Question

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?.

Answer 1

The width of the rectangle, for which area and length provided in polynomial equation form is,

$(x^43x^3+8x^2+4x)$

What is the area of a rectangle?

Area of a rectangle is the product of the length of the rectangle and the width of the rectangle. It can be given as,

$A=a\times b$

Here, (a)is the length of rectangle and (b) is the width of the rectangle

The area of the given rectangle is,

$A=(x^4 +4x^3 +3x^2 - 4x - 4),$

The length of the given rectangle is,

$L=x^3+ 5x^2 +8x+ 4).$

As, the area of a rectangle is the product of the length and the width of the rectangle. Let suppose the width of the rectangle is f(x). Therefore,

$(x^4 +4x^3 +3x^2 - 4x - 4)=(x^3+ 5x^2 +8x+ 4)\times f(x)$

Solve it further as,

$f(x)=(x^4 +4x^3 +3x^2 - 4x - 4)-(x^3+ 5x^2 +8x+ 4) \\f(x)=(x^4 +4x^3 +3x^2 - 4x - 4-x^3+ 5x^2 +8x+ 4$

Separate the like terms with same power of variable as,

$f(x)=x^4 +4x^3-x^3 +3x^2 + 5x^2- 4x +8x- 4 + 4\\f(x)=x^43x^3+8x^2+4x$

Hence, the width of the rectangle, for which area and length provided in polynomial equation form is,

$(x^43x^3+8x^2+4x)$

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