Question

the rectangle below has an area of 12y^2 + 21y^5. the width of the rectangle is equal to the greatest common monomial factor of 12y^2 and 21y^5. what is the length and width of the rectangle? width = length =

Explanation:

Step1: Find GCF of coefficients

Find GCF of 12 and 21. Factors of 12 are 1, 2, 3, 4, 6, 12. Factors of 21 are 1, 3, 7, 21. GCF is 3.

Step2: Find GCF of variable - parts

For $y^2$ and $y^5$, using rule $y^m$ and $y^n$ ($m\lt n$), GCF is $y^2$.

Step3: Determine width

The greatest - common monomial factor of $12y^2$ and $21y^5$ is $3y^2$, so the width $w = 3y^2$.

Step4: Calculate length

We know that the area of a rectangle $A=l\times w$, and $A = 12y^2+21y^5$, $w = 3y^2$. Then $l=\frac{12y^2 + 21y^5}{3y^2}$. Using the distributive property of division $\frac{a + b}{c}=\frac{a}{c}+\frac{b}{c}$, we have $l=\frac{12y^2}{3y^2}+\frac{21y^5}{3y^2}=4 + 7y^3$.

Answer:

Width = $3y^2$
Length = $4 + 7y^3$