Question

The rectangle below has an area of 12y^2+21y^512y 2 +21y 5 12, y, squared, plus, 21, y, start superscript, 5, end superscript. The width of the rectangle is equal to the greatest common monomial factor of 12y^212y 2 12, y, squared and 21y^521y 5 21, y, start superscript, 5, end superscript. What is the length and width of the rectangle? \text{Width} =Width=start text, W, i, d, t, h, end text, equals

Answer 1

4y^2

Explanation:

Answer 2

This means that the width = 3y²

The length = 4y+7y³

Explanation:

Given

Area of the rectangle = 12y^2+21y^5

Width of the rectangle is the greatest common monomial of 12y^2 and 21y^5

First let us get the greatest common monomial of 12y^2 and 1^5

12y^2 = (3 * y * y) * 4

21y^5 = (3 * y*y)*y*y*y* 7

Hence the greatest common monomial is 3 * y * y = 3y²

Factor out 3y² from the area function;

12y^2+21y^5

3y² (4y+7y³) = Area of the rectangle

Length * Width = 3y² (4y+7y³)

On comparing both sides of the equation

This means that the width = 3y²

The length = 4y+7y³