Expand Using the Binomial Theorem (p-q)^4

${(p - q)}^{4}$

Step 1

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

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

Step 2

Expand the summation.

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

Step 3

Simplify the exponents for each term of the expansion.

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

Step 4

Simplify each term.

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

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

Apply the product rule to $- q$ .

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

Rewrite using the commutative property of multiplication.

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

Anything raised to $0$ is $1$ .

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

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

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

Anything raised to $0$ is $1$ .

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

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

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

Simplify.

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

Rewrite using the commutative property of multiplication.

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

Multiply $4$ by $- 1$ .

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

Apply the product rule to $- q$ .

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

Rewrite using the commutative property of multiplication.

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

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

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

Multiply $6$ by $1$ .

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

Simplify.

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

Apply the product rule to $- q$ .

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

Rewrite using the commutative property of multiplication.

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

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

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

Multiply $4$ by $- 1$ .

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

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

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

Anything raised to $0$ is $1$ .

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

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

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

Apply the product rule to $- q$ .

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

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

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

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

$p^{4} - 4 p^{3} q + 6 p^{2} q^{2} - 4 p q^{3} + q^{4}$

Expand Using the Binomial Theorem (p-q)^4

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