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

Step 1: The roots of $x^2 + x + 1 = 0$ are $\omega$ and $\omega^2$, where $\omega^3 = 1$. Since $P(x)$ is divisible by $x^2 + x + 1$, we must have $P(\omega) = 0$ and $P(\omega^2) = 0$.
Step 2: Substitute $x = \omega$: $P(\omega) = f(\omega^3) + \omega g(\omega^3) = f(1) + \omega g(1) = 0$.
Step 3: Substitute $x = \omega^2$: $P(\omega^2) = f((\omega^2)^3) + \omega^2 g((\omega^2)^3) = f(1) + \omega^2 g(1) = 0$.
Step 4: Subtracting the two equations: $(\omega - \omega^2)g(1) = 0$. Since $\omega \neq \omega^2$, $g(1) = 0$. Substituting $g(1) = 0$ back gives $f(1) = 0$.
Step 5: $P(1) = f(1^3) + 1 \cdot g(1^3) = f(1) + g(1) = 0 + 0 = 0$.