Question

If ${{x}^{51}}+51$ is divided by x+1, the remainder is
[a] 0
[b] 1
[c] 49
[d] 50

Answer 1

Hint: Recall remainder theorem. According to the remainder theorem, the remainder when p(x) is divided by x-a is given by p(a). Hence find the value of ${{x}^{51}}+51$ at x =-1. The value is the remainder when ${{x}^{51}}+51$ is divided by x+1

Complete step-by-step answer:
We have $p\left( x \right)={{x}^{51}}+51$
Let r(x) be the remainder when p(x) is divided by x+1. Since r is the remainder, we have $\deg \left( g\left( x \right) \right)>\deg \left( r\left( x \right) \right)$
Hence, we have
$\deg \left( r\left( x \right) \right)<1\Rightarrow \deg \left( r\left( x \right) \right)\le 0$
Hence, r(x) is a constant term.
Now, we have
$p\left( x \right)=\left( x+1 \right)g\left( x \right)+r$, where g(x) is some polynomial in x.
Put x = -1, we get
$p\left( -1 \right)=\left( -1+1 \right)g\left( -1 \right)+r=0\times g\left( -1 \right)+r=r$
Hence, the remainder is equal to p(-1)
Now, we have $p\left( x \right)={{x}^{51}}+51$
Substituting x = -1 in the expression of p(x), we get
$p\left( -1 \right)={{\left( -1 \right)}^{51}}+51$
We know that ${{\left( -1 \right)}^{x}}=\left\{ \begin{matrix}
   1\text{ if }x\text{ is even} \\
   -1\text{ if }x\text{ is odd} \\
\end{matrix} \right.$
Now since 51 is odd, we have ${{\left( -1 \right)}^{51}}=-1$
Hence, we have
$p\left( -1 \right)=-1+51=50$
But p(-1) = r
Hence, we have
$r=50$
Hence the remainder when p(x) is divided by x+1 is 50
Hence option [d] is correct.

Note: [1] Do not explicitly divide p(x) by x+1 and find the remainder as it is going to be very long since the quotient will have approximately 50 terms. Hence the long division method will be very long and cumbersome.