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? length width $55x^{6}$ $22x^{4}$

Explanation:

Step1: Find GCF of coefficients

Find GCF of 55 and 22. Prime - factorize: 55 = 5×11, 22 = 2×11. So GCF(55, 22)=11.

Step2: Find GCF of variables

For \(x^6\) and \(x^4\), using the rule \(GCF(x^m,x^n)=x^{\min(m,n)}\), we have \(GCF(x^6,x^4)=x^4\).

Step3: Determine width

The greatest common monomial factor of \(55x^6\) and \(22x^4\) is \(11x^4\), so the width \(w = 11x^4\).

Step4: Calculate length

We know that the area of a rectangle \(A=l\times w\), and \(A = 55x^6+22x^4\), \(w = 11x^4\). Then \(l=\frac{A}{w}=\frac{55x^6 + 22x^4}{11x^4}\). Using the distributive property of division \(\frac{a + b}{c}=\frac{a}{c}+\frac{b}{c}\), we get \(l=\frac{55x^6}{11x^4}+\frac{22x^4}{11x^4}=5x^2 + 2\).

Answer:

Width: \(11x^4\), Length: \(5x^2 + 2\)