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

Name: PDF Generator App | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

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 $P(x)$ ?

Similar Questions

The terms of a $G.P.$ are positive. If each term is equal to the sum of two terms that follow it, then the common ratio is

If the first term of a $G.P. a_1, a_2, a_3......$ is unity such that $4a_2 + 5a_3$ is least, then the common ratio of $G.P.$ is

The $6^{th}$ term of a $G.P.$ is $32$ and its $8^{th}$ term is $128$, then the common ratio of the $G.P.$ is

If the sum of the series $1 + \frac{2}{x} + \frac{4}{{{x^2}}} + \frac{8}{{{x^3}}} + ....\infty $ is a finite number, then

If $S$ is the sum to infinity of a $G.P.$, whose first term is $a$, then the sum of the first $n$ terms is