Question

The rectangle below has an area of 6n^4+20n^3+14n^26n 4 +20n 3 +14n 2 6, n, start superscript, 4, end superscript, plus, 20, n, cubed, plus, 14, n, squared. The width of the rectangle is equal to the greatest common monomial factor of 6n^4, 20n^3,6n 4 ,20n 3 ,6, n, start superscript, 4, end superscript, comma, 20, n, cubed, comma and 14n^214n 2 14, n, squared. What is the length and width of the rectangle?

Answer 1

Given:

Area of rectangle =
`6n^4+20n^3+14n^2`

Width of the rectangle is equal to the greatest common monomial factor of
`6n^4, 20n^3,14n^2`.

To find:

Length and width of the rectangle.

Solution:

Width of the rectangle is equal to the greatest common monomial factor of
`6n^4, 20n^3,14n^2` is

`6n^4=2* 3* n* n* n* n`

`20n^3=2* 2* 5* n* n* n`

`14n^2=2* 7* n* n`

Now,

`GCF(6n^4, 20n^3,14n^2)=2* n* n=2n^2`

So, width of the rectangle is
`2n^2`.

Area of rectangle is

`Area=6n^4+20n^3+14n^2`

Taking out GCF, we get

`Area=2n^2(3n^2+10n+7)`

We know that, area of a rectangle is the product of its length and width.

Since, width of the rectangle is
`2n^2`, therefore length of the rectangle is
`(3n^2+10n+7)`.