Answer:
Width: [tex]5k^2[/tex],
Length: [tex]3k^2+7k+4[/tex]
Step-by-step explanation:
Consider the question: The rectangle below has an area of [tex]15k^4+35k^3+20k^2[/tex] square meters
. The width of the rectangle (in meters) is equal to the greatest common monomial factor of [tex]15k^4, 35k^3[/tex] and [tex]20k^2[/tex]. What is the length and width of the rectangle ?
First of all, we will find the greatest common monomial factor of [tex]15k^4, 35k^3[/tex] and [tex]20k^2[/tex] 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 [tex](k^4,k^3\text{ and }k^2)[/tex] is [tex]k^2[/tex].
Since the greatest common monomial factor of [tex]15k^4, 35k^3[/tex] and [tex]20k^2[/tex] is [tex]5k^2[/tex], therefore, the width of the given rectangle is [tex]5k^2[/tex] meters.
Since area of rectangle is product of length and width of rectangle, so we will divide area of given rectangle by width [tex](5k^2)[/tex] to find the length of rectangle:
[tex]\text{Length of rectangle}=\frac{15k^4+35k^3+20k^2}{5k^2}[/tex]
[tex]\text{Length of rectangle}=\frac{5k^2(3k^2+7k+4)}{5k^2}[/tex]
Cancel out common factors:
[tex]\text{Length of rectangle}=3k^2+7k+4[/tex]
Therefore, the length of the given rectangle is [tex]3k^2+7k+4[/tex] meters.
