
Find the remainder when $ \left( {{x^{101}}\, + \,101} \right) $ is divided by $ \left( {x\, + \,1} \right) $
Answer
494.1k+ views
Hint: The remainder is the polynomial "left over" after dividing one polynomial by another in polynomial algebra. When a dividend and a divisor are given, the modulo operation produces such a remainder. A remainder is also what remains after subtracting one number from another, though this is more precisely referred to as the difference.
Complete step-by-step answer:
Remainder theorem states that, when the polynomial $ {{p}}\left( {{x}} \right) $ is divided, by a degree of one or more than one, the linear polynomial $ {{x - a}} $ , $ {{p}}\left( {{a}} \right) $ is the remainder obtained, where $ {{a}} $ represents a real number.
Let us consider the given polynomial.
Let $ {{p}}\left( {{x}} \right)\,\,{{ = }}\,\,{{{x}}^{{{101}}}}\,\,{{ + }}\,\,{{101}} $
If we divide $ {{p}}\left( {{x}} \right) $ by $ \left( {{{x}}\,{{ + }}\,{{1}}} \right) $ , then $ {{p}}\left( {{{ - 1}}} \right) $ is the remainder obtained.
Now, let us substitute the value $ - 1 $ in place of $ {{x}} $ . Hence, we obtain
$ {{p}}\left( {{{ - 1}}} \right)\,\,{{ = }}\,\,{\left( {{{ - 1}}} \right)^{{{101}}}}\,{{ + }}\,{{101}} $
On further solving we get,
$ {{p}}\left( {{{ - 1}}} \right)\,\,{{ = }}\,\,\left( {{{ - 1}}} \right)\,{{ + }}\,{{101}} $
Solving the above equation, we get,
$ {{p}}\left( {{{ - 1}}} \right)\,\,{{ = }}\,100 $
Hence, the remainder when $ \left( {{x^{101}}\, + \,101} \right) $ is divided by $ \left( {x\, + \,1} \right) $ is $ 100 $
So, the correct answer is “100”.
Note: The Remainder Theorem is a polynomial division technique based on Euclidean division. If we divide a polynomial $ {{p}}\left( {{x}} \right) $ by a factor $ {{x - a}} $ , we get a smaller polynomial and a remainder, according to this theorem. This remainder is actually a $ {{p}}\left( {{x}} \right) $ value at $ {{x}}\,{{ = }}\,{{a}} $ , precisely $ {{p(a)}} $ . So, if and only if $ {{p}}\left( {{a}} \right)\,\,{{ = }}\,\,{{0}} $ , $ {{x - a}} $ is the divisor of $ {{p}}\left( {{x}} \right) $ . It is used to factor polynomials of different degrees in an elegant way.
Complete step-by-step answer:
Remainder theorem states that, when the polynomial $ {{p}}\left( {{x}} \right) $ is divided, by a degree of one or more than one, the linear polynomial $ {{x - a}} $ , $ {{p}}\left( {{a}} \right) $ is the remainder obtained, where $ {{a}} $ represents a real number.
Let us consider the given polynomial.
Let $ {{p}}\left( {{x}} \right)\,\,{{ = }}\,\,{{{x}}^{{{101}}}}\,\,{{ + }}\,\,{{101}} $
If we divide $ {{p}}\left( {{x}} \right) $ by $ \left( {{{x}}\,{{ + }}\,{{1}}} \right) $ , then $ {{p}}\left( {{{ - 1}}} \right) $ is the remainder obtained.
Now, let us substitute the value $ - 1 $ in place of $ {{x}} $ . Hence, we obtain
$ {{p}}\left( {{{ - 1}}} \right)\,\,{{ = }}\,\,{\left( {{{ - 1}}} \right)^{{{101}}}}\,{{ + }}\,{{101}} $
On further solving we get,
$ {{p}}\left( {{{ - 1}}} \right)\,\,{{ = }}\,\,\left( {{{ - 1}}} \right)\,{{ + }}\,{{101}} $
Solving the above equation, we get,
$ {{p}}\left( {{{ - 1}}} \right)\,\,{{ = }}\,100 $
Hence, the remainder when $ \left( {{x^{101}}\, + \,101} \right) $ is divided by $ \left( {x\, + \,1} \right) $ is $ 100 $
So, the correct answer is “100”.
Note: The Remainder Theorem is a polynomial division technique based on Euclidean division. If we divide a polynomial $ {{p}}\left( {{x}} \right) $ by a factor $ {{x - a}} $ , we get a smaller polynomial and a remainder, according to this theorem. This remainder is actually a $ {{p}}\left( {{x}} \right) $ value at $ {{x}}\,{{ = }}\,{{a}} $ , precisely $ {{p(a)}} $ . So, if and only if $ {{p}}\left( {{a}} \right)\,\,{{ = }}\,\,{{0}} $ , $ {{x - a}} $ is the divisor of $ {{p}}\left( {{x}} \right) $ . It is used to factor polynomials of different degrees in an elegant way.
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