Algebra Examples

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

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

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

Expand the summation.

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

Simplify the exponents for each term of the expansion.

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

Simplify the polynomial result.

$32 - 80 x + 80 x^{2} - 40 x^{3} + 10 x^{4} - x^{5}$