ajuda por favor ninguem sabe resolver alguem ?? :/
Anexos:
![](https://pt-static.z-dn.net/files/d8e/d5b89fdaddb9be44ae409eadd4597db7.jpg)
Soluções para a tarefa
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jubisla:
voce poderia me explicar?
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Olá Jubista,
se o polinõmio
![2 x^{3}+ x^{2} -22x+24 2 x^{3}+ x^{2} -22x+24](https://tex.z-dn.net/?f=2+x%5E%7B3%7D%2B+x%5E%7B2%7D+-22x%2B24+)
é produto dos fatores
![(x+4)*(x-2)*P(x_3) (x+4)*(x-2)*P(x_3)](https://tex.z-dn.net/?f=%28x%2B4%29%2A%28x-2%29%2AP%28x_3%29)
Podemos então fazer:
![P(x_3)*(x+4)*(x-2)=2 x^{3}+ x^{2} -22x+24\\
P(x_3)* x^{2} -2x+4x-8=2 x^{3}+ x^{2} -22x+24\\
P(x_3)* x^{2} +2x-8=2 x^{3}+ x^{2} -22x+24\\\\
P(x_3)= \dfrac{2 x^{3}+ x^{2} -22x+24 }{ x^{2} +2x-8}\\\\\\
Utilizando~o~metodo~da~chave,~teremos:\\\\
~~~2 x^{3}+ x^{2} -22x+24~~|~~ x^{2} +2x-8\\
-2x^{3}-4 x^{2} +16x~~~~~~~~~~~~~2x-3\\
---------\\
~~~~~0-3 x^{2} ~-6x~+24\\
~~~~~~~+3 x^{2}+6x~+24\\
~~~~~~~--------\\
~~~~~~~~~~~0~~~~~~0~~~~~~~0 P(x_3)*(x+4)*(x-2)=2 x^{3}+ x^{2} -22x+24\\
P(x_3)* x^{2} -2x+4x-8=2 x^{3}+ x^{2} -22x+24\\
P(x_3)* x^{2} +2x-8=2 x^{3}+ x^{2} -22x+24\\\\
P(x_3)= \dfrac{2 x^{3}+ x^{2} -22x+24 }{ x^{2} +2x-8}\\\\\\
Utilizando~o~metodo~da~chave,~teremos:\\\\
~~~2 x^{3}+ x^{2} -22x+24~~|~~ x^{2} +2x-8\\
-2x^{3}-4 x^{2} +16x~~~~~~~~~~~~~2x-3\\
---------\\
~~~~~0-3 x^{2} ~-6x~+24\\
~~~~~~~+3 x^{2}+6x~+24\\
~~~~~~~--------\\
~~~~~~~~~~~0~~~~~~0~~~~~~~0](https://tex.z-dn.net/?f=P%28x_3%29%2A%28x%2B4%29%2A%28x-2%29%3D2+x%5E%7B3%7D%2B+x%5E%7B2%7D+-22x%2B24%5C%5C%0AP%28x_3%29%2A+x%5E%7B2%7D+-2x%2B4x-8%3D2+x%5E%7B3%7D%2B+x%5E%7B2%7D+-22x%2B24%5C%5C%0AP%28x_3%29%2A+x%5E%7B2%7D+%2B2x-8%3D2+x%5E%7B3%7D%2B+x%5E%7B2%7D+-22x%2B24%5C%5C%5C%5C%0AP%28x_3%29%3D+%5Cdfrac%7B2+x%5E%7B3%7D%2B+x%5E%7B2%7D+-22x%2B24+%7D%7B+x%5E%7B2%7D+%2B2x-8%7D%5C%5C%5C%5C%5C%5C%0AUtilizando%7Eo%7Emetodo%7Eda%7Echave%2C%7Eteremos%3A%5C%5C%5C%5C%0A%7E%7E%7E2+x%5E%7B3%7D%2B+x%5E%7B2%7D+-22x%2B24%7E%7E%7C%7E%7E+x%5E%7B2%7D+%2B2x-8%5C%5C%0A-2x%5E%7B3%7D-4+x%5E%7B2%7D+%2B16x%7E%7E%7E%7E%7E%7E%7E%7E%7E%7E%7E%7E%7E2x-3%5C%5C%0A---------%5C%5C%0A%7E%7E%7E%7E%7E0-3+x%5E%7B2%7D+%7E-6x%7E%2B24%5C%5C%0A%7E%7E%7E%7E%7E%7E%7E%2B3+x%5E%7B2%7D%2B6x%7E%2B24%5C%5C%0A%7E%7E%7E%7E%7E%7E%7E--------%5C%5C%0A%7E%7E%7E%7E%7E%7E%7E%7E%7E%7E%7E0%7E%7E%7E%7E%7E%7E0%7E%7E%7E%7E%7E%7E%7E0++++)
Portanto, o 3º fator é:
![\boxed{P(x_3)=2x-3} \boxed{P(x_3)=2x-3}](https://tex.z-dn.net/?f=%5Cboxed%7BP%28x_3%29%3D2x-3%7D)
Espero ter ajudado e tenha ótimos estudos =))
se o polinõmio
é produto dos fatores
Podemos então fazer:
Portanto, o 3º fator é:
Espero ter ajudado e tenha ótimos estudos =))
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