Question

Consider \(f(x)=x^{4+x^{3}+x^{2}+x+1\), where \(x\) is a positive integer greater than \(1\). What will be the remainder if \(f(x^{5})\) is divided by \(f(x)\)?}

• A. 5
• B. 6
• C. 7
• D. a monomial in \(x\)
• E. a polynomial in \(x\)

Hint:

Using roots: if \(g(x)\) is divided by \(f(x)\), the remainder \(R(x)\) satisfies \(R(\alpha)=g(\alpha)\) for each root \(\alpha\) of \(f\). Constant values at enough points force a constant remainder.

Answer 1

The Correct Option is A

Step 1: Observe \(f(x)=\dfrac{x^{5}-1}{x-1}\). The roots of \(f(x)\) are the non-trivial 5th roots of unity \(\omega,\omega^{2},\omega^{3},\omega^{4}\) with \(\omega^{5}=1\) and \(\omega\neq1\). Step 2: Let the remainder upon division be \(R(x)\) with \(\deg R\le4\). For any root \(\omega\) of \(f(x)\), \[ R(\omega)=f(\omega^{5})=f(1)=1+1+1+1+1=5. \] So \(R(x)\) takes the constant value \(5\) at four distinct points; hence \(R(x)\equiv 5\).
Therefore the remainder is \(\boxed{5}\).