Respuesta :
Given:
Area of rectangle = [tex]6n^4+20n^3+14n^2[/tex]
Width of the rectangle is equal to the greatest common monomial factor of [tex]6n^4, 20n^3,14n^2[/tex].
To find:
Length and width of the rectangle.
Solution:
Width of the rectangle is equal to the greatest common monomial factor of [tex]6n^4, 20n^3,14n^2[/tex] is
[tex]6n^4=2\times 3\times n\times n\times n\times n[/tex]
[tex]20n^3=2\times 2\times 5\times n\times n\times n[/tex]
[tex]14n^2=2\times 7\times n\times n[/tex]
Now,
[tex]GCF(6n^4, 20n^3,14n^2)=2\times n\times n=2n^2[/tex]
So, width of the rectangle is [tex]2n^2[/tex].
Area of rectangle is
[tex]Area=6n^4+20n^3+14n^2[/tex]
Taking out GCF, we get
[tex]Area=2n^2(3n^2+10n+7)[/tex]
We know that, area of a rectangle is the product of its length and width.
Since, width of the rectangle is [tex]2n^2[/tex], therefore length of the rectangle is [tex](3n^2+10n+7)[/tex].
