Given the polynomial $Q(x) = ax^3 - bx^2 - cx + 6$, we are provided with the following information:

1. $(x + 2)$ is a factor of $Q(x)$.
2. $Q(x)$ passes through the points $(-1, 10)$ and $(2, 4)$.

Step 1: Use the fact that $(x + 2)$ is a factor.

Since $(x + 2)$ is a factor of $Q(x)$, by the Factor Theorem, we know: $Q(-2) = 0$ Substitute $x = -2$ into the polynomial: $Q(-2) = a(-2)^3 - b(-2)^2 - c(-2) + 6 = 0$ This simplifies to: $-8a - 4b + 2c + 6 = 0 \quad \text{(Equation 1)}$

Step 2: Use the fact that $Q(-1) = 10$.

Substitute $x = -1$ into the polynomial: $Q(-1) = a(-1)^3 - b(-1)^2 - c(-1) + 6 = 10$ This simplifies to: $-a - b + c + 6 = 10$ $-a - b + c = 4 \quad \text{(Equation 2)}$

Step 3: Use the fact that $Q(2) = 4$.

Substitute $x = 2$ into the polynomial: $Q(2) = a(2)^3 - b(2)^2 - c(2) + 6 = 4$ This simplifies to: $8a - 4b - 2c + 6 = 4$ $8a - 4b - 2c = -2 \quad \text{(Equation 3)}$

Step 4: Solve the system of equations.

We now have the following system of linear equations:

1. $-8a - 4b + 2c + 6 = 0$ $\Rightarrow -8a - 4b + 2c = -6$
2. $-a - b + c = 4$
3. $8a - 4b - 2c = -2$

Step 4.1: Simplify the equations:

Equation 1: $-8a - 4b + 2c = -6$

Dividing by 2: $-4a - 2b + c = -3 \quad \text{(Equation 1 simplified)}$

Equation 3: $8a - 4b - 2c = -2$

Dividing by 2: $4a - 2b - c = -1 \quad \text{(Equation 3 simplified)}$

Step 4.2: Add the simplified Equation 1 and Equation 3:

$(-4a - 2b + c) + (4a - 2b - c) = -3 + (-1)$ Simplifying gives: $-4b = -4 \Rightarrow b = 1$

Step 4.3: Substitute $b = 1$ into Equation 2:

$-a - 1 + c = 4 \Rightarrow -a + c = 5 \quad \text{(Equation 2 simplified)}$ So, $c = a + 5$.

Step 4.4: Substitute $b = 1$ and $c = a + 5$ into the simplified Equation 1:

$-4a - 2(1) + (a + 5) = -3$ Simplifying gives: $-4a - 2 + a + 5 = -3 \Rightarrow -3a + 3 = -3$ This simplifies to: $-3a = -6 \Rightarrow a = 2$

Now, substitute $a = 2$ back into the equations to find $c$: $c = 2 + 5 = 7$

Thus, the values are: $a = 2, \quad b = 1, \quad c = 7$

Step 5: Factorize $Q(x)$ completely.

Substitute $a = 2$, $b = 1$, and $c = 7$ into the original polynomial: $Q(x) = 2x^3 - x^2 - 7x + 6$

Given $(x + 2)$ is a factor: $Q(x) = (x + 2)(2x^2 - 5x + 3)$

Now, factorize the quadratic $2x^2 - 5x + 3$: The quadratic can be factorized as: $2x^2 - 5x + 3 = (2x - 3)(x - 1)$

Thus, the complete factorization of $Q(x)$ is: $Q(x) = (x + 2)(2x - 3)(x - 1)$

Final Answer:

• Values: $a = 2$, $b = 1$, $c = 7$
• Factorization: $Q(x) = (x + 2)(2x - 3)(x - 1)$

Would you like a more detailed explanation on any step? Here are some related questions:

1. How do you apply the Factor Theorem to factorize polynomials?
2. What are the steps to solve a system of linear equations?
3. How can you verify the factorization of a cubic polynomial?
4. What techniques are there for factoring quadratic equations?
5. How do transformations affect the graph of a polynomial function?

Tip: When factorizing polynomials, always check if simpler factors (like binomials) can be factored out first before attempting more complex methods.