Expand Using the Binomial Theorem (2y-9)^4

${(2 y - 9)}^{4}$

Step 1

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

$\sum_{k = 0}^{4} \frac{4!}{(4 - k)! k!} \cdot {(2 y)}^{4 - k} \cdot {(- 9)}^{k}$

Step 2

Expand the summation.

$\frac{4!}{(4 - 0)! 0!} \cdot {(2 y)}^{4 - 0} \cdot {(- 9)}^{0} + \frac{4!}{(4 - 1)! 1!} \cdot {(2 y)}^{4 - 1} \cdot {(- 9)}^{1} + \frac{4!}{(4 - 2)! 2!} \cdot {(2 y)}^{4 - 2} \cdot {(- 9)}^{2} + \frac{4!}{(4 - 3)! 3!} \cdot {(2 y)}^{4 - 3} \cdot {(- 9)}^{3} + \frac{4!}{(4 - 4)! 4!} \cdot {(2 y)}^{4 - 4} \cdot {(- 9)}^{4}$

Step 3

Simplify the exponents for each term of the expansion.

$1 \cdot {(2 y)}^{4} \cdot {(- 9)}^{0} + 4 \cdot {(2 y)}^{3} \cdot {(- 9)}^{1} + 6 \cdot {(2 y)}^{2} \cdot {(- 9)}^{2} + 4 \cdot {(2 y)}^{1} \cdot {(- 9)}^{3} + 1 \cdot {(2 y)}^{0} \cdot {(- 9)}^{4}$

Step 4

Simplify each term.

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

${(2 y)}^{4} \cdot {(- 9)}^{0} + 4 \cdot {(2 y)}^{3} \cdot {(- 9)}^{1} + 6 \cdot {(2 y)}^{2} \cdot {(- 9)}^{2} + 4 \cdot {(2 y)}^{1} \cdot {(- 9)}^{3} + 1 \cdot {(2 y)}^{0} \cdot {(- 9)}^{4}$

Apply the product rule to $2 y$ .

$2^{4} y^{4} \cdot {(- 9)}^{0} + 4 \cdot {(2 y)}^{3} \cdot {(- 9)}^{1} + 6 \cdot {(2 y)}^{2} \cdot {(- 9)}^{2} + 4 \cdot {(2 y)}^{1} \cdot {(- 9)}^{3} + 1 \cdot {(2 y)}^{0} \cdot {(- 9)}^{4}$

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

$16 y^{4} \cdot {(- 9)}^{0} + 4 \cdot {(2 y)}^{3} \cdot {(- 9)}^{1} + 6 \cdot {(2 y)}^{2} \cdot {(- 9)}^{2} + 4 \cdot {(2 y)}^{1} \cdot {(- 9)}^{3} + 1 \cdot {(2 y)}^{0} \cdot {(- 9)}^{4}$

Anything raised to $0$ is $1$ .

$16 y^{4} \cdot 1 + 4 \cdot {(2 y)}^{3} \cdot {(- 9)}^{1} + 6 \cdot {(2 y)}^{2} \cdot {(- 9)}^{2} + 4 \cdot {(2 y)}^{1} \cdot {(- 9)}^{3} + 1 \cdot {(2 y)}^{0} \cdot {(- 9)}^{4}$

Multiply $16$ by $1$ .

$16 y^{4} + 4 \cdot {(2 y)}^{3} \cdot {(- 9)}^{1} + 6 \cdot {(2 y)}^{2} \cdot {(- 9)}^{2} + 4 \cdot {(2 y)}^{1} \cdot {(- 9)}^{3} + 1 \cdot {(2 y)}^{0} \cdot {(- 9)}^{4}$

Apply the product rule to $2 y$ .

$16 y^{4} + 4 \cdot (2^{3} y^{3}) \cdot {(- 9)}^{1} + 6 \cdot {(2 y)}^{2} \cdot {(- 9)}^{2} + 4 \cdot {(2 y)}^{1} \cdot {(- 9)}^{3} + 1 \cdot {(2 y)}^{0} \cdot {(- 9)}^{4}$

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

$16 y^{4} + 4 \cdot (8 y^{3}) \cdot {(- 9)}^{1} + 6 \cdot {(2 y)}^{2} \cdot {(- 9)}^{2} + 4 \cdot {(2 y)}^{1} \cdot {(- 9)}^{3} + 1 \cdot {(2 y)}^{0} \cdot {(- 9)}^{4}$

Multiply $8$ by $4$ .

$16 y^{4} + 32 y^{3} \cdot {(- 9)}^{1} + 6 \cdot {(2 y)}^{2} \cdot {(- 9)}^{2} + 4 \cdot {(2 y)}^{1} \cdot {(- 9)}^{3} + 1 \cdot {(2 y)}^{0} \cdot {(- 9)}^{4}$

Evaluate the exponent.

$16 y^{4} + 32 y^{3} \cdot - 9 + 6 \cdot {(2 y)}^{2} \cdot {(- 9)}^{2} + 4 \cdot {(2 y)}^{1} \cdot {(- 9)}^{3} + 1 \cdot {(2 y)}^{0} \cdot {(- 9)}^{4}$

Multiply $- 9$ by $32$ .

$16 y^{4} - 288 y^{3} + 6 \cdot {(2 y)}^{2} \cdot {(- 9)}^{2} + 4 \cdot {(2 y)}^{1} \cdot {(- 9)}^{3} + 1 \cdot {(2 y)}^{0} \cdot {(- 9)}^{4}$

Apply the product rule to $2 y$ .

$16 y^{4} - 288 y^{3} + 6 \cdot (2^{2} y^{2}) \cdot {(- 9)}^{2} + 4 \cdot {(2 y)}^{1} \cdot {(- 9)}^{3} + 1 \cdot {(2 y)}^{0} \cdot {(- 9)}^{4}$

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

$16 y^{4} - 288 y^{3} + 6 \cdot (4 y^{2}) \cdot {(- 9)}^{2} + 4 \cdot {(2 y)}^{1} \cdot {(- 9)}^{3} + 1 \cdot {(2 y)}^{0} \cdot {(- 9)}^{4}$

Multiply $4$ by $6$ .

$16 y^{4} - 288 y^{3} + 24 y^{2} \cdot {(- 9)}^{2} + 4 \cdot {(2 y)}^{1} \cdot {(- 9)}^{3} + 1 \cdot {(2 y)}^{0} \cdot {(- 9)}^{4}$

Raise $- 9$ to the power of $2$ .

$16 y^{4} - 288 y^{3} + 24 y^{2} \cdot 81 + 4 \cdot {(2 y)}^{1} \cdot {(- 9)}^{3} + 1 \cdot {(2 y)}^{0} \cdot {(- 9)}^{4}$

Multiply $81$ by $24$ .

$16 y^{4} - 288 y^{3} + 1944 y^{2} + 4 \cdot {(2 y)}^{1} \cdot {(- 9)}^{3} + 1 \cdot {(2 y)}^{0} \cdot {(- 9)}^{4}$

Simplify.

$16 y^{4} - 288 y^{3} + 1944 y^{2} + 4 \cdot (2 y) \cdot {(- 9)}^{3} + 1 \cdot {(2 y)}^{0} \cdot {(- 9)}^{4}$

Multiply $2$ by $4$ .

$16 y^{4} - 288 y^{3} + 1944 y^{2} + 8 y \cdot {(- 9)}^{3} + 1 \cdot {(2 y)}^{0} \cdot {(- 9)}^{4}$

Raise $- 9$ to the power of $3$ .

$16 y^{4} - 288 y^{3} + 1944 y^{2} + 8 y \cdot - 729 + 1 \cdot {(2 y)}^{0} \cdot {(- 9)}^{4}$

Multiply $- 729$ by $8$ .

$16 y^{4} - 288 y^{3} + 1944 y^{2} - 5832 y + 1 \cdot {(2 y)}^{0} \cdot {(- 9)}^{4}$

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

$16 y^{4} - 288 y^{3} + 1944 y^{2} - 5832 y + {(2 y)}^{0} \cdot {(- 9)}^{4}$

Apply the product rule to $2 y$ .

$16 y^{4} - 288 y^{3} + 1944 y^{2} - 5832 y + 2^{0} y^{0} \cdot {(- 9)}^{4}$

Anything raised to $0$ is $1$ .

$16 y^{4} - 288 y^{3} + 1944 y^{2} - 5832 y + 1 y^{0} \cdot {(- 9)}^{4}$

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

$16 y^{4} - 288 y^{3} + 1944 y^{2} - 5832 y + y^{0} \cdot {(- 9)}^{4}$

Anything raised to $0$ is $1$ .

$16 y^{4} - 288 y^{3} + 1944 y^{2} - 5832 y + 1 \cdot {(- 9)}^{4}$

Multiply ${(- 9)}^{4}$ by $1$ .

$16 y^{4} - 288 y^{3} + 1944 y^{2} - 5832 y + {(- 9)}^{4}$

Raise $- 9$ to the power of $4$ .

$16 y^{4} - 288 y^{3} + 1944 y^{2} - 5832 y + 6561$

Expand Using the Binomial Theorem (2y-9)^4

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