The rectangle below has an area of 8x^5+12x^3+20x^28x 5 +12x 3 +20x 2 8, x, start superscript, 5, end superscript, plus, 12, x, cubed, plus, 20, x, squared. The width of the rectangle is equal to the greatest common monomial factor of 8x^58x 5 8, x, start superscript, 5, end superscript, 12x^312x 3 12, x, cubed, and 20x^220x 2 20, x, squared.

The question is incomplete. Here is the complete question.

The rectangle bleow has an area of [tex]8x^{5}+12x^{3}+20x^{2}[/tex]. The width of the rectangle is equal to the greatest common monomial factor of [tex]8x^{5},12x^{3},20x^{2}[/tex]. What is the length and width of the rectangle?

Answer: width = [tex]4x^{2}[/tex]

              length = [tex]2x^{3}+3x+5[/tex]

Step-by-step explanation: Greatest common factor is the largest number that will divide into that number without rest, i.e., it's a number that will result in an exact division. The same can be applied to a polynomial.

To find the greatest common factor:

1) Write each in prime factored form:

2.2.2.x.x.x.x.x + 2.2.3.x.x.x + 2.2.5.x.x

2) Identify the common factor among the terms:

For this polynomial, the repetitive factor is [tex]4x^{2}[/tex]

Therefore, the width of the rectangle is:

w = [tex]4x^{2}[/tex]

Area of a rectangle is the multiplication of width and length, so:

[tex]A=w*l\\l=\frac{A}{w}[/tex]

To calculate length, we will have to divide polynomials:

[tex]l=\frac{8x^{5}+12x^{3}+20x^{2}}{4x^{2}}[/tex]

[tex]l = 2x^{3}+3x+5[/tex]

Width and length of the rectangle are 4x² and [tex]2x^{3}+3x+5[/tex], respectively.

Heeeeeeeeeeeeeeeeeeeelp asap due today at 7

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