Expand Using the Binomial Theorem (2x-2)^5

${(2 x - 2)}^{5}$

Step 1

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

$\sum_{k = 0}^{5} \frac{5!}{(5 - k)! k!} \cdot {(2 x)}^{5 - k} \cdot {(- 2)}^{k}$

Step 2

Expand the summation.

$\frac{5!}{(5 - 0)! 0!} \cdot {(2 x)}^{5 - 0} \cdot {(- 2)}^{0} + \frac{5!}{(5 - 1)! 1!} \cdot {(2 x)}^{5 - 1} \cdot {(- 2)}^{1} + \dots + \frac{5!}{(5 - 5)! 5!} \cdot {(2 x)}^{5 - 5} \cdot {(- 2)}^{5}$

Step 3

Simplify the exponents for each term of the expansion.

$1 \cdot {(2 x)}^{5} \cdot {(- 2)}^{0} + 5 \cdot {(2 x)}^{4} \cdot {(- 2)}^{1} + \dots + 1 \cdot {(2 x)}^{0} \cdot {(- 2)}^{5}$

Step 4

Simplify the polynomial result.

$32 x^{5} - 160 x^{4} + 320 x^{3} - 320 x^{2} + 160 x - 32$

Expand Using the Binomial Theorem (2x-2)^5

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