Question

If $ {{x}^{n}}-1 $ is divisible by $ x-k $ then least positive integral value of $ k $ is \[\]
A.1 \[\]
B.2 \[\]
C.3 \[\]
D,4 \[\]

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-k $ d and have $ p\left( x \right)=\left( x-k \right)q\left( x \right) $ . We put $ x=k $ to find $ p\left( k \right)=0 $ . We find for what values of $ k $ , $ p\left( k \right)=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 factors of $ \left( x \right) $ . Let us assume $ p\left( x \right)={{x}^{n}}-1 $ and $ d\left( x \right)=x-k $ . Since we are given $ d\left( x \right)=x-k $ is factor of $ p\left( x \right)={{x}^{n}}-1 $ then we have $ r\left( x \right)=0 $ and ;
\[\begin{align}
  & p\left( x \right)=q\left( x \right)d\left( x \right) \\
 & \Rightarrow p\left( x \right)=\left( x-k \right)q\left( x \right) \\
\end{align}\]
Let us put $ x=k $ in the above step to have;
\[\Rightarrow p\left( x \right)=0\cdot q\left( x \right)=0\]
So $ x=k $ is a zero of the polynomial $ p\left( x \right)={{x}^{n}}+1 $ . So we have;
 $ \begin{align}
  & p\left( k \right)={{k}^{n}}-1=0 \\
 & \Rightarrow {{k}^{n}}=1 \\
\end{align} $
 The solutions of the above equations are $ n, $ $ {{n}^{\text{th}}} $ roots of unity. The only positive integral solution is 1 and hence the correct option is A. \[\]


Note:
We note that zeroes or roots of a polynomial are the value of $ x $ for which the polynomial returned zero. If the degree of the polynomial is $ n $ , then there are $ n $ zeroes may be real or complex. The roots of the polynomial $ {{x}^{n}}-1 $ are given by $ {{e}^{i\dfrac{\pi }{n}}}=\cos \dfrac{\pi }{n}+i\sin \dfrac{\pi }{n} $ for $ n=1,2,3...n $ . 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 $ {{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=1,b=x $ . 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 $ , $ x-1 $ is a factor of $ {{x}^{n}}-1 $ and hence $ x=1 $ is zero of the polynomial.