Expand Using the Binomial Theorem (2m-p)^4

${(2 m - p)}^{4}$

Step 1

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

$\sum_{k = 0}^{4} \frac{4!}{(4 - k)! k!} \cdot {(2 m)}^{4 - k} \cdot {(- p)}^{k}$

Step 2

Expand the summation.

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

Step 3

Simplify the exponents for each term of the expansion.

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

Step 4

Simplify each term.

Multiply ${(2 m)}^{4}$ by $1$ .

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

Apply the product rule to $2 m$ .

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

Raise $2$ to the power of $4$ .

$16 m^{4} \cdot {(- p)}^{0} + 4 \cdot {(2 m)}^{3} \cdot {(- p)}^{1} + 6 \cdot {(2 m)}^{2} \cdot {(- p)}^{2} + 4 \cdot {(2 m)}^{1} \cdot {(- p)}^{3} + 1 \cdot {(2 m)}^{0} \cdot {(- p)}^{4}$

Apply the product rule to $- p$ .

$16 m^{4} \cdot ({(- 1)}^{0} p^{0}) + 4 \cdot {(2 m)}^{3} \cdot {(- p)}^{1} + 6 \cdot {(2 m)}^{2} \cdot {(- p)}^{2} + 4 \cdot {(2 m)}^{1} \cdot {(- p)}^{3} + 1 \cdot {(2 m)}^{0} \cdot {(- p)}^{4}$

Rewrite using the commutative property of multiplication.

$16 \cdot {(- 1)}^{0} m^{4} p^{0} + 4 \cdot {(2 m)}^{3} \cdot {(- p)}^{1} + 6 \cdot {(2 m)}^{2} \cdot {(- p)}^{2} + 4 \cdot {(2 m)}^{1} \cdot {(- p)}^{3} + 1 \cdot {(2 m)}^{0} \cdot {(- p)}^{4}$

Anything raised to $0$ is $1$ .

$16 \cdot 1 m^{4} p^{0} + 4 \cdot {(2 m)}^{3} \cdot {(- p)}^{1} + 6 \cdot {(2 m)}^{2} \cdot {(- p)}^{2} + 4 \cdot {(2 m)}^{1} \cdot {(- p)}^{3} + 1 \cdot {(2 m)}^{0} \cdot {(- p)}^{4}$

Multiply $16$ by $1$ .

$16 m^{4} p^{0} + 4 \cdot {(2 m)}^{3} \cdot {(- p)}^{1} + 6 \cdot {(2 m)}^{2} \cdot {(- p)}^{2} + 4 \cdot {(2 m)}^{1} \cdot {(- p)}^{3} + 1 \cdot {(2 m)}^{0} \cdot {(- p)}^{4}$

Anything raised to $0$ is $1$ .

$16 m^{4} \cdot 1 + 4 \cdot {(2 m)}^{3} \cdot {(- p)}^{1} + 6 \cdot {(2 m)}^{2} \cdot {(- p)}^{2} + 4 \cdot {(2 m)}^{1} \cdot {(- p)}^{3} + 1 \cdot {(2 m)}^{0} \cdot {(- p)}^{4}$

Multiply $16$ by $1$ .

$16 m^{4} + 4 \cdot {(2 m)}^{3} \cdot {(- p)}^{1} + 6 \cdot {(2 m)}^{2} \cdot {(- p)}^{2} + 4 \cdot {(2 m)}^{1} \cdot {(- p)}^{3} + 1 \cdot {(2 m)}^{0} \cdot {(- p)}^{4}$

Apply the product rule to $2 m$ .

$16 m^{4} + 4 \cdot (2^{3} m^{3}) \cdot {(- p)}^{1} + 6 \cdot {(2 m)}^{2} \cdot {(- p)}^{2} + 4 \cdot {(2 m)}^{1} \cdot {(- p)}^{3} + 1 \cdot {(2 m)}^{0} \cdot {(- p)}^{4}$

Raise $2$ to the power of $3$ .

$16 m^{4} + 4 \cdot (8 m^{3}) \cdot {(- p)}^{1} + 6 \cdot {(2 m)}^{2} \cdot {(- p)}^{2} + 4 \cdot {(2 m)}^{1} \cdot {(- p)}^{3} + 1 \cdot {(2 m)}^{0} \cdot {(- p)}^{4}$

Multiply $8$ by $4$ .

$16 m^{4} + 32 m^{3} \cdot {(- p)}^{1} + 6 \cdot {(2 m)}^{2} \cdot {(- p)}^{2} + 4 \cdot {(2 m)}^{1} \cdot {(- p)}^{3} + 1 \cdot {(2 m)}^{0} \cdot {(- p)}^{4}$

Simplify.

$16 m^{4} + 32 m^{3} \cdot (- p) + 6 \cdot {(2 m)}^{2} \cdot {(- p)}^{2} + 4 \cdot {(2 m)}^{1} \cdot {(- p)}^{3} + 1 \cdot {(2 m)}^{0} \cdot {(- p)}^{4}$

Rewrite using the commutative property of multiplication.

$16 m^{4} + 32 \cdot - 1 m^{3} p + 6 \cdot {(2 m)}^{2} \cdot {(- p)}^{2} + 4 \cdot {(2 m)}^{1} \cdot {(- p)}^{3} + 1 \cdot {(2 m)}^{0} \cdot {(- p)}^{4}$

Multiply $32$ by $- 1$ .

$16 m^{4} - 32 m^{3} p + 6 \cdot {(2 m)}^{2} \cdot {(- p)}^{2} + 4 \cdot {(2 m)}^{1} \cdot {(- p)}^{3} + 1 \cdot {(2 m)}^{0} \cdot {(- p)}^{4}$

Apply the product rule to $2 m$ .

$16 m^{4} - 32 m^{3} p + 6 \cdot (2^{2} m^{2}) \cdot {(- p)}^{2} + 4 \cdot {(2 m)}^{1} \cdot {(- p)}^{3} + 1 \cdot {(2 m)}^{0} \cdot {(- p)}^{4}$

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

$16 m^{4} - 32 m^{3} p + 6 \cdot (4 m^{2}) \cdot {(- p)}^{2} + 4 \cdot {(2 m)}^{1} \cdot {(- p)}^{3} + 1 \cdot {(2 m)}^{0} \cdot {(- p)}^{4}$

Multiply $4$ by $6$ .

$16 m^{4} - 32 m^{3} p + 24 m^{2} \cdot {(- p)}^{2} + 4 \cdot {(2 m)}^{1} \cdot {(- p)}^{3} + 1 \cdot {(2 m)}^{0} \cdot {(- p)}^{4}$

Apply the product rule to $- p$ .

$16 m^{4} - 32 m^{3} p + 24 m^{2} \cdot ({(- 1)}^{2} p^{2}) + 4 \cdot {(2 m)}^{1} \cdot {(- p)}^{3} + 1 \cdot {(2 m)}^{0} \cdot {(- p)}^{4}$

Rewrite using the commutative property of multiplication.

$16 m^{4} - 32 m^{3} p + 24 \cdot {(- 1)}^{2} m^{2} p^{2} + 4 \cdot {(2 m)}^{1} \cdot {(- p)}^{3} + 1 \cdot {(2 m)}^{0} \cdot {(- p)}^{4}$

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

$16 m^{4} - 32 m^{3} p + 24 \cdot 1 m^{2} p^{2} + 4 \cdot {(2 m)}^{1} \cdot {(- p)}^{3} + 1 \cdot {(2 m)}^{0} \cdot {(- p)}^{4}$

Multiply $24$ by $1$ .

$16 m^{4} - 32 m^{3} p + 24 m^{2} p^{2} + 4 \cdot {(2 m)}^{1} \cdot {(- p)}^{3} + 1 \cdot {(2 m)}^{0} \cdot {(- p)}^{4}$

Simplify.

$16 m^{4} - 32 m^{3} p + 24 m^{2} p^{2} + 4 \cdot (2 m) \cdot {(- p)}^{3} + 1 \cdot {(2 m)}^{0} \cdot {(- p)}^{4}$

Multiply $2$ by $4$ .

$16 m^{4} - 32 m^{3} p + 24 m^{2} p^{2} + 8 m \cdot {(- p)}^{3} + 1 \cdot {(2 m)}^{0} \cdot {(- p)}^{4}$

Apply the product rule to $- p$ .

$16 m^{4} - 32 m^{3} p + 24 m^{2} p^{2} + 8 m \cdot ({(- 1)}^{3} p^{3}) + 1 \cdot {(2 m)}^{0} \cdot {(- p)}^{4}$

Rewrite using the commutative property of multiplication.

$16 m^{4} - 32 m^{3} p + 24 m^{2} p^{2} + 8 \cdot {(- 1)}^{3} m p^{3} + 1 \cdot {(2 m)}^{0} \cdot {(- p)}^{4}$

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

$16 m^{4} - 32 m^{3} p + 24 m^{2} p^{2} + 8 \cdot - 1 m p^{3} + 1 \cdot {(2 m)}^{0} \cdot {(- p)}^{4}$

Multiply $8$ by $- 1$ .

$16 m^{4} - 32 m^{3} p + 24 m^{2} p^{2} - 8 m p^{3} + 1 \cdot {(2 m)}^{0} \cdot {(- p)}^{4}$

Multiply ${(2 m)}^{0}$ by $1$ .

$16 m^{4} - 32 m^{3} p + 24 m^{2} p^{2} - 8 m p^{3} + {(2 m)}^{0} \cdot {(- p)}^{4}$

Apply the product rule to $2 m$ .

$16 m^{4} - 32 m^{3} p + 24 m^{2} p^{2} - 8 m p^{3} + 2^{0} m^{0} \cdot {(- p)}^{4}$

Anything raised to $0$ is $1$ .

$16 m^{4} - 32 m^{3} p + 24 m^{2} p^{2} - 8 m p^{3} + 1 m^{0} \cdot {(- p)}^{4}$

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

$16 m^{4} - 32 m^{3} p + 24 m^{2} p^{2} - 8 m p^{3} + m^{0} \cdot {(- p)}^{4}$

Anything raised to $0$ is $1$ .

$16 m^{4} - 32 m^{3} p + 24 m^{2} p^{2} - 8 m p^{3} + 1 \cdot {(- p)}^{4}$

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

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

Apply the product rule to $- p$ .

$16 m^{4} - 32 m^{3} p + 24 m^{2} p^{2} - 8 m p^{3} + {(- 1)}^{4} p^{4}$

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

$16 m^{4} - 32 m^{3} p + 24 m^{2} p^{2} - 8 m p^{3} + 1 p^{4}$

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

$16 m^{4} - 32 m^{3} p + 24 m^{2} p^{2} - 8 m p^{3} + p^{4}$

Expand Using the Binomial Theorem (2m-p)^4

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