Question

If $f(x)$ and $g(x)$ are two polynomials such that the polynomial $P(x) = f(x^3) + x g(x^3)$ is divisible by $(x^2+x+1)$ then $P(1)$ is equal to ______.

Answer 1

0

Answer 2

To solve the problem, we use the property that if a polynomial $P(x)$ is divisible by another polynomial $Q(x)$, then all roots of $Q(x)$ are also roots of $P(x)$.

The polynomial $P(x) = f(x^3) + x g(x^3)$ is divisible by $(x^2+x+1)$. The roots of the quadratic equation $x^2+x+1=0$ are the non-real cube roots of unity, denoted by $\omega$ and $\omega^2$. This means that $P(\omega) = 0$ and $P(\omega^2) = 0$.

We know the following properties of cube roots of unity:

1. $\omega^3 = 1$
2. $\omega^2 + \omega + 1 = 0$

Step 1: Evaluate $P(\omega)$

Substitute $x = \omega$ into the expression for $P(x)$:

$P(\omega) = f(\omega^3) + \omega g(\omega^3)$

Since $\omega^3 = 1$, we have:

$P(\omega) = f(1) + \omega g(1)$

As $P(\omega) = 0$, we get our first equation:

$f(1) + \omega g(1) = 0 \quad \text{(Equation 1)}$

Step 2: Evaluate $P(\omega^2)$

Substitute $x = \omega^2$ into the expression for $P(x)$:

$P(\omega^2) = f((\omega^2)^3) + \omega^2 g((\omega^2)^3)$

Since $(\omega^2)^3 = \omega^6 = (\omega^3)^2 = 1^2 = 1$, we have:

$P(\omega^2) = f(1) + \omega^2 g(1)$

As $P(\omega^2) = 0$, we get our second equation:

$f(1) + \omega^2 g(1) = 0 \quad \text{(Equation 2)}$

Step 3: Solve the system of equations for $f(1)$ and $g(1)$

We have the system of equations:

1. $f(1) + \omega g(1) = 0$
2. $f(1) + \omega^2 g(1) = 0$

Subtract Equation 1 from Equation 2:

$(f(1) + \omega^2 g(1)) - (f(1) + \omega g(1)) = 0 - 0$

$\omega^2 g(1) - \omega g(1) = 0$

$g(1)(\omega^2 - \omega) = 0$

Since $\omega$ is a non-real complex number, $\omega^2 \neq \omega$, which means $(\omega^2 - \omega) \neq 0$. For the product $g(1)(\omega^2 - \omega)$ to be zero, $g(1)$ must be zero. So, $g(1) = 0$.

Substitute $g(1) = 0$ back into Equation 1:

$f(1) + \omega (0) = 0$

$f(1) = 0$

Step 4: Calculate $P(1)$

The question asks for the value of $P(1)$. Substitute $x=1$ into the expression for $P(x)$:

$P(1) = f(1^3) + 1 \cdot g(1^3)$

$P(1) = f(1) + g(1)$

Using the values we found: $f(1) = 0$ and $g(1) = 0$.

$P(1) = 0 + 0 = 0$.