Question

The number of distinct solutions of the equation \[ x^{11}-x^7+x^4-1=0 \] is

• A. \(9\)
• B. \(11\)
• C. \(10\)
• D. \(8\)

Hint:

When a polynomial is written as a product of factors, count the roots of each factor and subtract the common roots to get the number of distinct solutions.

Answer 1

The Correct Option is C

Step 1: Factorize the given equation.
Given equation is \[ x^{11}-x^7+x^4-1=0 \] Group the terms as \[ (x^{11}-x^7)+(x^4-1)=0 \] Taking common factor \(x^7\) from the first group, \[ x^7(x^4-1)+(x^4-1)=0 \] Now take \((x^4-1)\) common, \[ (x^4-1)(x^7+1)=0 \]

Step 2: Count distinct roots of \(x^4-1=0\).
From \[ x^4-1=0 \] we get \[ x^4=1 \] This equation has \(4\) distinct roots.

Step 3: Count distinct roots of \(x^7+1=0\).
From \[ x^7+1=0 \] we get \[ x^7=-1 \] This equation has \(7\) distinct roots.

Step 4: Check common roots.
The root \(x=-1\) satisfies both \[ x^4-1=0 \] and \[ x^7+1=0 \] So, one root is common and must be counted only once.
Hence, total number of distinct solutions is \[ 4+7-1=10 \]

Step 5: Final conclusion.
Therefore, \[ \boxed{10} \]