Expand Using the Binomial Theorem (2x+4)^3

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

Step 1

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

$\sum_{k = 0}^{3} \frac{3!}{(3 - k)! k!} \cdot {(2 x)}^{3 - k} \cdot {(4)}^{k}$

Step 2

Expand the summation.

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

Step 3

Simplify the exponents for each term of the expansion.

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

Step 4

Simplify each term.

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

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

Apply the product rule to $2 x$ .

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

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

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

Anything raised to $0$ is $1$ .

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

Multiply $8$ by $1$ .

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

Apply the product rule to $2 x$ .

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

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

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

Multiply $4$ by $3$ .

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

Evaluate the exponent.

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

Multiply $4$ by $12$ .

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

Simplify.

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

Multiply $2$ by $3$ .

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

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

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

Multiply $16$ by $6$ .

$8 x^{3} + 48 x^{2} + 96 x + 1 \cdot {(2 x)}^{0} \cdot {(4)}^{3}$

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

$8 x^{3} + 48 x^{2} + 96 x + {(2 x)}^{0} \cdot {(4)}^{3}$

Apply the product rule to $2 x$ .

$8 x^{3} + 48 x^{2} + 96 x + 2^{0} x^{0} \cdot {(4)}^{3}$

Anything raised to $0$ is $1$ .

$8 x^{3} + 48 x^{2} + 96 x + 1 x^{0} \cdot {(4)}^{3}$

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

$8 x^{3} + 48 x^{2} + 96 x + x^{0} \cdot {(4)}^{3}$

Anything raised to $0$ is $1$ .

$8 x^{3} + 48 x^{2} + 96 x + 1 \cdot {(4)}^{3}$

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

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

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

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

Expand Using the Binomial Theorem (2x+4)^3

Related Questions

Support LCMGCF.com by disabling your adblocker.