If `A = 6x^(4) + 5x^(3) - 14x^(2) + 2x + 2 and B = 3x^(2) - 2x - 1`, then the remainder when `A divide B` is

Options

A. 5x+2
B. 2x+3
C. 3x+2
D. 4x+3

Explanation

To find the remainder when polynomial \( A \) is divided by polynomial \( B \), we will use polynomial long division. Let's go through the steps systematically.

### Step 1: Set up the division
We have:
- \( A = 6x^4 + 5x^3 - 14x^2 + 2x + 2 \)
- \( B = 3x^2 - 2x - 1 \)

We will divide \( A \) by \( B \).

### Step 2: Divide the leading terms
Divide the leading term of \( A \) (which is \( 6x^4 \)) by the leading term of \( B \) (which is \( 3x^2 \)):
\[ \frac{6x^4}{3x^2} = 2x^2 \]
This is the first term of our quotient.

### Step 3: Multiply and subtract
Now, multiply \( B \) by \( 2x^2 \):
\[ 2x^2 \cdot (3x^2 - 2x - 1) = 6x^4 - 4x^3 - 2x^2 \]
Now, subtract this from \( A \):
\[ (6x^4 + 5x^3 - 14x^2 + 2x + 2) - (6x^4 - 4x^3 - 2x^2) = (5x^3 + 4x^3) + (-14x^2 + 2x^2) + 2 \]
This simplifies to:
\[ 9x^3 - 12x^2 + 2x + 2 \]

### Step 4: Repeat the process
Now, divide the leading term of the new polynomial \( 9x^3 \) by the leading term of \( B \):
\[ \frac{9x^3}{3x^2} = 3x \]
Multiply \( B \) by \( 3x \):
\[ 3x \cdot (3x^2 - 2x - 1) = 9x^3 - 6x^2 - 3x \]
Subtract this from the previous result:
\[ (9x^3 - 12x^2 + 2x + 2) - (9x^3 - 6x^2 - 3x) = (-12x^2 + 6x^2) + (2x + 3x) + 2 \]
This simplifies to:
\[ -6x^2 + 5x + 2 \]

### Step 5: Continue dividing
Now, divide the leading term of the new polynomial \( -6x^2 \) by the leading term of \( B \):
\[ \frac{-6x^2}{3x^2} = -2 \]
Multiply \( B \) by \( -2 \):
\[ -2 \cdot (3x^2 - 2x - 1) = -6x^2 + 4x + 2 \]
Subtract this from the previous result:
\[ (-6x^2 + 5x + 2) - (-6x^2 + 4x + 2) = (5x - 4x) + (2 - 2) \]
This simplifies to:
\[ x \]

### Step 6: Final remainder
Since the degree of the new polynomial \( x \) is less than the degree of \( B \), we stop here. The remainder when \( A \) is divided by \( B \) is:
\[ \text{Remainder} = x \]

### Final Answer
The remainder when \( A \) is divided by \( B \) is \( x \).