Expand Using the Binomial Theorem (p-q)^2

${(p - q)}^{2}$

Step 1

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

$\sum_{k = 0}^{2} \frac{2!}{(2 - k)! k!} \cdot {(p)}^{2 - k} \cdot {(- q)}^{k}$

Step 2

Expand the summation.

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

Step 3

Simplify the exponents for each term of the expansion.

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

Step 4

Simplify each term.

Multiply ${(p)}^{2}$ by $1$ .

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

Apply the product rule to $- q$ .

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

Rewrite using the commutative property of multiplication.

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

Anything raised to $0$ is $1$ .

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

Multiply $p^{2}$ by $1$ .

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

Anything raised to $0$ is $1$ .

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

Multiply $p^{2}$ by $1$ .

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

Simplify.

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

Simplify.

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

Rewrite using the commutative property of multiplication.

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

Multiply $2$ by $- 1$ .

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

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

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

Anything raised to $0$ is $1$ .

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

Multiply ${(- q)}^{2}$ by $1$ .

$p^{2} - 2 p q + {(- q)}^{2}$

Apply the product rule to $- q$ .

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

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

$p^{2} - 2 p q + 1 q^{2}$

Multiply $q^{2}$ by $1$ .

$p^{2} - 2 p q + q^{2}$

Expand Using the Binomial Theorem (p-q)^2

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