Question

If \( f(x) \) and \( g(x) \) are two polynomials such that \( \phi(x) = f(x^3) + xg(x^3) \) is divisible by \( x^2 + x + 1 \), then:

• A. \( \phi(x) \) is divisible by \( (x - 1) \)
• B. none of \( f(x) \) and \( g(x) \) is divisible by \( (x - 1) \)
• C. \( g(x) \) is divisible by \( (x - 1) \) but \( f(x) \) is not divisible by \( (x - 1) \)
• D. \( f(x) \) is divisible by \( (x - 1) \) but \( g(x) \) is not divisible by \( (x - 1) \)

Hint:

Use the property that if a polynomial \( P(x) \) is divisible by \( (x - a) \), then \( P(a) = 0 \). The roots of \( x^2 + x + 1 = 0 \) are crucial here.

Answer 1

The Correct Option is A

Step 1: Finding the roots of \( x^2 + x + 1 = 0 \).
The roots are \( \omega \) and \( \omega^2 \), where \( \omega^3 = 1 \) and \( 1 + \omega + \omega^2 = 0 \).\n\n
Step 2: Applying the roots to \( \phi(x) \).
\n\( \phi(\omega) = f(\omega^3) + \omega g(\omega^3) = f(1) + \omega g(1) = 0 \)\n\( \phi(\omega^2) = f((\omega^2)^3) + \omega^2 g((\omega^2)^3) = f(1) + \omega^2 g(1) = 0 \)\n\n
Step 3: Solving for \( f(1) \) and \( g(1) \).
\nSubtracting the equations yields \( (\omega - \omega^2) g(1) = 0 \Rightarrow g(1) = 0 \).
\nSubstituting \( g(1) = 0 \) gives \( f(1) = 0 \).
\n\n
Step 4: Divisibility by \( (x - 1) \).
\nSince \( f(1) = 0 \), \( f(x) \) is divisible by \( (x - 1) \).
\nSince \( g(1) = 0 \), \( g(x) \) is divisible by \( (x - 1) \).
\n\n
Step 5: Checking divisibility of \( \phi(x) \) by \( (x - 1) \).\n\( \phi(1) = f(1^3) + 1 \cdot g(1^3) = f(1) + g(1) = 0 + 0 = 0 \).\nTherefore, \( \phi(x) \) is divisible by \( (x - 1) \).