Question

The polynomial $ x+1 $ is a factor of $ {{x}^{n}}+1 $ only if \[\]
 A $ n $ is an odd integer \[\]
 B. $ n $ is an even integer \[\]
C. $ n $ is a negative integer \[\]
 D. $ n $ is a positive integer\[\]

Answer 1

Hint: We recall the Euclidean division of polynomial $ p\left( x \right)=q\left( x \right)d\left( x \right)+r\left( x \right) $ where the divisor polynomial $ d\left( x \right) $ can be a factor if and only if $ r\left( x \right)=0 $ . We take $ p\left( x \right)={{x}^{n}}+1 $ and $ d\left( x \right)=x+1 $ d and have $ p\left( x \right)=\left( x+1 \right)q\left( x \right) $ . We put $ x=-1 $ to find $ p\left( -1 \right)=0 $ . We find for what values of $ n $ , $ p\left( -1 \right)={{\left( -1 \right)}^{n}}+1=0 $ \[\]

Complete step by step answer:
We know that when we divide a divided polynomial $ p\left( x \right) $ with degree $ n $ by some divisor polynomial $ d\left( x \right) $ with degree $ m\le n $ then we get the quotient polynomial $ q\left( x \right) $ of degree $ n-m $ and the remainder polynomial as $ r\left( x \right) $ of degree either equal to $ m $ or $ m-1 $ .We use Euclidean division of polynomial and have;
\[p\left( x \right)=q\left( x \right)d\left( x \right)+r\left( x \right)\]
If the remainder polynomial $ r\left( x \right)=0 $ then $ d\left( x \right) $ becomes a factor of $ \left( x \right) $ . Let us assume $ p\left( x \right)={{x}^{n}}+1 $ and $ d\left( x \right)=x+1 $ . Since we are given $ d\left( x \right)=x+1 $ is factor of $ p\left( x \right)={{x}^{n}}+1 $ then we have $ r\left( x \right)=0 $ and ;
\[\begin{align}
  & p\left( x \right)=q\left( x \right)d\left( x \right) \\
 & \Rightarrow p\left( x \right)=\left( x+1 \right)q\left( x \right) \\
\end{align}\]
Let us put $ x=-1 $ in the above step to have;
\[\Rightarrow p\left( x \right)=0\cdot q\left( x \right)=0\]
So $ x=-1 $ is a zero of the polynomial $ p\left( x \right)={{x}^{n}}+1 $ . So we have;
 $ p\left( -1 \right)={{\left( -1 \right)}^{n}}+1=0 $
We see in above equation that the only way we can make $ p\left( -1 \right)={{\left( -1 \right)}^{n}}+1=0 $ when $ {{\left( -1 \right)}^{n}}=-1. $ We know that $ {{\left( -1 \right)}^{x}}=1 $ if $ n $ is even and $ {{\left( -1 \right)}^{n}}=-1 $ if $ n $ is odd. So $ n $ has to be odd and hence the correct option is A.\[\]
If we take $ n $ as even we get $ p\left( -1 \right)=1+1=2\ne 0 $ and hence option B is incorrect. Since $ n $ cannot be negative and positive integers include both even and odd numbers, option C and D are not correct. So only correct option is A. \[\]

Note:
 We can alternatively conclude $ x=-1 $ is a zero of $ p\left( x \right) $ from the factor theorem which states that a polynomial $ p\left( x \right) $ has a factor $ \left( x-a \right) $ if and only if $ p\left( a \right)=0 $ in other words $ a $ is a zero of $ p\left( x \right) $ . We can alternatively solve using the algebraic identity for odd $ n $ $ {{a}^{n}}+{{b}^{n}}=\left( a-b \right)\left( {{a}^{n-1}}-{{a}^{n-2}}b+...-a{{b}^{n-2}}+{{b}^{n-1}} \right) $ with $ a=x,b=1 $ . We shall find that $ \left( {{1}^{n}}+{{x}^{n}} \right)=\left( 1+x \right)\left( 1-x+{{x}^{2}}+...+{{x}^{n-1}} \right) $ . Since $ 1-x+{{x}^{2}}+...+{{x}^{n-1}}\ne 0 $ when $ x=-1 $ that makes $ x+1 $ a factor of $ {{x}^{n}}+1 $ with odd $ n $ .