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The area of a rectangle is (x4 + 4x3 + 3x2 – 4x – 4), and the length of the rectangle is (x3 + 5x2 + 8x + 4). If area = length × width, what is the width of the rectangle

Respuesta :

thespeedo2346oqvsen
length x width = area

-> width = area / length
= (x^4 + 4x^3 + 3x^2 - 4x - 4) / (x^3 + 5x^2 + 8x +4)
= (x+2)^2*(x-1)(x+1) / (x+2)^2*(x+1)
= x-1

-> width = x-1

Answer:

[tex]x-1[/tex]

Step-by-step explanation:

Given : The area of a rectangle is [tex]x^4 + 4x^3 + 3x^2-4x - 4[/tex].

            Length of rectangle = [tex]x^3 + 5x^2 + 8x + 4[/tex]

To Find: Width of rectangle

Solution:

Area of rectangle = [tex]Length \times width[/tex]

Substituting the values:

[tex]x^4 + 4x^3 + 3x^2-4x - 4=(x^3 + 5x^2 + 8x + 4) \times Width[/tex]

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

[tex]x-1=Width[/tex]

Thus the width of the given rectangle is [tex]x-1[/tex]

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