Expand Using Pascal's Triangle (2t+4)^3

${(2 t + 4)}^{3}$

Step 1

Pascal\'s Triangle can be displayed as such:

$1$

$1 - 1$

$1 - 2 - 1$

$1 - 3 - 3 - 1$

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 ${(2 t + 4)}^{3}$ , $n = 3$ so the coefficients of the expansion will correspond with line $4$ .

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 $1 - 3 - 3 - 1$ .

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

Step 3

Substitute the actual values of $a$ $2 t$ and $b$ $4$ into the expression.

$1 {(2 t)}^{3} {(4)}^{0} + 3 {(2 t)}^{2} {(4)}^{1} + 3 {(2 t)}^{1} {(4)}^{2} + 1 {(2 t)}^{0} {(4)}^{3}$

Step 4

Simplify each term.

Multiply ${(2 t)}^{3}$ by $1$ .

${(2 t)}^{3} {(4)}^{0} + 3 {(2 t)}^{2} {(4)}^{1} + 3 {(2 t)}^{1} {(4)}^{2} + 1 {(2 t)}^{0} {(4)}^{3}$

Apply the product rule to $2 t$ .

$2^{3} t^{3} {(4)}^{0} + 3 {(2 t)}^{2} {(4)}^{1} + 3 {(2 t)}^{1} {(4)}^{2} + 1 {(2 t)}^{0} {(4)}^{3}$

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

$8 t^{3} {(4)}^{0} + 3 {(2 t)}^{2} {(4)}^{1} + 3 {(2 t)}^{1} {(4)}^{2} + 1 {(2 t)}^{0} {(4)}^{3}$

Anything raised to $0$ is $1$ .

$8 t^{3} \cdot 1 + 3 {(2 t)}^{2} {(4)}^{1} + 3 {(2 t)}^{1} {(4)}^{2} + 1 {(2 t)}^{0} {(4)}^{3}$

Multiply $8$ by $1$ .

$8 t^{3} + 3 {(2 t)}^{2} {(4)}^{1} + 3 {(2 t)}^{1} {(4)}^{2} + 1 {(2 t)}^{0} {(4)}^{3}$

Apply the product rule to $2 t$ .

$8 t^{3} + 3 (2^{2} t^{2}) {(4)}^{1} + 3 {(2 t)}^{1} {(4)}^{2} + 1 {(2 t)}^{0} {(4)}^{3}$

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

$8 t^{3} + 3 (4 t^{2}) {(4)}^{1} + 3 {(2 t)}^{1} {(4)}^{2} + 1 {(2 t)}^{0} {(4)}^{3}$

Multiply $4$ by $3$ .

$8 t^{3} + 12 t^{2} {(4)}^{1} + 3 {(2 t)}^{1} {(4)}^{2} + 1 {(2 t)}^{0} {(4)}^{3}$

Evaluate the exponent.

$8 t^{3} + 12 t^{2} \cdot 4 + 3 {(2 t)}^{1} {(4)}^{2} + 1 {(2 t)}^{0} {(4)}^{3}$

Multiply $4$ by $12$ .

$8 t^{3} + 48 t^{2} + 3 {(2 t)}^{1} {(4)}^{2} + 1 {(2 t)}^{0} {(4)}^{3}$

Simplify.

$8 t^{3} + 48 t^{2} + 3 (2 t) {(4)}^{2} + 1 {(2 t)}^{0} {(4)}^{3}$

Multiply $2$ by $3$ .

$8 t^{3} + 48 t^{2} + 6 t {(4)}^{2} + 1 {(2 t)}^{0} {(4)}^{3}$

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

$8 t^{3} + 48 t^{2} + 6 t \cdot 16 + 1 {(2 t)}^{0} {(4)}^{3}$

Multiply $16$ by $6$ .

$8 t^{3} + 48 t^{2} + 96 t + 1 {(2 t)}^{0} {(4)}^{3}$

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

$8 t^{3} + 48 t^{2} + 96 t + {(2 t)}^{0} {(4)}^{3}$

Apply the product rule to $2 t$ .

$8 t^{3} + 48 t^{2} + 96 t + 2^{0} t^{0} {(4)}^{3}$

Anything raised to $0$ is $1$ .

$8 t^{3} + 48 t^{2} + 96 t + 1 t^{0} {(4)}^{3}$

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

$8 t^{3} + 48 t^{2} + 96 t + t^{0} {(4)}^{3}$

Anything raised to $0$ is $1$ .

$8 t^{3} + 48 t^{2} + 96 t + 1 {(4)}^{3}$

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

$8 t^{3} + 48 t^{2} + 96 t + {(4)}^{3}$

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

$8 t^{3} + 48 t^{2} + 96 t + 64$

Expand Using Pascal's Triangle (2t+4)^3

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