Question

In the following exercises, expand each binomial ${\left(p\hspace{0.17em}-\hspace{0.17em}1\right)}^5$.

Short answer

The solution of the exercise is $p^5-5p^4+10p^3-10p^2+5p-1$.

Step-by-step solution

Step 1. Given Information.

Given Binomial expression is ${\left(p\hspace{0.17em}-\hspace{0.17em}1\right)}^5$.

Step 2. Binomial expansion formula .

The binomial expansion formula is :-

${\left(a+b\right)}^n=\sum _{i=0}^n\left(\genfrac{}{}{0ex}{}{n}{i}\right)a^{\left(n-i\right)}{b}^i$

Step 3. Solve Binomial Expansion

Expanding using Binomial Theorem:-

${\left(p\hspace{0.17em}-\hspace{0.17em}1\right)}^5=\sum _{i=0}^5\left(\genfrac{}{}{0ex}{}{5}{i}\right)p^{\left(5-i\right)}{\left(-1\right)}^i$

$\sum _{i=0}^5\left(\genfrac{}{}{0ex}{}{5}{i}\right)p^{\left(5-i\right)}{\left(-1\right)}^i=\frac{5!}{0!\left(5-0\right)!}{p}^5{\left(-1\right)}^0+\frac{5!}{1!\left(5-1\right)!}{p}^4{\left(-1\right)}^1+\frac{5!}{2!\left(5-2\right)!}{p}^3{\left(-1\right)}^2+\frac{5!}{3!\left(5-3\right)!}{p}^2{\left(-1\right)}^3+\frac{5!}{4!\left(5-4\right)!}{p}^1{\left(-1\right)}^4+\frac{5!}{5!\left(5-5\right)!}{p}^0{\left(-1\right)}^5$

Simplify the expression:-

$p^5-5p^4+10p^3-10p^2+5p-1$