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

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

Step 1

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

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

Step 2

Expand the summation.

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

Step 3

Simplify the exponents for each term of the expansion.

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

Step 4

Simplify each term.

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

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

Apply the product rule to $- z$ .

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

Rewrite using the commutative property of multiplication.

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

Anything raised to $0$ is $1$ .

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

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

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

Anything raised to $0$ is $1$ .

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

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

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

Simplify.

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

Rewrite using the commutative property of multiplication.

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

Multiply $4$ by $- 1$ .

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

Apply the product rule to $- z$ .

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

Rewrite using the commutative property of multiplication.

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

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

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

Multiply $6$ by $1$ .

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

Simplify.

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

Apply the product rule to $- z$ .

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

Rewrite using the commutative property of multiplication.

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

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

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

Multiply $4$ by $- 1$ .

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

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

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

Anything raised to $0$ is $1$ .

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

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

$y^{4} - 4 y^{3} z + 6 y^{2} z^{2} - 4 y z^{3} + {(- z)}^{4}$

Apply the product rule to $- z$ .

$y^{4} - 4 y^{3} z + 6 y^{2} z^{2} - 4 y z^{3} + {(- 1)}^{4} z^{4}$

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

$y^{4} - 4 y^{3} z + 6 y^{2} z^{2} - 4 y z^{3} + 1 z^{4}$

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

$y^{4} - 4 y^{3} z + 6 y^{2} z^{2} - 4 y z^{3} + z^{4}$

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

Related Questions

Support LCMGCF.com by disabling your adblocker.