Let P(x)=1+x+x^2+x^3+x^4+x^5. What is the rem | vedclass

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Question

(b)

We have,

$P(x)=1+x+x^2+x^3+x^4+x^5$

$P(x)=\frac{1-x^6}{1-x}$

${\left[\because a+a r+a r^2+\ldots+a r^n=\frac{a\left(-r^n\right)}{1-r}\

Vedclass · Accepted Answer

$6$

Vedclass · Answer

(b)

We have,

$P(x)=1+x+x^2+x^3+x^4+x^5$

$P(x)=\frac{1-x^6}{1-x}$

${\left[\because a+a r+a r^2+\ldots+a r^n=\frac{a\left(-r^night)}{1-r}ight]}$

It has $5$ roots let $\alpha_1, \alpha_2, \alpha_3, \alpha_4, \alpha_5$ they are $6$ th roots of unity

Now,

$P\left(x^{12}ight)=1+x^{12}+x^{24}+x^{36}+x^{48}+x^{60}$

$	herefore P\left(x^{12}ight) =P(x) \cdot Q(x)+R(x)$

Here, $R(x)$ is a polynomial of maximum degree $4 .$

Put $x=\alpha_1, \alpha_2, \alpha_3, \alpha_4, \alpha_5$; we get

$R\left(\alpha_1ight)=6=R\left(\alpha_2ight) =R\left(\alpha_3ight)$

$=R\left(\alpha_4ight)=R\left(\alpha_5ight)$

$	herefore R(x)-6=0$ has 6 roots, which contradicts that $R(x)$ is maximum of degree $4 .$

$	herefore$ So it is an identity.

$\because \quad R(x)=6$

Let $P(x)=1+x+x^2+x^3+x^4+x^5$. What is the remainder when $P\left(x^{12}\right)$ is divided by $P(x)$ ?

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