Expand Using the Binomial Theorem (np-1)^4

${(n p - 1)}^{4}$

Step 1

Use the binomial expansion theorem to find each term. The binomial theorem states ${(a + b)}^{n} = \sum_{k = 0}^{n} n C k \cdot (a^{n - k} b^{k})$ .

$\sum_{k = 0}^{4} \frac{4!}{(4 - k)! k!} \cdot {(n p)}^{4 - k} \cdot {(- 1)}^{k}$

Step 2

Expand the summation.

$\frac{4!}{(4 - 0)! 0!} \cdot {(n p)}^{4 - 0} \cdot {(- 1)}^{0} + \frac{4!}{(4 - 1)! 1!} \cdot {(n p)}^{4 - 1} \cdot {(- 1)}^{1} + \frac{4!}{(4 - 2)! 2!} \cdot {(n p)}^{4 - 2} \cdot {(- 1)}^{2} + \frac{4!}{(4 - 3)! 3!} \cdot {(n p)}^{4 - 3} \cdot {(- 1)}^{3} + \frac{4!}{(4 - 4)! 4!} \cdot {(n p)}^{4 - 4} \cdot {(- 1)}^{4}$

Step 3

Simplify the exponents for each term of the expansion.

$1 \cdot {(n p)}^{4} \cdot {(- 1)}^{0} + 4 \cdot {(n p)}^{3} \cdot {(- 1)}^{1} + 6 \cdot {(n p)}^{2} \cdot {(- 1)}^{2} + 4 \cdot {(n p)}^{1} \cdot {(- 1)}^{3} + 1 \cdot {(n p)}^{0} \cdot {(- 1)}^{4}$

Step 4

Simplify each term.

Multiply ${(n p)}^{4}$ by $1$ .

${(n p)}^{4} \cdot {(- 1)}^{0} + 4 \cdot {(n p)}^{3} \cdot {(- 1)}^{1} + 6 \cdot {(n p)}^{2} \cdot {(- 1)}^{2} + 4 \cdot {(n p)}^{1} \cdot {(- 1)}^{3} + 1 \cdot {(n p)}^{0} \cdot {(- 1)}^{4}$

Apply the product rule to $n p$ .

$n^{4} p^{4} \cdot {(- 1)}^{0} + 4 \cdot {(n p)}^{3} \cdot {(- 1)}^{1} + 6 \cdot {(n p)}^{2} \cdot {(- 1)}^{2} + 4 \cdot {(n p)}^{1} \cdot {(- 1)}^{3} + 1 \cdot {(n p)}^{0} \cdot {(- 1)}^{4}$

Anything raised to $0$ is $1$ .

$n^{4} p^{4} \cdot 1 + 4 \cdot {(n p)}^{3} \cdot {(- 1)}^{1} + 6 \cdot {(n p)}^{2} \cdot {(- 1)}^{2} + 4 \cdot {(n p)}^{1} \cdot {(- 1)}^{3} + 1 \cdot {(n p)}^{0} \cdot {(- 1)}^{4}$

Multiply $n^{4}$ by $1$ .

$n^{4} p^{4} + 4 \cdot {(n p)}^{3} \cdot {(- 1)}^{1} + 6 \cdot {(n p)}^{2} \cdot {(- 1)}^{2} + 4 \cdot {(n p)}^{1} \cdot {(- 1)}^{3} + 1 \cdot {(n p)}^{0} \cdot {(- 1)}^{4}$

Apply the product rule to $n p$ .

$n^{4} p^{4} + 4 \cdot (n^{3} p^{3}) \cdot {(- 1)}^{1} + 6 \cdot {(n p)}^{2} \cdot {(- 1)}^{2} + 4 \cdot {(n p)}^{1} \cdot {(- 1)}^{3} + 1 \cdot {(n p)}^{0} \cdot {(- 1)}^{4}$

Evaluate the exponent.

$n^{4} p^{4} + 4 n^{3} p^{3} \cdot - 1 + 6 \cdot {(n p)}^{2} \cdot {(- 1)}^{2} + 4 \cdot {(n p)}^{1} \cdot {(- 1)}^{3} + 1 \cdot {(n p)}^{0} \cdot {(- 1)}^{4}$

Multiply $- 1$ by $4$ .

$n^{4} p^{4} - 4 n^{3} p^{3} + 6 \cdot {(n p)}^{2} \cdot {(- 1)}^{2} + 4 \cdot {(n p)}^{1} \cdot {(- 1)}^{3} + 1 \cdot {(n p)}^{0} \cdot {(- 1)}^{4}$

Apply the product rule to $n p$ .

$n^{4} p^{4} - 4 n^{3} p^{3} + 6 \cdot (n^{2} p^{2}) \cdot {(- 1)}^{2} + 4 \cdot {(n p)}^{1} \cdot {(- 1)}^{3} + 1 \cdot {(n p)}^{0} \cdot {(- 1)}^{4}$

Raise $- 1$ to the power of $2$ .

$n^{4} p^{4} - 4 n^{3} p^{3} + 6 n^{2} p^{2} \cdot 1 + 4 \cdot {(n p)}^{1} \cdot {(- 1)}^{3} + 1 \cdot {(n p)}^{0} \cdot {(- 1)}^{4}$

Multiply $6$ by $1$ .

$n^{4} p^{4} - 4 n^{3} p^{3} + 6 n^{2} p^{2} + 4 \cdot {(n p)}^{1} \cdot {(- 1)}^{3} + 1 \cdot {(n p)}^{0} \cdot {(- 1)}^{4}$

Simplify.

$n^{4} p^{4} - 4 n^{3} p^{3} + 6 n^{2} p^{2} + 4 \cdot (n p) \cdot {(- 1)}^{3} + 1 \cdot {(n p)}^{0} \cdot {(- 1)}^{4}$

Raise $- 1$ to the power of $3$ .

$n^{4} p^{4} - 4 n^{3} p^{3} + 6 n^{2} p^{2} + 4 n p \cdot - 1 + 1 \cdot {(n p)}^{0} \cdot {(- 1)}^{4}$

Multiply $- 1$ by $4$ .

$n^{4} p^{4} - 4 n^{3} p^{3} + 6 n^{2} p^{2} - 4 n p + 1 \cdot {(n p)}^{0} \cdot {(- 1)}^{4}$

Multiply ${(n p)}^{0}$ by $1$ .

$n^{4} p^{4} - 4 n^{3} p^{3} + 6 n^{2} p^{2} - 4 n p + {(n p)}^{0} \cdot {(- 1)}^{4}$

Apply the product rule to $n p$ .

$n^{4} p^{4} - 4 n^{3} p^{3} + 6 n^{2} p^{2} - 4 n p + n^{0} p^{0} \cdot {(- 1)}^{4}$

Anything raised to $0$ is $1$ .

$n^{4} p^{4} - 4 n^{3} p^{3} + 6 n^{2} p^{2} - 4 n p + 1 p^{0} \cdot {(- 1)}^{4}$

Multiply $p^{0}$ by $1$ .

$n^{4} p^{4} - 4 n^{3} p^{3} + 6 n^{2} p^{2} - 4 n p + p^{0} \cdot {(- 1)}^{4}$

Anything raised to $0$ is $1$ .

$n^{4} p^{4} - 4 n^{3} p^{3} + 6 n^{2} p^{2} - 4 n p + 1 \cdot {(- 1)}^{4}$

Multiply ${(- 1)}^{4}$ by $1$ .

$n^{4} p^{4} - 4 n^{3} p^{3} + 6 n^{2} p^{2} - 4 n p + {(- 1)}^{4}$

Raise $- 1$ to the power of $4$ .

$n^{4} p^{4} - 4 n^{3} p^{3} + 6 n^{2} p^{2} - 4 n p + 1$

Expand Using the Binomial Theorem (np-1)^4

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