Expand Using the Binomial Theorem (2z+5k)^4

${(2 z + 5 k)}^{4}$

Step 1

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

$\sum_{k = 0}^{4} \frac{4!}{(4 - k)! k!} \cdot {(2 z)}^{4 - k} \cdot {(5 k)}^{k}$

Step 2

Expand the summation.

$\frac{4!}{(4 - 0)! 0!} \cdot {(2 z)}^{4 - 0} \cdot {(5 k)}^{0} + \frac{4!}{(4 - 1)! 1!} \cdot {(2 z)}^{4 - 1} \cdot {(5 k)}^{1} + \frac{4!}{(4 - 2)! 2!} \cdot {(2 z)}^{4 - 2} \cdot {(5 k)}^{2} + \frac{4!}{(4 - 3)! 3!} \cdot {(2 z)}^{4 - 3} \cdot {(5 k)}^{3} + \frac{4!}{(4 - 4)! 4!} \cdot {(2 z)}^{4 - 4} \cdot {(5 k)}^{4}$

Step 3

Simplify the exponents for each term of the expansion.

$1 \cdot {(2 z)}^{4} \cdot {(5 k)}^{0} + 4 \cdot {(2 z)}^{3} \cdot {(5 k)}^{1} + 6 \cdot {(2 z)}^{2} \cdot {(5 k)}^{2} + 4 \cdot {(2 z)}^{1} \cdot {(5 k)}^{3} + 1 \cdot {(2 z)}^{0} \cdot {(5 k)}^{4}$

Step 4

Simplify each term.

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

${(2 z)}^{4} \cdot {(5 k)}^{0} + 4 \cdot {(2 z)}^{3} \cdot {(5 k)}^{1} + 6 \cdot {(2 z)}^{2} \cdot {(5 k)}^{2} + 4 \cdot {(2 z)}^{1} \cdot {(5 k)}^{3} + 1 \cdot {(2 z)}^{0} \cdot {(5 k)}^{4}$

Apply the product rule to $2 z$ .

$2^{4} z^{4} \cdot {(5 k)}^{0} + 4 \cdot {(2 z)}^{3} \cdot {(5 k)}^{1} + 6 \cdot {(2 z)}^{2} \cdot {(5 k)}^{2} + 4 \cdot {(2 z)}^{1} \cdot {(5 k)}^{3} + 1 \cdot {(2 z)}^{0} \cdot {(5 k)}^{4}$

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

$16 z^{4} \cdot {(5 k)}^{0} + 4 \cdot {(2 z)}^{3} \cdot {(5 k)}^{1} + 6 \cdot {(2 z)}^{2} \cdot {(5 k)}^{2} + 4 \cdot {(2 z)}^{1} \cdot {(5 k)}^{3} + 1 \cdot {(2 z)}^{0} \cdot {(5 k)}^{4}$

Apply the product rule to $5 k$ .

$16 z^{4} \cdot (5^{0} k^{0}) + 4 \cdot {(2 z)}^{3} \cdot {(5 k)}^{1} + 6 \cdot {(2 z)}^{2} \cdot {(5 k)}^{2} + 4 \cdot {(2 z)}^{1} \cdot {(5 k)}^{3} + 1 \cdot {(2 z)}^{0} \cdot {(5 k)}^{4}$

Rewrite using the commutative property of multiplication.

$16 \cdot 5^{0} z^{4} k^{0} + 4 \cdot {(2 z)}^{3} \cdot {(5 k)}^{1} + 6 \cdot {(2 z)}^{2} \cdot {(5 k)}^{2} + 4 \cdot {(2 z)}^{1} \cdot {(5 k)}^{3} + 1 \cdot {(2 z)}^{0} \cdot {(5 k)}^{4}$

Anything raised to $0$ is $1$ .

$16 \cdot 1 z^{4} k^{0} + 4 \cdot {(2 z)}^{3} \cdot {(5 k)}^{1} + 6 \cdot {(2 z)}^{2} \cdot {(5 k)}^{2} + 4 \cdot {(2 z)}^{1} \cdot {(5 k)}^{3} + 1 \cdot {(2 z)}^{0} \cdot {(5 k)}^{4}$

Multiply $16$ by $1$ .

$16 z^{4} k^{0} + 4 \cdot {(2 z)}^{3} \cdot {(5 k)}^{1} + 6 \cdot {(2 z)}^{2} \cdot {(5 k)}^{2} + 4 \cdot {(2 z)}^{1} \cdot {(5 k)}^{3} + 1 \cdot {(2 z)}^{0} \cdot {(5 k)}^{4}$

Anything raised to $0$ is $1$ .

$16 z^{4} \cdot 1 + 4 \cdot {(2 z)}^{3} \cdot {(5 k)}^{1} + 6 \cdot {(2 z)}^{2} \cdot {(5 k)}^{2} + 4 \cdot {(2 z)}^{1} \cdot {(5 k)}^{3} + 1 \cdot {(2 z)}^{0} \cdot {(5 k)}^{4}$

Multiply $16$ by $1$ .

$16 z^{4} + 4 \cdot {(2 z)}^{3} \cdot {(5 k)}^{1} + 6 \cdot {(2 z)}^{2} \cdot {(5 k)}^{2} + 4 \cdot {(2 z)}^{1} \cdot {(5 k)}^{3} + 1 \cdot {(2 z)}^{0} \cdot {(5 k)}^{4}$

Apply the product rule to $2 z$ .

$16 z^{4} + 4 \cdot (2^{3} z^{3}) \cdot {(5 k)}^{1} + 6 \cdot {(2 z)}^{2} \cdot {(5 k)}^{2} + 4 \cdot {(2 z)}^{1} \cdot {(5 k)}^{3} + 1 \cdot {(2 z)}^{0} \cdot {(5 k)}^{4}$

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

$16 z^{4} + 4 \cdot (8 z^{3}) \cdot {(5 k)}^{1} + 6 \cdot {(2 z)}^{2} \cdot {(5 k)}^{2} + 4 \cdot {(2 z)}^{1} \cdot {(5 k)}^{3} + 1 \cdot {(2 z)}^{0} \cdot {(5 k)}^{4}$

Multiply $8$ by $4$ .

$16 z^{4} + 32 z^{3} \cdot {(5 k)}^{1} + 6 \cdot {(2 z)}^{2} \cdot {(5 k)}^{2} + 4 \cdot {(2 z)}^{1} \cdot {(5 k)}^{3} + 1 \cdot {(2 z)}^{0} \cdot {(5 k)}^{4}$

Simplify.

$16 z^{4} + 32 z^{3} \cdot (5 k) + 6 \cdot {(2 z)}^{2} \cdot {(5 k)}^{2} + 4 \cdot {(2 z)}^{1} \cdot {(5 k)}^{3} + 1 \cdot {(2 z)}^{0} \cdot {(5 k)}^{4}$

Rewrite using the commutative property of multiplication.

$16 z^{4} + 32 \cdot 5 z^{3} k + 6 \cdot {(2 z)}^{2} \cdot {(5 k)}^{2} + 4 \cdot {(2 z)}^{1} \cdot {(5 k)}^{3} + 1 \cdot {(2 z)}^{0} \cdot {(5 k)}^{4}$

Multiply $32$ by $5$ .

$16 z^{4} + 160 z^{3} k + 6 \cdot {(2 z)}^{2} \cdot {(5 k)}^{2} + 4 \cdot {(2 z)}^{1} \cdot {(5 k)}^{3} + 1 \cdot {(2 z)}^{0} \cdot {(5 k)}^{4}$

Apply the product rule to $2 z$ .

$16 z^{4} + 160 z^{3} k + 6 \cdot (2^{2} z^{2}) \cdot {(5 k)}^{2} + 4 \cdot {(2 z)}^{1} \cdot {(5 k)}^{3} + 1 \cdot {(2 z)}^{0} \cdot {(5 k)}^{4}$

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

$16 z^{4} + 160 z^{3} k + 6 \cdot (4 z^{2}) \cdot {(5 k)}^{2} + 4 \cdot {(2 z)}^{1} \cdot {(5 k)}^{3} + 1 \cdot {(2 z)}^{0} \cdot {(5 k)}^{4}$

Multiply $4$ by $6$ .

$16 z^{4} + 160 z^{3} k + 24 z^{2} \cdot {(5 k)}^{2} + 4 \cdot {(2 z)}^{1} \cdot {(5 k)}^{3} + 1 \cdot {(2 z)}^{0} \cdot {(5 k)}^{4}$

Apply the product rule to $5 k$ .

$16 z^{4} + 160 z^{3} k + 24 z^{2} \cdot (5^{2} k^{2}) + 4 \cdot {(2 z)}^{1} \cdot {(5 k)}^{3} + 1 \cdot {(2 z)}^{0} \cdot {(5 k)}^{4}$

Rewrite using the commutative property of multiplication.

$16 z^{4} + 160 z^{3} k + 24 \cdot 5^{2} z^{2} k^{2} + 4 \cdot {(2 z)}^{1} \cdot {(5 k)}^{3} + 1 \cdot {(2 z)}^{0} \cdot {(5 k)}^{4}$

Raise $5$ to the power of $2$ .

$16 z^{4} + 160 z^{3} k + 24 \cdot 25 z^{2} k^{2} + 4 \cdot {(2 z)}^{1} \cdot {(5 k)}^{3} + 1 \cdot {(2 z)}^{0} \cdot {(5 k)}^{4}$

Multiply $24$ by $25$ .

$16 z^{4} + 160 z^{3} k + 600 z^{2} k^{2} + 4 \cdot {(2 z)}^{1} \cdot {(5 k)}^{3} + 1 \cdot {(2 z)}^{0} \cdot {(5 k)}^{4}$

Simplify.

$16 z^{4} + 160 z^{3} k + 600 z^{2} k^{2} + 4 \cdot (2 z) \cdot {(5 k)}^{3} + 1 \cdot {(2 z)}^{0} \cdot {(5 k)}^{4}$

Multiply $2$ by $4$ .

$16 z^{4} + 160 z^{3} k + 600 z^{2} k^{2} + 8 z \cdot {(5 k)}^{3} + 1 \cdot {(2 z)}^{0} \cdot {(5 k)}^{4}$

Apply the product rule to $5 k$ .

$16 z^{4} + 160 z^{3} k + 600 z^{2} k^{2} + 8 z \cdot (5^{3} k^{3}) + 1 \cdot {(2 z)}^{0} \cdot {(5 k)}^{4}$

Rewrite using the commutative property of multiplication.

$16 z^{4} + 160 z^{3} k + 600 z^{2} k^{2} + 8 \cdot 5^{3} z k^{3} + 1 \cdot {(2 z)}^{0} \cdot {(5 k)}^{4}$

Raise $5$ to the power of $3$ .

$16 z^{4} + 160 z^{3} k + 600 z^{2} k^{2} + 8 \cdot 125 z k^{3} + 1 \cdot {(2 z)}^{0} \cdot {(5 k)}^{4}$

Multiply $8$ by $125$ .

$16 z^{4} + 160 z^{3} k + 600 z^{2} k^{2} + 1000 z k^{3} + 1 \cdot {(2 z)}^{0} \cdot {(5 k)}^{4}$

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

$16 z^{4} + 160 z^{3} k + 600 z^{2} k^{2} + 1000 z k^{3} + {(2 z)}^{0} \cdot {(5 k)}^{4}$

Apply the product rule to $2 z$ .

$16 z^{4} + 160 z^{3} k + 600 z^{2} k^{2} + 1000 z k^{3} + 2^{0} z^{0} \cdot {(5 k)}^{4}$

Anything raised to $0$ is $1$ .

$16 z^{4} + 160 z^{3} k + 600 z^{2} k^{2} + 1000 z k^{3} + 1 z^{0} \cdot {(5 k)}^{4}$

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

$16 z^{4} + 160 z^{3} k + 600 z^{2} k^{2} + 1000 z k^{3} + z^{0} \cdot {(5 k)}^{4}$

Anything raised to $0$ is $1$ .

$16 z^{4} + 160 z^{3} k + 600 z^{2} k^{2} + 1000 z k^{3} + 1 \cdot {(5 k)}^{4}$

Multiply ${(5 k)}^{4}$ by $1$ .

$16 z^{4} + 160 z^{3} k + 600 z^{2} k^{2} + 1000 z k^{3} + {(5 k)}^{4}$

Apply the product rule to $5 k$ .

$16 z^{4} + 160 z^{3} k + 600 z^{2} k^{2} + 1000 z k^{3} + 5^{4} k^{4}$

Raise $5$ to the power of $4$ .

$16 z^{4} + 160 z^{3} k + 600 z^{2} k^{2} + 1000 z k^{3} + 625 k^{4}$

Expand Using the Binomial Theorem (2z+5k)^4

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