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

${(2 q - 3)}^{4}$

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 ${(2 q - 3)}^{4}$ , $n = 4$ 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 $a$ $2 q$ and $b$ $- 3$ into the expression.

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

Step 4

Simplify each term.

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

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

Apply the product rule to $2 q$ .

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

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

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

Anything raised to $0$ is $1$ .

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

Multiply $16$ by $1$ .

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

Apply the product rule to $2 q$ .

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

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

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

Multiply $8$ by $4$ .

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

Evaluate the exponent.

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

Multiply $- 3$ by $32$ .

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

Apply the product rule to $2 q$ .

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

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

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

Multiply $4$ by $6$ .

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

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

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

Multiply $9$ by $24$ .

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

Simplify.

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

Multiply $2$ by $4$ .

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

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

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

Multiply $- 27$ by $8$ .

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

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

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

Apply the product rule to $2 q$ .

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

Anything raised to $0$ is $1$ .

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

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

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

Anything raised to $0$ is $1$ .

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

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

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

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

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

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

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