Understanding the Factor Theorem

The problem states that $x^2 - 1$ is a factor of the polynomial $P(x) = ax^4 + bx^3 + cx^2 + dx + e$.

We know that $x^2 - 1$ can be factored as $(x - 1)(x + 1)$.

According to the Factor Theorem, if a polynomial $(x - k)$ is a factor of $P(x)$, then $P(k) = 0$. Since both $(x - 1)$ and $(x + 1)$ are factors of $P(x)$, we must have $P(1) = 0$ and $P(-1) = 0$.

Applying the Factor Theorem

1. Calculate P(1):

   Substitute $x = 1$ into the polynomial:

   $P(1) = a(1)^4 + b(1)^3 + c(1)^2 + d(1) + e$

   $P(1) = a + b + c + d + e$

   Since $P(1) = 0$, we get the first equation:

   $a + b + c + d + e = 0 \quad (1)$

2. Calculate P(-1):

   Substitute $x = -1$ into the polynomial:

   $P(-1) = a(-1)^4 + b(-1)^3 + c(-1)^2 + d(-1) + e$

   $P(-1) = a(1) + b(-1) + c(1) + d(-1) + e$

   $P(-1) = a - b + c - d + e$

   Since $P(-1) = 0$, we get the second equation:

   $a - b + c - d + e = 0 \quad (2)$

Deriving the Relation Between Coefficients

1. Add Equation (1) and Equation (2):

   $(a + b + c + d + e) + (a - b + c - d + e) = 0 + 0$

   $2a + 2c + 2e = 0$

   Divide by 2:

   $a + c + e = 0 \quad (3)$

2. Subtract Equation (2) from Equation (1):

   $(a + b + c + d + e) - (a - b + c - d + e) = 0 - 0$

   $a + b + c + d + e - a + b - c + d - e = 0$

   $2b + 2d = 0$

   Divide by 2:

   $b + d = 0 \quad (4)$

3. Combine results:

   From equation (3), we have $a + c + e = 0$.

   From equation (4), we have $b + d = 0$.

   We need to find which option satisfies these conditions. Let's check Option 1: $a + c + e = b + d$.

   Substituting our results ($a + c + e = 0$ and $b + d = 0$), we get $0 = 0$. This is true.

Conclusion

The relation $a + c + e = b + d$ holds true because both sides are equal to zero, derived directly from the condition that $x^2 - 1$ is a factor of the given polynomial.