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

$6n^4=2\times 3\times n\times n\times n\times n$

$20n^3=2\times 2\times 5\times n\times n\times n$

$14n^2=2\times 7\times n\times n$

Now,

$GCF(6n^4, 20n^3,14n^2)=2\times n\times n=2n^2$

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 $(3n^2+10n+7)$.