Question

the rectangle below has an area of 55x^6 + 22x^4. the width of the rectangle is equal to the greatest common monomial factor of 55x^6 and 22x^4. what is the length and width of the rectangle? width = length =

Explanation:

Step1: Find GCF of coefficients

Find the greatest - common factor of 55 and 22. The factors of 55 are 1, 5, 11, 55. The factors of 22 are 1, 2, 11, 22. So the GCF of 55 and 22 is 11.

Step2: Find GCF of variable parts

For the variable parts $x^{6}$ and $x^{4}$, using the rule of exponents for GCF of powers of the same base ($GCF(x^{m},x^{n})=x^{\min(m,n)}$), the GCF of $x^{6}$ and $x^{4}$ is $x^{4}$.

Step3: Determine the width

The greatest - common monomial factor of $55x^{6}$ and $22x^{4}$ is the product of the GCF of the coefficients and the GCF of the variable parts. So the width $w = 11x^{4}$.

Step4: Determine the length

We know that the area of a rectangle $A=l\times w$, and $A = 55x^{6}+22x^{4}$, $w = 11x^{4}$. Then $l=\frac{55x^{6}+22x^{4}}{11x^{4}}$. Using the distributive property of division over addition $\frac{a + b}{c}=\frac{a}{c}+\frac{b}{c}$, we have $l=\frac{55x^{6}}{11x^{4}}+\frac{22x^{4}}{11x^{4}}$. Simplifying each term: $\frac{55x^{6}}{11x^{4}} = 5x^{2}$ and $\frac{22x^{4}}{11x^{4}}=2$. So $l = 5x^{2}+2$.

Answer:

Width = $11x^{4}$
Length = $5x^{2}+2$