Question

find the product.
(2m - p)^4

(2m - p)^4 = \square
(simplify your answer.)

Explanation:

Step1: Recall Binomial Theorem

The binomial theorem states that \((a + b)^n=\sum_{k = 0}^{n}\binom{n}{k}a^{n - k}b^{k}\), where \(\binom{n}{k}=\frac{n!}{k!(n - k)!}\) and \(n!=n\times(n - 1)\times\cdots\times1\). For \((2m-p)^4\), we have \(a = 2m\), \(b=-p\), and \(n = 4\).

Step2: Calculate each term

• For \(k = 0\): \(\binom{4}{0}(2m)^{4}(-p)^{0}=\frac{4!}{0!4!}(16m^{4})(1)=16m^{4}\)
• For \(k = 1\): \(\binom{4}{1}(2m)^{3}(-p)^{1}=\frac{4!}{1!3!}(8m^{3})(-p)=4\times8m^{3}\times(-p)=-32m^{3}p\)
• For \(k = 2\): \(\binom{4}{2}(2m)^{2}(-p)^{2}=\frac{4!}{2!2!}(4m^{2})(p^{2})=6\times4m^{2}p^{2}=24m^{2}p^{2}\)
• For \(k = 3\): \(\binom{4}{3}(2m)^{1}(-p)^{3}=\frac{4!}{3!1!}(2m)(-p^{3})=4\times2m\times(-p^{3})=-8mp^{3}\)
• For \(k = 4\): \(\binom{4}{4}(2m)^{0}(-p)^{4}=\frac{4!}{4!0!}(1)(p^{4})=p^{4}\)

Step3: Sum the terms

Add all the terms together: \(16m^{4}-32m^{3}p + 24m^{2}p^{2}-8mp^{3}+p^{4}\)

Answer:

\(16m^{4}-32m^{3}p + 24m^{2}p^{2}-8mp^{3}+p^{4}\)