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The polynomial \(x^{101} + Ax + B\)  is divisible by \(x^2 + x + 1\) for some real numbers \(A\) and \(B\). Find \(A+B\)

 Mar 20, 2020
 #1
avatar+109497 
0

Where is the original question? Did you have any answers? You should have provided a link to it.

 Mar 20, 2020
 #2
avatar+24932 
+3

The polynomial \(x^{101} + Ax + B\)  is divisible by \(x^2 + x + 1\) for some real numbers \(A\) and \(B\).
Find
\(A+B\)

 

\(\begin{array}{|rcll|} \hline \mathbf{x^2+x+1} &=& \mathbf{0} \\ x &=& \dfrac{-1\pm \sqrt{1-4*1} }{2} \\ x &=& \dfrac{-1\pm i\sqrt{3} }{2} \\ \hline \mathbf{x_1} = \mathbf{-\dfrac{1}{2}+ \dfrac{i\sqrt{3}}{2}} && \mathbf{x_2} = \mathbf{-\left(\dfrac{1}{2}+ \dfrac{i\sqrt{3}}{2}\right)} \\ \hline \end{array} \)

 

\(\begin{array}{|lrcll|} \hline & x^{101} + Ax + B &=& q(x)(x^2 + x + 1) \\ \hline x_1 = -\dfrac{1}{2}+ \dfrac{i\sqrt{3}}{2}: & \left(-\dfrac{1}{2}+ \dfrac{i\sqrt{3}}{2}\right)^{101} +A\left( -\dfrac{1}{2}+ \dfrac{i\sqrt{3}}{2}\right) + B &=& q(x_1)\times 0 \\ &\mathbf{ \left(-\dfrac{1}{2}+ \dfrac{i\sqrt{3}}{2}\right)^{101} +A\left( -\dfrac{1}{2}+ \dfrac{i\sqrt{3}}{2}\right) + B } &=& \mathbf{ 0 } \qquad (1) \\ \hline x_2 = -\left(\dfrac{1}{2}+ \dfrac{i\sqrt{3}}{2}\right): & \left[-\left(\dfrac{1}{2}+ \dfrac{i\sqrt{3}}{2}\right)\right]^{101} - A\left( \dfrac{1}{2}+ \dfrac{i\sqrt{3}}{2}\right) + B &=& q(x_2)\times 0 \\ &\mathbf{ -\left(\dfrac{1}{2}+ \dfrac{i\sqrt{3}}{2}\right)^{101} - A\left( \dfrac{1}{2}+ \dfrac{i\sqrt{3}}{2}\right) + B }&=& \mathbf{ 0 } \qquad (2) \\ \hline \end{array}\)

 

\(\begin{array}{|rcll|} \hline \left(-\dfrac{1}{2}+ \dfrac{i\sqrt{3}}{2}\right)^{1} &=& -\dfrac{1}{2}+ \dfrac{i\sqrt{3}}{2} \\ \left(-\dfrac{1}{2}+ \dfrac{i\sqrt{3}}{2}\right)^{2} &=& -\dfrac{1}{2}- \dfrac{i\sqrt{3}}{2} \\ \left(-\dfrac{1}{2}+ \dfrac{i\sqrt{3}}{2}\right)^{3} &=& 1 \\ \hline \left(-\dfrac{1}{2}+ \dfrac{i\sqrt{3}}{2}\right)^{4}&=&\left(-\dfrac{1}{2}+ \dfrac{i\sqrt{3}}{2}\right)^{1} \\ \left(-\dfrac{1}{2}+ \dfrac{i\sqrt{3}}{2}\right)^{5}&=&\left(-\dfrac{1}{2}+ \dfrac{i\sqrt{3}}{2}\right)^{2} \\ \left(-\dfrac{1}{2}+ \dfrac{i\sqrt{3}}{2}\right)^{6}&=&\left(-\dfrac{1}{2}+ \dfrac{i\sqrt{3}}{2}\right)^{3} \\ \ldots \\ \hline \left(-\dfrac{1}{2}+ \dfrac{i\sqrt{3}}{2}\right)^{101} &=& \left(-\dfrac{1}{2}+ \dfrac{i\sqrt{3}}{2}\right)^{33*3+2} \\ &=& \left(-\dfrac{1}{2}+ \dfrac{i\sqrt{3}}{2}\right)^{2}\\ \left(-\dfrac{1}{2}+ \dfrac{i\sqrt{3}}{2}\right)^{101} &=& -\dfrac{1}{2}- \dfrac{i\sqrt{3}}{2} \\ \mathbf{ \left(-\dfrac{1}{2}+ \dfrac{i\sqrt{3}}{2}\right)^{101} } &=& \mathbf{ -\left(\dfrac{1}{2}+\dfrac{i\sqrt{3}}{2}\right) } \\ \hline \end{array} \begin{array}{|rcll|} \hline \left(+\dfrac{1}{2}+ \dfrac{i\sqrt{3}}{2}\right)^{1} &=& \dfrac{1}{2}+ \dfrac{i\sqrt{3}}{2} \\ \left(+\dfrac{1}{2}+ \dfrac{i\sqrt{3}}{2}\right)^{2} &=& -\dfrac{1}{2}+ \dfrac{i\sqrt{3}}{2} \\ \left(+\dfrac{1}{2}+ \dfrac{i\sqrt{3}}{2}\right)^{3} &=& -1 \\ \left(+\dfrac{1}{2}+ \dfrac{i\sqrt{3}}{2}\right)^{4} &=& -\dfrac{1}{2}- \dfrac{i\sqrt{3}}{2} \\ \left(+\dfrac{1}{2}+ \dfrac{i\sqrt{3}}{2}\right)^{5} &=& \dfrac{1}{2}- \dfrac{i\sqrt{3}}{2} \\ \left(+\dfrac{1}{2}+ \dfrac{i\sqrt{3}}{2}\right)^{6} &=& 1 \\ \hline \left(\dfrac{1}{2}+ \dfrac{i\sqrt{3}}{2}\right)^{7}&=&\left(-\dfrac{1}{2}+ \dfrac{i\sqrt{3}}{2}\right)^{1} \\ \left(\dfrac{1}{2}+ \dfrac{i\sqrt{3}}{2}\right)^{8}&=&\left(-\dfrac{1}{2}+ \dfrac{i\sqrt{3}}{2}\right)^{2} \\ \left(\dfrac{1}{2}+ \dfrac{i\sqrt{3}}{2}\right)^{9}&=&\left(-\dfrac{1}{2}+ \dfrac{i\sqrt{3}}{2}\right)^{3} \\ \left(\dfrac{1}{2}+ \dfrac{i\sqrt{3}}{2}\right)^{10}&=&\left(-\dfrac{1}{2}+ \dfrac{i\sqrt{3}}{2}\right)^{4} \\ \left(\dfrac{1}{2}+ \dfrac{i\sqrt{3}}{2}\right)^{11}&=&\left(-\dfrac{1}{2}+ \dfrac{i\sqrt{3}}{2}\right)^{5} \\ \left(\dfrac{1}{2}+ \dfrac{i\sqrt{3}}{2}\right)^{12}&=&\left(-\dfrac{1}{2}+ \dfrac{i\sqrt{3}}{2}\right)^{6} \\ \ldots \\ \hline \left(\dfrac{1}{2}+ \dfrac{i\sqrt{3}}{2}\right)^{101} &=& \left(\dfrac{1}{2}+ \dfrac{i\sqrt{3}}{2}\right)^{16*6+5} \\ &=& \left(\dfrac{1}{2}+ \dfrac{i\sqrt{3}}{5}\right)^{5}\\ \mathbf{ \left(\dfrac{1}{2}+ \dfrac{i\sqrt{3}}{2}\right)^{101} } &=& \mathbf{ \dfrac{1}{2}- \dfrac{i\sqrt{3}}{2}} \\ \hline \end{array}\)

 

\(\begin{array}{|lrcll|} \hline (1): & \mathbf{ \left(-\dfrac{1}{2}+ \dfrac{i\sqrt{3}}{2}\right)^{101} +A\left( -\dfrac{1}{2}+ \dfrac{i\sqrt{3}}{2}\right) + B } &=& \mathbf{ 0 } \quad | \quad \mathbf{ \left(-\dfrac{1}{2}+ \dfrac{i\sqrt{3}}{2}\right)^{101} = -\left(\dfrac{1}{2}+\dfrac{i\sqrt{3}}{2}\right) } \\ & -\left(\dfrac{1}{2}+\dfrac{i\sqrt{3}}{2}\right) +A\left( -\dfrac{1}{2}+ \dfrac{i\sqrt{3}}{2}\right) + B &=& 0 \qquad (3) \\ \hline (2): & \mathbf{ -\left(\dfrac{1}{2}+ \dfrac{i\sqrt{3}}{2}\right)^{101} - A\left( \dfrac{1}{2}+ \dfrac{i\sqrt{3}}{2}\right) + B }&=& \mathbf{ 0 } \quad | \quad \mathbf{ \left(\dfrac{1}{2}+ \dfrac{i\sqrt{3}}{2}\right)^{101} = \dfrac{1}{2}- \dfrac{i\sqrt{3}}{2}} \\ & -\left(\dfrac{1}{2}- \dfrac{i\sqrt{3}}{2}\right) - A\left( \dfrac{1}{2}+ \dfrac{i\sqrt{3}}{2}\right) + B &=& 0 \qquad (4) \\ \hline (3)-(4): & -\left(\dfrac{1}{2}+\dfrac{i\sqrt{3}}{2}\right) +A\left( -\dfrac{1}{2}+ \dfrac{i\sqrt{3}}{2}\right) + B \\ & -\Bigg(-\left(\dfrac{1}{2}- \dfrac{i\sqrt{3}}{2}\right) - A\left( \dfrac{1}{2}+ \dfrac{i\sqrt{3}}{2}\right) + B\Bigg) &=& 0 \\\\ & -\left(\dfrac{1}{2}+\dfrac{i\sqrt{3}}{2}\right) +A\left( -\dfrac{1}{2}+ \dfrac{i\sqrt{3}}{2}\right) + B \\ & +\left(\dfrac{1}{2}- \dfrac{i\sqrt{3}}{2}\right) + A\left( \dfrac{1}{2}+ \dfrac{i\sqrt{3}}{2}\right) - B &=& 0 \\\\ & -\left(\dfrac{1}{2}+\dfrac{i\sqrt{3}}{2}\right) +A\left( -\dfrac{1}{2}+ \dfrac{i\sqrt{3}}{2}\right) \\ & +\left(\dfrac{1}{2}- \dfrac{i\sqrt{3}}{2}\right) + A\left( \dfrac{1}{2}+ \dfrac{i\sqrt{3}}{2}\right) &=& 0 \\\\ & -\dfrac{1}{2}-\dfrac{i\sqrt{3}}{2} -\dfrac{A}{2}+ \dfrac{Ai\sqrt{3}}{2} \\ & +\dfrac{1}{2}- \dfrac{i\sqrt{3}}{2}+\dfrac{A}{2}+ \dfrac{Ai\sqrt{3}}{2} &=& 0 \\\\ & -\dfrac{i\sqrt{3}}{2}+ \dfrac{Ai\sqrt{3}}{2} - \dfrac{i\sqrt{3}}{2}+ \dfrac{Ai\sqrt{3}}{2} &=& 0 \\\\ & -i\sqrt{3} + Ai\sqrt{3} &=& 0 \\\\ & Ai\sqrt{3} &=& i\sqrt{3} \\\\ & \mathbf{ A } &=& \mathbf{1} \\ \hline \end{array}\)

 

\(\begin{array}{|lrcll|} \hline (4): & -\left(\dfrac{1}{2}- \dfrac{i\sqrt{3}}{2}\right) - A\left( \dfrac{1}{2}+ \dfrac{i\sqrt{3}}{2}\right) + B &=& 0 \quad | \quad \mathbf{A=1} \\\\ & -\left(\dfrac{1}{2}- \dfrac{i\sqrt{3}}{2}\right) - \left( \dfrac{1}{2}+ \dfrac{i\sqrt{3}}{2}\right) + B &=& 0 \\\\ & -\dfrac{1}{2}+ \dfrac{i\sqrt{3}}{2} - \dfrac{1}{2}- \dfrac{i\sqrt{3}}{2} + B &=& 0 \\\\ & -\dfrac{1}{2} - \dfrac{1}{2} + B &=& 0 \\\\ & -1 + B &=& 0 \\\\ & \mathbf{ B } &=& \mathbf{1} \\ \hline \end{array}\)

 

\(A+B = 1+1\\ \mathbf{A+B = 2} \)

 

laugh

 Mar 20, 2020

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