Let the width of the classroom be \( w \) feet. The length of the classroom, being six feet shorter than the width, can be expressed as \( w - 6 \) feet.
The original area, \( A \), of the classroom can be calculated using the formula for area of a rectangle:
\[ A = \text{length} \times \text{width} = (w - 6) \times w = w^2 - 6w \]
Now, if Mrs. Lordes increases both dimensions by 2 feet, the new width becomes \( w + 2 \) and the new length becomes \( (w - 6) + 2 = w - 4 \).
The new area \( N \) is then:
\[ N = (w - 4)(w + 2) \]
We can expand this equation:
\[ N = w^2 + 2w - 4w - 8 = w^2 - 2w - 8 \]
So the new area \( N \) can be represented by the equation:
\[ N = w^2 - 2w - 8 \]
Thus, the correct answer is:
D. \( N = w^2 - 2w - 8 \)