}^{2}}={{a}^{2}}-2ab+{{b}^{2}} \\
\end{align}\]
By using this formula in the above equation then we get
\[\Rightarrow {{x}^{2}}+4x+4={{\omega }^{2}}-2{{\omega }^{3}}+{{\omega }^{4}}\]
We know that the standard results for the cube root of unity as
\[{{\omega }^{3}}=1\]
\[1+\omega +{{\omega }^{2}}=0\]
By using these standard results in the above equation then we get
\[\begin{align}
  & \Rightarrow {{x}^{2}}+4x+4={{\omega }^{2}}-2+\omega \\
 & \Rightarrow {{x}^{2}}+4x+4=-2-1 \\
 & \Rightarrow {{x}^{2}}+4x+7=0............equation(i) \\
\end{align}\]
We are asked to find the value of \[{{x}^{4}}+3{{x}^{3}}+2{{x}^{2}}-11x-6\]
Let us assume that the given expression as
\[\Rightarrow f\left( x \right)={{x}^{4}}+3{{x}^{3}}+2{{x}^{2}}-11x-6\]
Now, let us divide the polynomial \[f\left( x \right)\] with \[g\left( x \right)={{x}^{2}}+4x+7\] by using the long division method
\[{{x}^{2}}+4x+7\overline{\left){{{x}^{4}}+3{{x}^{3}}+2{{x}^{2}}-11x-6}\right.}\]
We know that in the long division method we eliminate the first term of the dividend by multiplying the certain term for divisor so that the multiplied term will be the quotient
By using the above process let us multiply the divisor with \[{{x}^{2}}\] then we get
\[{{x}^{2}}+4x+7\overset{{{x}^{2}}}{\overline{\left){\begin{align}
  & {{x}^{4}}+3{{x}^{3}}+2{{x}^{2}}-11x-6 \\
 & -\left( {{x}^{4}}+4{{x}^{3}}+7{{x}^{2}} \right) \\
 & =\left( -{{x}^{3}}-5{{x}^{2}}-11x-6 \right) \\
\end{align}}\right.}}\]
Here, we can see that we can continue the division further.
By using the same process of long division that is multiplying the divisor with \[\left( -x \right)\] then we get
\[{{x}^{2}}+4x+7\overset{{{x}^{2}}-x}{\overline{\left){\begin{align}
  & {{x}^{4}}+3{{x}^{3}}+2{{x}^{2}}-11x-6 \\
 & -\left( {{x}^{4}}+4{{x}^{3}}+7{{x}^{2}} \right) \\
 & =\left( -{{x}^{3}}-5{{x}^{2}}-11x-6 \right) \\
 & -\left( -{{x}^{3}}-4{{x}^{2}}-7x \right) \\
 & =\left( -{{x}^{2}}-4x-6 \right) \\
\end{align}}\right.}}\]
Here, we can see that we can continue the division further.
By using the same process of long division that is multiplying the divisor with \[\left( -1 \right)\] then we get
\[{{x}^{2}}+4x+7\overset{{{x}^{2}}-x-1}{\overline{\left){\begin{align}
  & {{x}^{4}}+3{{x}^{3}}+2{{x}^{2}}-11x-6 \\
 & -\left( {{x}^{4}}+4{{x}^{3}}+7{{x}^{2}} \right) \\
 & =\left( -{{x}^{3}}-5{{x}^{2}}-11x-6 \right) \\
 & -\left( -{{x}^{3}}-4{{x}^{2}}-7x \right) \\
 & =\left( -{{x}^{2}}-4x-6 \right) \\
 & -\left( -{{x}^{2}}-4x-7 \right) \\
 & =1 \\
\end{align}}\right.}}\]
We know that the formula of division that is
\[\text{Dividend}=\text{Divisor}\times \text{Quotient}+\text{Remainder}\]
By using this formula to above division then we get
\[\Rightarrow \left( {{x}^{4}}+3{{x}^{3}}+2{{x}^{2}}-11x-6 \right)=\left( {{x}^{2}}+4x+7 \right)\times \left( {{x}^{2}}-x-1 \right)+1\]
Now, by substituting the required values form equation (i) in above equation then we get
\[\begin{align}
  & \Rightarrow \left( {{x}^{4}}+3{{x}^{3}}+2{{x}^{2}}-11x-6 \right)=0\times \left( {{x}^{2}}-x-1 \right)+1 \\
 & \Rightarrow \left( {{x}^{4}}+3{{x}^{3}}+2{{x}^{2}}-11x-6 \right)=1 \\
\end{align}\]
Therefore, we can conclude that the value of \[{{x}^{4}}+3{{x}^{3}}+2{{x}^{2}}-11x-6\] is 1


Note:
 Students may do mistakes by using the long process.
We are given that the value of \[x\] as \[x=\omega -{{\omega }^{2}}-2\]
By substituting the value of \[x\] in the required equation will require a lot of calculations which is very difficult to solve the problem.
So, we need to use the short method. So that we need to form an equation of \[x\] that becomes 0 then we can calculate the value of the required value by dividing the required expression with the expression we find.
So, we can calculate the required value by using the division lemma.