Question

The rectangle below has an area of 15k4+35k3+20k2 Square meters

The width of the rectangle is equal to the greatest common monomial factor of 70y^870y 8 70, y, start superscript, 8, end superscript and 30y^630y 6 30, y, start superscript, 6, end superscript.

What is the length and width of the rectangle ?

Answer 1

Width=5k^2

Length=3k^2+7k+4

Explanation:

Answer 2

Width:
`5k^2`,

Length:
`3k^2+7k+4`

Explanation:

Consider the question: The rectangle below has an area of
`15k^4+35k^3+20k^2` square meters

. The width of the rectangle (in meters) is equal to the greatest common monomial factor of
`15k^4, 35k^3` and
`20k^2`. What is the length and width of the rectangle ?

First of all, we will find the greatest common monomial factor of
`15k^4, 35k^3` and
`20k^2` to determine the width as:

The greatest common factor of number part is 15, 20 and 35 is 5. The greatest common factor of variable part
`(k^4,k^3\text{ and }k^2)` is
`k^2`.

Since the greatest common monomial factor of
`15k^4, 35k^3` and
`20k^2` is
`5k^2`, therefore, the width of the given rectangle is
`5k^2` meters.

Since area of rectangle is product of length and width of rectangle, so we will divide area of given rectangle by width
`(5k^2)` to find the length of rectangle:

`\text{Length of rectangle}=(15k^4+35k^3+20k^2)/(5k^2)`

`\text{Length of rectangle}=(5k^2(3k^2+7k+4))/(5k^2)`

Cancel out common factors:

`\text{Length of rectangle}=3k^2+7k+4`

Therefore, the length of the given rectangle is
`3k^2+7k+4` meters.