Question

the rectangle below has an area of 15k^4 + 35k^3+ 20k^2. the width of the rectangle 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? length width 15k^4 35k^3 20k^2 width =

Explanation:

Step1: Find GCF of coefficients

Find GCF of 15, 35, 20. Prime - factor: $15=3\times5$, $35 = 5\times7$, $20=2^{2}\times5$. GCF is 5.

Step2: Find GCF of variable parts

For $k^{4},k^{3},k^{2}$, the GCF of the variable part is $k^{2}$ (using rule $k^{m},k^{n}$ with $m>n$, GCF is $k^{n}$).

Step3: Determine the width

The greatest common monomial factor of $15k^{4},35k^{3},20k^{2}$ is $5k^{2}$, so the width $w = 5k^{2}$.

Step4: Calculate the length

We know that the area of a rectangle $A=l\times w$. Given $A = 15k^{4}+35k^{3}+20k^{2}$ and $w = 5k^{2}$. Then $l=\frac{A}{w}=\frac{15k^{4}+35k^{3}+20k^{2}}{5k^{2}}=\frac{15k^{4}}{5k^{2}}+\frac{35k^{3}}{5k^{2}}+\frac{20k^{2}}{5k^{2}}=3k^{2}+7k + 4$.

Answer:

Width: $5k^{2}$
Length: $3k^{2}+7k + 4$