Question

greatest common monomial factor of $15k^{4}$, $35k^{3}$, and $20k^{2}$. what is the length and width of the rectangle? width = length = related content taking common factor from binomial 4:58 taking common factor: area model 5:27 taking common factor from trinomial

Explanation:

Step1: Find the GCF of coefficients

Find the GCF of \(15\), \(35\), and \(20\).
Factors of \(15 = 3\times5\)
Factors of \(35=5\times7\)
Factors of \(20 = 2^{2}\times5\)
The GCF of \(15\), \(35\), and \(20\) is \(5\).

Step2: Find the GCF of variable parts

For the variable \(k\), consider the exponents \(4\), \(3\), and \(2\).
Using the rule \(GCF(k^{m},k^{n},k^{p})=k^{\min(m,n,p)}\), where \(m = 4\), \(n = 3\), \(p=2\). So \(GCF(k^{4},k^{3},k^{2})=k^{2}\)

Step3: Calculate the GCF of monomials

Multiply the GCF of coefficients and variable parts.
\(GCF(15k^{4},35k^{3},20k^{2})=5k^{2}\)
Since the width is the GCF (common side of the rectangles), width \( = 5k^{2}\)

Step4: Calculate the length

Divide each monomial by the GCF.
\(\frac{15k^{4}}{5k^{2}}=3k^{2}\), \(\frac{35k^{3}}{5k^{2}} = 7k\), \(\frac{20k^{2}}{5k^{2}}=4\)
Length \(=3k^{2}+7k + 4\)

Answer:

Width \(=5k^{2}\)
Length \(=3k^{2}+7k + 4\)