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

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

Step 1

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

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

Step 2

Expand the summation.

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

Step 3

Simplify the exponents for each term of the expansion.

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

Step 4

Simplify each term.

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

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

Anything raised to $0$ is $1$ .

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

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

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

Evaluate the exponent.

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

Multiply $- 2$ by $4$ .

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

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

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

Multiply $4$ by $6$ .

$y^{4} - 8 y^{3} + 24 y^{2} + 4 \cdot {(y)}^{1} \cdot {(- 2)}^{3} + 1 \cdot {(y)}^{0} \cdot {(- 2)}^{4}$

Simplify.

$y^{4} - 8 y^{3} + 24 y^{2} + 4 \cdot y \cdot {(- 2)}^{3} + 1 \cdot {(y)}^{0} \cdot {(- 2)}^{4}$

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

$y^{4} - 8 y^{3} + 24 y^{2} + 4 y \cdot - 8 + 1 \cdot {(y)}^{0} \cdot {(- 2)}^{4}$

Multiply $- 8$ by $4$ .

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

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

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

Anything raised to $0$ is $1$ .

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

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

$y^{4} - 8 y^{3} + 24 y^{2} - 32 y + {(- 2)}^{4}$

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

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

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

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