We are tasked with finding the values of $a$ and $b$ such that the polynomial

$P(x) = x^4 + ax^3 - 7x^2 + 8x + b$

is divisible by both $x + 2$ and $x + 3$. To solve this, we will use the Remainder Theorem, which states that if a polynomial $P(x)$ is divisible by a factor like $x + c$, then $P(-c) = 0$.

Given that the polynomial is divisible by $x + 2$ and $x + 3$, we have:

$P(-2) = 0 \quad \text{and} \quad P(-3) = 0$

Step 1: Apply the condition $P(-2) = 0$

Substitute $x = -2$ into $P(x)$:

$(-2)^4 + a(-2)^3 - 7(-2)^2 + 8(-2) + b = 0$

This simplifies to:

$16 - 8a - 28 - 16 + b = 0$

$-28 - 8a + b = 0 \quad \Rightarrow \quad b = 8a + 28 \quad \text{(Equation 1)}$

Step 2: Apply the condition $P(-3) = 0$

Substitute $x = -3$ into $P(x)$:

$(-3)^4 + a(-3)^3 - 7(-3)^2 + 8(-3) + b = 0$

This simplifies to:

$81 - 27a - 63 - 24 + b = 0$

$-6 - 27a + b = 0 \quad \Rightarrow \quad b = 27a + 6 \quad \text{(Equation 2)}$

Step 3: Solve the system of equations

From Equation 1 and Equation 2, we now have the system:

$b = 8a + 28 \quad \text{(Equation 1)}$ $b = 27a + 6 \quad \text{(Equation 2)}$

Set these two expressions for $b$ equal to each other:

$8a + 28 = 27a + 6$

Solve for $a$:

$28 - 6 = 27a - 8a$ $22 = 19a$ $a = \frac{22}{19}$

Step 4: Solve for $b$

Now substitute $a = \frac{22}{19}$ into either Equation 1 or Equation 2. Using Equation 1:

$b = 8\left(\frac{22}{19}\right) + 28$ $b = \frac{176}{19} + \frac{532}{19} = \frac{708}{19}$

Thus, the values of $a$ and $b$ are:

$a = \frac{22}{19}, \quad b = \frac{708}{19}$

Would you like further details or explanations?

Here are 5 related questions:

1. How do you apply the remainder theorem in polynomial division?
2. What is the factor theorem, and how is it different from the remainder theorem?
3. How can synthetic division help verify if a polynomial is divisible by a binomial?
4. What are the steps to solve a system of linear equations with fractions?
5. How does polynomial long division differ from factoring polynomials?

Tip: When solving for unknowns in polynomials, always use the remainder or factor theorem for efficient problem-solving.