Algebra Examples

{(x + 2)}^{6}

${(x + 2)}^{6}$

Step 1

Use the binomial expansion theorem to find each term. The binomial theorem states

{(a + b)}^{n} = n \sum k = 0 n C k \cdot (a^{n - k} b^{k})

${(a + b)}^{n} = \sum_{k = 0}^{n} n C k \cdot (a^{n - k} b^{k})$ .

6 \sum k = 0 \frac{6!}{(6 - k)! k!} \cdot {(x)}^{6 - k} \cdot {(2)}^{k}

$\sum_{k = 0}^{6} \frac{6!}{(6 - k)! k!} \cdot {(x)}^{6 - k} \cdot {(2)}^{k}$

Step 2

Expand the summation.

\frac{6!}{(6 - 0)! 0!} {(x)}^{6 - 0} \cdot {(2)}^{0} + \frac{6!}{(6 - 1)! 1!} {(x)}^{6 - 1} \cdot {(2)}^{1} + \dots + \frac{6!}{(6 - 6)! 6!} {(x)}^{6 - 6} \cdot {(2)}^{6}

$\frac{6!}{(6 - 0)! 0!} {(x)}^{6 - 0} \cdot {(2)}^{0} + \frac{6!}{(6 - 1)! 1!} {(x)}^{6 - 1} \cdot {(2)}^{1} + \dots + \frac{6!}{(6 - 6)! 6!} {(x)}^{6 - 6} \cdot {(2)}^{6}$

Step 3

Simplify the exponents for each term of the expansion.

1 \cdot {(x)}^{6} \cdot {(2)}^{0} + 6 \cdot {(x)}^{5} \cdot {(2)}^{1} + \dots + 1 \cdot {(x)}^{0} \cdot {(2)}^{6}

$1 \cdot {(x)}^{6} \cdot {(2)}^{0} + 6 \cdot {(x)}^{5} \cdot {(2)}^{1} + \dots + 1 \cdot {(x)}^{0} \cdot {(2)}^{6}$

Step 4

Simplify the polynomial result.

x^{6} + 12 x^{5} + 60 x^{4} + 160 x^{3} + 240 x^{2} + 192 x + 64

$x^{6} + 12 x^{5} + 60 x^{4} + 160 x^{3} + 240 x^{2} + 192 x + 64$