If `(5x^(2)+14x +2)^(2)-(4x^(2) -5x +7)^(2)` is divided by `x^(2)+x+1`, then what is the remainder?

Options

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

Answer

To solve the problem, we need to find the remainder when the expression \((5x^2 + 14x + 2)^2 - (4x^2 - 5x + 7)^2\) is divided by \(x^2 + x + 1\).

### Step-by-Step Solution:

1. **Identify the Expression**:
We have the expression:
\[ (5x^2 + 14x + 2)^2 - (4x^2 - 5x + 7)^2 \]

2. **Use the Difference of Squares Formula**:
Recall that \(a^2 - b^2 = (a + b)(a - b)\).
Let \(a = 5x^2 + 14x + 2\) and \(b = 4x^2 - 5x + 7\).
Then, we can rewrite the expression as:
\[ (a + b)(a - b) \]

3. **Calculate \(a + b\)**:
\[ a + b = (5x^2 + 14x + 2) + (4x^2 - 5x + 7) = (5x^2 + 4x^2) + (14x - 5x) + (2 + 7) = 9x^2 + 9x + 9 = 9(x^2 + x + 1) \]

4. **Calculate \(a - b\)**:
\[ a - b = (5x^2 + 14x + 2) - (4x^2 - 5x + 7) = (5x^2 - 4x^2) + (14x + 5x) + (2 - 7) = x^2 + 19x - 5 \]

5. **Combine the Results**:
Now we have:
\[ (a + b)(a - b) = 9(x^2 + x + 1)(x^2 + 19x - 5) \]

6. **Dividing by \(x^2 + x + 1\)**:
When we divide \(9(x^2 + x + 1)(x^2 + 19x - 5)\) by \(x^2 + x + 1\), the \(x^2 + x + 1\) terms cancel out, leaving us with:
\[ 9(x^2 + 19x - 5) \]

7. **Finding the Remainder**:
Since \(9(x^2 + 19x - 5)\) is a polynomial of degree 2, we need to find the remainder when this polynomial is divided by \(x^2 + x + 1\).
However, since \(x^2 + x + 1\) is a factor of the original expression, the remainder will be \(0\).

### Final Answer:
The remainder when \((5x^2 + 14x + 2)^2 - (4x^2 - 5x + 7)^2\) is divided by \(x^2 + x + 1\) is:
\[ \text{Remainder} = 0 \]