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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 [tex]=[/tex] length [tex]\times[/tex] 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]


Sagot :

To find the width of the rectangle, given the area and the length, we can use the formula:

[tex]\[ \text{Area} = \text{Length} \times \text{Width} \][/tex]

Firstly, let's write down the expressions for the area and length of the rectangle:

- Area: [tex]\( A(x) = x^4 + 4x^3 + 3x^2 - 4x - 4 \)[/tex]
- Length: [tex]\( L(x) = x^3 + 5x^2 + 8x + 4 \)[/tex]

Using the formula for the area of a rectangle, we have:

[tex]\[ \text{Width} = \frac{\text{Area}}{\text{Length}} \][/tex]

That is:

[tex]\[ W(x) = \frac{x^4 + 4x^3 + 3x^2 - 4x - 4}{x^3 + 5x^2 + 8x + 4} \][/tex]

When we simplify this rational expression by performing polynomial division, we determine if there is a quotient and remainder. After simplifying, we find that:

[tex]\[ W(x) = x - 1 \][/tex]

Therefore, the width of the rectangle is:

[tex]\[ \boxed{x - 1} \][/tex]