The rectangle below has an area of 12y^2+21y^5.
The width of the rectangle is equal to the greatest common monomial factor 12y^2 and 21y^5
What is the length and width of the rectangle?
Width= ?
Length=?

Respuesta :

Answer:

The  length of the rectangle is[tex](4+7y^3)[/tex] and its  width is[tex]3y^2[/tex]

Step-by-step explanation:

Given:

The Area of the rectangle =[tex]12y^2+21y^5[/tex].

Width is the greatest common factor of  [tex]12y^2[/tex] and[tex]21y^5[/tex]

To Find:

The length and width of the rectangle = ?

Solution:

Step 1: Finding the greatest common factor of [tex]12y^2[/tex]and[tex]21y^5[/tex]

[tex]12 y^2[/tex]= 2 x 3 x 3 x y x y

[tex]21y^5[/tex] = 3 x 7 x y x y x y x y x y

Thus the greatest common factor is [tex]3y^2[/tex] which is the width

Step 2: Finding the length and width

The area of the rectangle  = length x breath

width x length=  [tex]12y^2+21y^5.[/tex]

Taking[tex]3y^2[/tex] commonly from each term, we get

width x length= [tex](3y^2)(4 + 7y^3)[/tex]