Question

The remainder when the polynomial \[x + {x^3} + {x^9} + {x^{27}} + {x^{81}} + {x^{2243}}\] is divided by \[{x^2} + 1\] is \[k\] then the value of \[\dfrac{k}{{5k + 1}} = \]
A.0
B.1
C.2
D.3
E.4
F.5
G.6
H.7
I.8
J.9

Answer 1

Hint: Here, we will first solve the divisor to find the value of \[{x^2}\]. Then we will substitute this value in each term of the given dividend. Then by solving it, we will get the value of remainder which when further substituted in \[\dfrac{k}{{5k + 1}}\] will give us the required value.

Complete step-by-step answer:
Here, the dividend is \[x + {x^3} + {x^9} + {x^{27}} + {x^{81}} + {x^{2243}}\].
Now, we have to divide this polynomial by \[{x^2} + 1\].
Now, the divisor can be written as:
\[{x^2} = - 1\]
Now, in the given polynomial, using the formula \[{a^{m + n}} = {a^m} \cdot {a^n}\] and substituting \[{x^2} = - 1\] , we will values of each term. Therefore,
\[{x^3} = {x^2} \cdot x = - x\]
\[\begin{array}{l}{x^9} = {x^8} \cdot x = {\left( {{x^2}} \right)^4} \cdot x\\ \Rightarrow {x^9} = {\left( { - 1} \right)^4} \cdot x = x\end{array}\]
\[\begin{array}{l}{x^{27}} = {x^{26}} \cdot x = {\left( {{x^2}} \right)^{13}} \cdot x\\ \Rightarrow {x^{27}} = {\left( { - 1} \right)^{13}} \cdot x = - x\end{array}\]
\[{x^{81}} = {x^{80}} \cdot x = {\left( {{x^2}} \right)^{40}} \cdot x = {\left( { - 1} \right)^{40}} \cdot x = x\]
\[{x^{2243}} = {x^{2242}} \cdot x = {\left( {{x^2}} \right)^{1121}} \cdot x = {\left( { - 1} \right)^{1121}} \cdot x = - x\]
Hence, substituting all these value in the polynomial \[x + {x^3} + {x^9} + {x^{27}} + {x^{81}} + {x^{2243}}\], we get
\[\dfrac{{x + {x^3} + {x^9} + {x^{27}} + {x^{81}} + {x^{2243}}}}{{{x^2} + 1}} = x - x + x - x + x - x = 0\]
Hence, the remainder is 0.
But it is given that the remainder is equal to \[k\]
Therefore, \[k = 0\]
Now, substituting this value in \[\dfrac{k}{{5k + 1}}\], we get,
\[\dfrac{k}{{5k + 1}} = \dfrac{0}{{5\left( 0 \right) + 1}} = 0\]
Therefore, the remainder when the polynomial \[x + {x^3} + {x^9} + {x^{27}} + {x^{81}} + {x^{2243}}\] is divided by \[{x^2} + 1\] is \[k\] then the value of \[\dfrac{k}{{5k + 1}} = 0\]
Hence, option A is the correct answer.

Note: An alternate way to solve this question is by using complex numbers.
Since, given polynomial to be divided is: \[x + {x^3} + {x^9} + {x^{27}} + {x^{81}} + {x^{2243}}\]
Now, we have to divide this polynomial by \[{x^2} + 1\]
Now, the divisor can be written as:
\[{x^2} = - 1\]
Hence, comparing this with complex numbers, i.e. \[{i^2} = - 1\]
We get, \[x = i\]
Hence, the given polynomial can be written as:
\[i + {i^3} + {i^9} + {i^{27}} + {i^{81}} + {i^{2243}}\]
Now, solving this in the same manner just the variable will be \[i\] instead of \[x\], we get,
\[i + {i^3} + {i^9} + {i^{27}} + {i^{81}} + {i^{2243}} = i - i + i - i + i - i = 0\]
Hence, the value of \[k = 0\]
Now, substituting this value in \[\dfrac{k}{{5k + 1}}\], we get,
\[\dfrac{k}{{5k + 1}} = \dfrac{0}{{5\left( 0 \right) + 1}}\]
\[ \Rightarrow \dfrac{k}{{5k + 1}} = \dfrac{0}{1} = 0\]
Therefore, the value of \[\dfrac{k}{{5k + 1}} = 0\]
Hence, option A is the correct answer.