Matematik

faktorisering af polynomier

19. juni 2020 af ty16 - Niveau: Universitet/Videregående

Hej. Jeg har brug for hjælp til følgende opgave.

Er tallet 1+2i en rod i polynomiet p(z) = z⁴ + 2z² + 8z +5 ( Afgør ved indsættelse )

Betyder dette, at jeg skal bruge divisionsalgoritmen og dividere 1+2i med p(z)??

eller skal jeg skrive 1+2i anderledes, altså skal det skrive z+2i?

Brugbart svar (1)

Svar #1
19. juni 2020 af mathon

$\small \small \small \begin{array}{lllll} \textup{Inds\ae ttelse:}&(1+2\textbf{\textit{i}})^4+2\cdot (1+2\textbf{\textit{i}})^2+8\cdot (1+2\textbf{\textit{i}})+5=\\\\& 1^4+4\cdot 1^3\cdot 2\textbf{\textit{i}}+6\cdot 1^2\cdot \left ( 2\textbf{\textit{i}} \right )^2+4\cdot 1\cdot \left ( 2\textbf{\textit{i}} \right )^3+\left ( 2\textbf{\textit{i}} \right )^4+\\\\&+2\cdot 1^2+2\cdot 4\textbf{\textit{i}}+2\cdot (2\textbf{\textit{i}})^2+8+16\textbf{\textit{i}}+5=\\\\\\& 1+8\textbf{\textit{i}}+24\textbf{\textit{i}}^2+32\textbf{\textit{i}}^3+16\textbf{\textit{i}}^4+2+8\textbf{\textit{i}}+8\textbf{\textit{i}}^2+8+16\textbf{\textit{i}}+5\\\\& \left (1-24+16+2-8+8+5 \right )+(8-32+8+16)\cdot\textbf{\textit{i}}=0 \\ \textup{dvs}\\&1+2\textbf{\textit{i}}\textup{ \textbf{er} rod i polynomiet }z^4+2z^2+8z+5 \end{array}$

Brugbart svar (0)

Svar #2
19. juni 2020 af mathon

detaljer:
$\small \begin{array}{lllll} \left ( a+b \right )^4=a^4+4\cdot a^3\cdot b+6\cdot a^2\cdot b^2+4\cdot a\cdot b^3+b^4\\\\ \textbf{\textit{i}}^2=-1\qquad \textbf{\textit{i}}^3=-\textbf{\textit{i}}\qquad \textbf{\textit{i}}^4=1 \end{array}$

Svar #3
19. juni 2020 af ty16

tak for hjælpen mathon

Brugbart svar (0)

Svar #4
21. juni 2020 af mathon

division:

$\small \begin{array}{llllll} \underline{z-(1+2\textbf{\textit{i}})}|z^4+2z^2+8z+5|\underline{z^3+(1+2\textbf{\textit{i}})z^2+ (-1+4\textbf{\textit{i}})z+ (-1+2\textbf{\textit{i}})}\\ \qquad \qquad \quad \; \, \underline{z^4-(1+2\textbf{\textit{i}})z^3}\\ \qquad \qquad \qquad \; \; \; \; \; \; (1+2\textbf{\textit{i}})z^3+2z^2+8z+5\\ \qquad \qquad \qquad \quad\; \;\underline{ (1+2\textbf{\textit{i}})z^3-(-3+4\textbf{\textit{i}})z^2}\\ \qquad \qquad \qquad \qquad \qquad\qquad\quad\, (-1+4\textbf{\textit{i}})z^2+8z+5\\ \qquad \qquad \qquad \qquad \qquad \qquad \quad\, \underline{(-1+4\textbf{\textit{i}})z^2-\left (-9+2\textbf{\textit{i}} \right )z}\\ \qquad \qquad \qquad \qquad \qquad \qquad\qquad \qquad \qquad \quad\; \, \, (-1+2\textbf{\textit{i}})z+5\\ \qquad \qquad \qquad \qquad \qquad \qquad\qquad \qquad \qquad \quad\; \, \, \underline {(-1+2\textbf{\textit{i}})z+5}\\ \qquad \qquad \qquad \qquad \qquad \qquad\qquad \qquad \qquad \quad \qquad \qquad0 \end{array}$

Brugbart svar (0)

Svar #5
22. juni 2020 af mathon

division:

$\small \small \begin{array}{llllll} \underline{z-(1-2\textbf{\textit{i}})}|z^3+(1+2\textbf{\textit{i}})z^2+(-1+4\textbf{\textit{i}})z+(-1+2\textbf{\textit{i}})|\underline{z^2+2z+1=(z+1)^2}\\ \qquad\qquad \quad\; \, \underline{z^3-(1-2\textbf{\textit{i}})z^2}\\ \qquad \qquad \qquad \qquad \qquad \; \,2z^2+(-1+4\textbf{\textit{i}})z+(-1+2\textbf{\textit{i}})\\ \qquad \qquad \qquad \qquad \qquad \; \,\underline{2z^2-(2-4\textbf{\textit{i}})z}\\ \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \;\, \, z+(-1+2\textbf{\textit{i}})\\ \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \; \, \, \underline{z-(1-2\textbf{\textit{i}})}\\ \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad\qquad \quad\, \, 0 \end{array}$

Brugbart svar (0)

Svar #6
22. juni 2020 af mathon

dvs
$\small \begin{array}{lllll} \textup{faktorisering:}&z^4+2z^2+8z+5=(z+1)^2\cdot (z-(1+2\textbf{\textit{i}}))\cdot (z-(1-2\textbf{\textit{i}})) \end{array}$

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