The area of a rectangle is [tex]\left(x^4 + 4x^3 + 3x^2 - 4x - 4\right)[/tex], and the length of the rectangle is [tex]\left(x^3 + 5x^2 + 8x + 4\right)[/tex]. If area = length × width, what is the width of the rectangle?

A. [tex]x + 1[/tex]
B. [tex]x - 9[/tex]
C. [tex]x + 4[/tex]
D. [tex]x + 1[/tex]



Answer :

To find the width of the rectangle given the area and the length, consider the relationship:
[tex]\[ \text{Area} = \text{Length} \times \text{Width} \][/tex]

Let’s denote the area of the rectangle by [tex]\( A \)[/tex], the length by [tex]\( L \)[/tex], and the width by [tex]\( W \)[/tex]:
[tex]\[ A = L \times W \][/tex]

Given the area [tex]\( A \)[/tex] is:
[tex]\[ A = x^4 + 4x^3 + 3x^2 - 4x - 4 \][/tex]

And the length [tex]\( L \)[/tex] is:
[tex]\[ L = x^3 + 5x^2 + 8x + 4 \][/tex]

We can find the width [tex]\( W \)[/tex] by rearranging the relationship to solve for [tex]\( W \)[/tex]:
[tex]\[ W = \frac{A}{L} \][/tex]

Substitute the given expressions for [tex]\( A \)[/tex] and [tex]\( L \)[/tex] into the equation:
[tex]\[ W = \frac{x^4 + 4x^3 + 3x^2 - 4x - 4}{x^3 + 5x^2 + 8x + 4} \][/tex]

Simplify this expression to find [tex]\( W \)[/tex]. After performing the polynomial division, we obtain:
[tex]\[ W = x - 1 \][/tex]

Hence, the width of the rectangle is:
[tex]\[ \boxed{x - 1} \][/tex]