If `x ^(2)+bx+b` is a factor of `x ^(3) +2x ^(2) +2x+c (c ne 0),` then `b-c` is :

Options

A. 2
B. `-1`
C. 0
D. `-2`

Answer

To solve the problem, we need to determine the values of \( b \) and \( c \) given that \( x^2 + bx + b \) is a factor of \( x^3 + 2x^2 + 2x + c \) (where \( c \neq 0 \)). We will do this by performing polynomial long division and finding the conditions under which the remainder is zero.

### Step-by-Step Solution:

1. **Set Up the Division:**
We need to divide the cubic polynomial \( x^3 + 2x^2 + 2x + c \) by the quadratic polynomial \( x^2 + bx + b \).

2. **Perform the First Division:**
- Divide the leading term of the cubic \( x^3 \) by the leading term of the quadratic \( x^2 \) to get \( x \).
- Multiply \( x \) by \( x^2 + bx + b \):
\[ x(x^2 + bx + b) = x^3 + bx^2 + bx \]
- Subtract this from the original cubic:
\[ (x^3 + 2x^2 + 2x + c) - (x^3 + bx^2 + bx) = (2 - b)x^2 + (2 - b)x + c \]

3. **Perform the Second Division:**
- Now, divide the leading term \( (2 - b)x^2 \) by \( x^2 \) to get \( 2 - b \).
- Multiply \( 2 - b \) by \( x^2 + bx + b \):
\[ (2 - b)(x^2 + bx + b) = (2 - b)x^2 + (2 - b)bx + (2 - b)b \]
- Subtract this from the previous result:
\[ ((2 - b)x^2 + (2 - b)x + c) - ((2 - b)x^2 + (2 - b)bx + (2 - b)b) = (2 - b - (2 - b)b)x + (c - (2 - b)b) \]
- This simplifies to:
\[ (2 - b)(1 - b)x + (c - (2 - b)b) \]

4. **Set the Remainder to Zero:**
For \( x^2 + bx + b \) to be a factor, the remainder must equal zero:
- The coefficient of \( x \) must be zero:
\[ (2 - b)(1 - b) = 0 \]
- This gives us two cases:
- Case 1: \( 2 - b = 0 \) → \( b = 2 \)
- Case 2: \( 1 - b = 0 \) → \( b = 1 \)

5. **Find Corresponding \( c \):**
- For \( b = 2 \):
\[ c - (2 - 2) \cdot 2 = 0 \implies c = 0 \quad (\text{not valid since } c \neq 0) \]
- For \( b = 1 \):
\[ c - (2 - 1) \cdot 1 = 0 \implies c - 1 = 0 \implies c = 1 \]

6. **Calculate \( b - c \):**
Now we have \( b = 1 \) and \( c = 1 \):
\[ b - c = 1 - 1 = 0 \]

### Final Answer:
Thus, the value of \( b - c \) is \( 0 \).