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Formation of Algebraic Expressions

Let Px = 1 ...

Question

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

0

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6

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1 + x

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1 + x + x^{2} + x^{3} + x^{4}

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Solution

The correct option is A $6$
As $P (x)$ is the sum of GP. = $\frac{1 - x^{6}}{1 - x}$

It has 5 roots , let $a_{1}, a_{2}, a_{3}, a_{4}, a_{5}$ , and they are the $6 t h$ roots of unity except unity.

Now $P (x^{12}) = 1 + x^{12} + x^{24} + x^{36} + x^{48} + x^{60} = P (x) . Q (x) + R (x)$ .

Here $R (x)$ is a remainder and a polynomial of maximum degree 4.

Put $x = a_{1}, a_{2} . . . . . . . . . . . . . . ., a_{5}$
We get,
$R (a_{1}) = 6$ , $R (a_{2}) = 6$ , $R (a_{3}) = 6$ , $R (a_{4}) = 6$ , $R (a_{5}) = 6$

i.e, $R (x) - 6 = 0$ has 6 roots.

Which contradict that $R (x)$ is maximum of degree 4.

So, it is an identity
Therefore, $R (x) = 6$ .

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