Question

the width of the rectangle is equal to the greatest common monomial factor of 8x^5, 12x^3, and 20x^2. what is the length and width of the rectangle? width = length =

Explanation:

Step1: Find GCF of coefficients

Find GCF of 8, 12, 20. Prime - factorize: \(8 = 2\times2\times2\), \(12=2\times2\times3\), \(20 = 2\times2\times5\). GCF of 8, 12, 20 is 4.

Step2: Find GCF of variable parts

For \(x^{5},x^{3},x^{2}\), using the rule of exponents for GCF of powers of the same base (\(x^{m},x^{n}\) with \(m\geq n\), GCF is \(x^{n}\)), GCF is \(x^{2}\).

Step3: Calculate the GCF of monomials

The GCF of \(8x^{5},12x^{3},20x^{2}\) is the product of the GCF of coefficients and variable - parts, so \(4x^{2}\). This is the width.

Step4: Calculate the length

The sum of the areas of the three sub - rectangles is \(8x^{5}+12x^{3}+20x^{2}\). Divide by the width \(4x^{2}\): \(\frac{8x^{5}+12x^{3}+20x^{2}}{4x^{2}}=\frac{8x^{5}}{4x^{2}}+\frac{12x^{3}}{4x^{2}}+\frac{20x^{2}}{4x^{2}} = 2x^{3}+3x + 5\).

Answer:

Width: \(4x^{2}\)
Length: \(2x^{3}+3x + 5\)