Expand Using Pascal's Triangle (2z-3k)^4

${(2 z - 3 k)}^{4}$

Step 1

Pascal\'s Triangle can be displayed as such:

$1$

$1 - 1$

$1 - 2 - 1$

$1 - 3 - 3 - 1$

$1 - 4 - 6 - 4 - 1$

The triangle can be used to calculate the coefficients of the expansion of ${(a + b)}^{n}$ by taking the exponent $n$ and adding $1$ . The coefficients will correspond with line $n + 1$ of the triangle. For ${(2 z - 3 k)}^{4}$ , $n = 4$ so the coefficients of the expansion will correspond with line $5$ .

Step 2

The expansion follows the rule ${(a + b)}^{n} = c_{0} a^{n} b^{0} + c_{1} a^{n - 1} b^{1} + c_{n - 1} a^{1} b^{n - 1} + c_{n} a^{0} b^{n}$ . The values of the coefficients, from the triangle, are $1 - 4 - 6 - 4 - 1$ .

$1 a^{4} b^{0} + 4 a^{3} b + 6 a^{2} b^{2} + 4 a b^{3} + 1 a^{0} b^{4}$

Step 3

Substitute the actual values of $a$ $2 z$ and $b$ $- 3 k$ into the expression.

$1 {(2 z)}^{4} {(- 3 k)}^{0} + 4 {(2 z)}^{3} {(- 3 k)}^{1} + 6 {(2 z)}^{2} {(- 3 k)}^{2} + 4 {(2 z)}^{1} {(- 3 k)}^{3} + 1 {(2 z)}^{0} {(- 3 k)}^{4}$

Step 4

Simplify each term.

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

${(2 z)}^{4} {(- 3 k)}^{0} + 4 {(2 z)}^{3} {(- 3 k)}^{1} + 6 {(2 z)}^{2} {(- 3 k)}^{2} + 4 {(2 z)}^{1} {(- 3 k)}^{3} + 1 {(2 z)}^{0} {(- 3 k)}^{4}$

Apply the product rule to $2 z$ .

$2^{4} z^{4} {(- 3 k)}^{0} + 4 {(2 z)}^{3} {(- 3 k)}^{1} + 6 {(2 z)}^{2} {(- 3 k)}^{2} + 4 {(2 z)}^{1} {(- 3 k)}^{3} + 1 {(2 z)}^{0} {(- 3 k)}^{4}$

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

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

Apply the product rule to $- 3 k$ .

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

Rewrite using the commutative property of multiplication.

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

Anything raised to $0$ is $1$ .

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

Multiply $16$ by $1$ .

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

Anything raised to $0$ is $1$ .

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

Multiply $16$ by $1$ .

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

Apply the product rule to $2 z$ .

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

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

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

Multiply $8$ by $4$ .

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

Simplify.

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

Rewrite using the commutative property of multiplication.

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

Multiply $32$ by $- 3$ .

$16 z^{4} - 96 z^{3} k + 6 {(2 z)}^{2} {(- 3 k)}^{2} + 4 {(2 z)}^{1} {(- 3 k)}^{3} + 1 {(2 z)}^{0} {(- 3 k)}^{4}$

Apply the product rule to $2 z$ .

$16 z^{4} - 96 z^{3} k + 6 (2^{2} z^{2}) {(- 3 k)}^{2} + 4 {(2 z)}^{1} {(- 3 k)}^{3} + 1 {(2 z)}^{0} {(- 3 k)}^{4}$

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

$16 z^{4} - 96 z^{3} k + 6 (4 z^{2}) {(- 3 k)}^{2} + 4 {(2 z)}^{1} {(- 3 k)}^{3} + 1 {(2 z)}^{0} {(- 3 k)}^{4}$

Multiply $4$ by $6$ .

$16 z^{4} - 96 z^{3} k + 24 z^{2} {(- 3 k)}^{2} + 4 {(2 z)}^{1} {(- 3 k)}^{3} + 1 {(2 z)}^{0} {(- 3 k)}^{4}$

Apply the product rule to $- 3 k$ .

$16 z^{4} - 96 z^{3} k + 24 z^{2} ({(- 3)}^{2} k^{2}) + 4 {(2 z)}^{1} {(- 3 k)}^{3} + 1 {(2 z)}^{0} {(- 3 k)}^{4}$

Rewrite using the commutative property of multiplication.

$16 z^{4} - 96 z^{3} k + 24 \cdot {(- 3)}^{2} z^{2} k^{2} + 4 {(2 z)}^{1} {(- 3 k)}^{3} + 1 {(2 z)}^{0} {(- 3 k)}^{4}$

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

$16 z^{4} - 96 z^{3} k + 24 \cdot 9 z^{2} k^{2} + 4 {(2 z)}^{1} {(- 3 k)}^{3} + 1 {(2 z)}^{0} {(- 3 k)}^{4}$

Multiply $24$ by $9$ .

$16 z^{4} - 96 z^{3} k + 216 z^{2} k^{2} + 4 {(2 z)}^{1} {(- 3 k)}^{3} + 1 {(2 z)}^{0} {(- 3 k)}^{4}$

Simplify.

$16 z^{4} - 96 z^{3} k + 216 z^{2} k^{2} + 4 (2 z) {(- 3 k)}^{3} + 1 {(2 z)}^{0} {(- 3 k)}^{4}$

Multiply $2$ by $4$ .

$16 z^{4} - 96 z^{3} k + 216 z^{2} k^{2} + 8 z {(- 3 k)}^{3} + 1 {(2 z)}^{0} {(- 3 k)}^{4}$

Apply the product rule to $- 3 k$ .

$16 z^{4} - 96 z^{3} k + 216 z^{2} k^{2} + 8 z ({(- 3)}^{3} k^{3}) + 1 {(2 z)}^{0} {(- 3 k)}^{4}$

Rewrite using the commutative property of multiplication.

$16 z^{4} - 96 z^{3} k + 216 z^{2} k^{2} + 8 \cdot {(- 3)}^{3} z k^{3} + 1 {(2 z)}^{0} {(- 3 k)}^{4}$

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

$16 z^{4} - 96 z^{3} k + 216 z^{2} k^{2} + 8 \cdot - 27 z k^{3} + 1 {(2 z)}^{0} {(- 3 k)}^{4}$

Multiply $8$ by $- 27$ .

$16 z^{4} - 96 z^{3} k + 216 z^{2} k^{2} - 216 z k^{3} + 1 {(2 z)}^{0} {(- 3 k)}^{4}$

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

$16 z^{4} - 96 z^{3} k + 216 z^{2} k^{2} - 216 z k^{3} + {(2 z)}^{0} {(- 3 k)}^{4}$

Apply the product rule to $2 z$ .

$16 z^{4} - 96 z^{3} k + 216 z^{2} k^{2} - 216 z k^{3} + 2^{0} z^{0} {(- 3 k)}^{4}$

Anything raised to $0$ is $1$ .

$16 z^{4} - 96 z^{3} k + 216 z^{2} k^{2} - 216 z k^{3} + 1 z^{0} {(- 3 k)}^{4}$

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

$16 z^{4} - 96 z^{3} k + 216 z^{2} k^{2} - 216 z k^{3} + z^{0} {(- 3 k)}^{4}$

Anything raised to $0$ is $1$ .

$16 z^{4} - 96 z^{3} k + 216 z^{2} k^{2} - 216 z k^{3} + 1 {(- 3 k)}^{4}$

Multiply ${(- 3 k)}^{4}$ by $1$ .

$16 z^{4} - 96 z^{3} k + 216 z^{2} k^{2} - 216 z k^{3} + {(- 3 k)}^{4}$

Apply the product rule to $- 3 k$ .

$16 z^{4} - 96 z^{3} k + 216 z^{2} k^{2} - 216 z k^{3} + {(- 3)}^{4} k^{4}$

Raise $- 3$ to the power of $4$ .

$16 z^{4} - 96 z^{3} k + 216 z^{2} k^{2} - 216 z k^{3} + 81 k^{4}$

Expand Using Pascal's Triangle (2z-3k)^4

Related Questions

Support LCMGCF.com by disabling your adblocker.