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

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

Step 1

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

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

Step 2

Expand the summation.

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

Step 3

Simplify the exponents for each term of the expansion.

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

Step 4

Simplify each term.

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

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

Apply the product rule to $2 x$ .

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

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

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

Anything raised to $0$ is $1$ .

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

Multiply $16$ by $1$ .

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

Apply the product rule to $2 x$ .

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

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

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

Multiply $8$ by $4$ .

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

Evaluate the exponent.

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

Multiply $- 3$ by $32$ .

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

Apply the product rule to $2 x$ .

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

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

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

Multiply $4$ by $6$ .

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

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

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

Multiply $9$ by $24$ .

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

Simplify.

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

Multiply $2$ by $4$ .

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

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

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

Multiply $- 27$ by $8$ .

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

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

$16 x^{4} - 96 x^{3} + 216 x^{2} - 216 x + {(2 x)}^{0} \cdot {(- 3)}^{4}$

Apply the product rule to $2 x$ .

$16 x^{4} - 96 x^{3} + 216 x^{2} - 216 x + 2^{0} x^{0} \cdot {(- 3)}^{4}$

Anything raised to $0$ is $1$ .

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

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

$16 x^{4} - 96 x^{3} + 216 x^{2} - 216 x + x^{0} \cdot {(- 3)}^{4}$

Anything raised to $0$ is $1$ .

$16 x^{4} - 96 x^{3} + 216 x^{2} - 216 x + 1 \cdot {(- 3)}^{4}$

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

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

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

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

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

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