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Elodia [21]
2 years ago
13

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?
Mathematics
1 answer:
Sindrei [870]2 years ago
5 0

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).

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Part 2:

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