If ${x^{51}} + 51$ is divided by $x + 1$, Then the remainder is
$
A)0 \\
B)1 \\
C)49 \\
D)50 \\
$
Last updated date: 22nd Mar 2023
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Total views: 306k
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Views today: 2.85k
Answer
306k+ views
Hint: Approach the solution by using remainder theorem concept.
Remainder theorem
If any polynomial f(x) is divided by f(x-h), then the remainder will be f (h)
Here given polynomial is ${x^{51}} + 51$ and divided by $x + 1$
So by using the remainder theorem we can say that
Remainder=$f( - 1)$ since the divisor is $x + 1$
Therefore we can write it as
$ \Rightarrow f(x) = {x^{51}} + 51 \\
\Rightarrow f( - 1) = {( - 1)^{51}} + 51 \\
\Rightarrow f( - 1) = - 1 + 51 \\
\therefore f( - 1) = 50 \\
\\
$
Hence the remainder is $50$
Note: Focus on the divisor of the polynomial which helps to give the remainder value.
Remainder theorem only works when a function is divided by a linear polynomial, which is of the form x + number or x - number.
Remainder theorem
If any polynomial f(x) is divided by f(x-h), then the remainder will be f (h)
Here given polynomial is ${x^{51}} + 51$ and divided by $x + 1$
So by using the remainder theorem we can say that
Remainder=$f( - 1)$ since the divisor is $x + 1$
Therefore we can write it as
$ \Rightarrow f(x) = {x^{51}} + 51 \\
\Rightarrow f( - 1) = {( - 1)^{51}} + 51 \\
\Rightarrow f( - 1) = - 1 + 51 \\
\therefore f( - 1) = 50 \\
\\
$
Hence the remainder is $50$
Note: Focus on the divisor of the polynomial which helps to give the remainder value.
Remainder theorem only works when a function is divided by a linear polynomial, which is of the form x + number or x - number.
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