Find the remainder obtained on dividing $p(x)=x^3+1$ by $x+1$.
Find the remainder obtained on dividing $p(x)=x^3+1$ by $x+1$.
By long division,
$\overset{\,\,\,\,\,\,\,\,\,\,\,\,\,{{x}^{2}}-x+1}{\mathop{x+1\sqrt{\begin{align}
& {{x}^{3}}+1 \\
& {{x}^{3}}\pm {{x}^{2}} \\
\end{align}}}}\,$
$\_\_\_\_\_\_\_\_\_\_\_\_\_$
$-{{x}^{2}}+1$
$\mp {{x}^{2}}\pm {{x}}$
$\_\_\_\_\_\_\_\_\_\_\_\_\_\_$
$x+1$
$- x \pm 1$
$\_\_\_\_\_\_\_\_\_\_\_\_\_\_$
$0$
So, we find that the remainder is $0$.
Here $p(x) = x^3 + 1$, and the root of $x + 1 = 0$ is $x = -1$. We see that
$p(-1) = (-1)^3 + 1$
$= -1 + 1 $
$= 0$
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