Given Q(x) = ax^3 - bx^2 - cx + 6. If (x + 2) is a factor of Q(x) and Q(x) passes through the points (-1,10) and (2,4), find the values of a, b and c. Hence, factorise Q(x) completely

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.

Learn how to factorize the cubic polynomial Q(x) = 2x^3 - x^2 - 7x + 6 using the Factor Theorem and solve a system of equations. This problem involves finding coefficients a, b, and c by using points and factors provided. Suitable for advanced high school students (Grades 10-12).

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