Resposta final para o problema
$x^{4}-x^{2}+1+\frac{-1}{x^2+1}$
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Solução explicada passo a passo
1
Dividimos polinômios, $x^6$ por $x^2+1$
$\begin{array}{l}\phantom{\phantom{;}x^{2}+1;}{\phantom{;}x^{4}\phantom{-;x^n}-x^{2}\phantom{-;x^n}+1\phantom{;}\phantom{;}}\\\phantom{;}x^{2}+1\overline{\smash{)}\phantom{;}x^{6}\phantom{-;x^n}\phantom{-;x^n}\phantom{-;x^n}\phantom{-;x^n}\phantom{-;x^n}\phantom{-;x^n}}\\\phantom{\phantom{;}x^{2}+1;}\underline{-x^{6}\phantom{-;x^n}-x^{4}\phantom{-;x^n}\phantom{-;x^n}\phantom{-;x^n}\phantom{-;x^n}}\\\phantom{-x^{6}-x^{4};}-x^{4}\phantom{-;x^n}\phantom{-;x^n}\phantom{-;x^n}\phantom{-;x^n}\\\phantom{\phantom{;}x^{2}+1-;x^n;}\underline{\phantom{;}x^{4}\phantom{-;x^n}+x^{2}\phantom{-;x^n}\phantom{-;x^n}}\\\phantom{;\phantom{;}x^{4}+x^{2}-;x^n;}\phantom{;}x^{2}\phantom{-;x^n}\phantom{-;x^n}\\\phantom{\phantom{;}x^{2}+1-;x^n-;x^n;}\underline{-x^{2}\phantom{-;x^n}-1\phantom{;}\phantom{;}}\\\phantom{;;-x^{2}-1\phantom{;}\phantom{;}-;x^n-;x^n;}-1\phantom{;}\phantom{;}\\\end{array}$
2
Da divisão, obtemos o seguinte polinômio como resultado
$x^{4}-x^{2}+1+\frac{-1}{x^2+1}$
Resposta final para o problema
$x^{4}-x^{2}+1+\frac{-1}{x^2+1}$