Question

If f(x) and g(x) are two polynomials such that the polynomial P(x) = f(x³) + x · g(x³) is divisible by x² + x + 1, then P(1) is equal to ________.

Hint:

Whenever a polynomial is divisible by $x^2+x+1$, use the properties of cube roots of unity ($\omega^3=1$ and $1+\omega+\omega^2=0$) to simplify.

Answer 1

Correct Answer: 0

To determine P(1), we know P(x) is divisible by x² + x + 1. This implies that the roots of x² + x + 1, namely the cube roots of unity other than 1, must also be roots of P(x)ω and ω² where ω = e^2πi/3. Therefore, we have:

P(ω) = f(ω³) + ω · g(ω³) = f(1) + ω · g(1) = 0

P(ω²) = f(ω⁶) + ω² · g(ω⁶) = f(1) + ω² · g(1) = 0 

The system of linear equations becomes:

1. f(1) + ω · g(1) = 0

2. f(1) + ω² · g(1) = 0

To solve for f(1) and g(1), subtract (2) from (1):

ω · g(1) - ω² · g(1) = 0

(ω - ω²) · g(1) = 0

Since ω ≠ ω², it follows that g(1) = 0. Substitute into (1):

f(1) + ω · 0 = 0

f(1) = 0

Thus, P(1) = f(1) + 1 · g(1) = 0 + 1 · 0 = 0.

Finally, verify that 0 falls within the given range of 0,0. Therefore, P(1) = 0 is validated.