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, start superscript, 3, end superscript, plus, 20, x, start superscript, 2, end superscript. 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, start superscript, 3, end superscript, and 20x^220x 2 20, x, start superscript, 2, end superscript. What is the length and width of the rectangle? \text{Width} = Width=W, i, d, t, h, equals \text{Length} = Length=L, e, n, g, t, h, equals

Answer:The width is: [tex]w(x)=4x^2[/tex]The length is [tex]l(x)=2x^2+3x+5[/tex]Step-by-step explanation:The given rectangle has area given algebraically by the function:[tex]a(x)=8x^5+12x^3+20x^2[/tex]The width of the rectangle is the greatest common factor of [tex]8x^5[/tex],  [tex]12x^3[/tex] and [tex]20x^2[/tex]That is the width is: [tex]w(x)=4x^2[/tex]We now divide the area by the width to obtain the length of the rectangle:[tex]l(x)=\frac{8x^5+12x^3+20x^2}{4x^2}[/tex]This simplifies to:[tex]l(x)=\frac{8x^5}{4x^2}+\frac{12x^3}{4x^2}+\frac{20x^2}{4x^2}[/tex][tex]l(x)=2x^2+3x+5[/tex]