Expand Using the Binomial Theorem (3x+2)^8

${(3 x + 2)}^{8}$

Step 1

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

$\sum_{k = 0}^{8} \frac{8!}{(8 - k)! k!} \cdot {(3 x)}^{8 - k} \cdot {(2)}^{k}$

Step 2

Expand the summation.

$\frac{8!}{(8 - 0)! 0!} \cdot {(3 x)}^{8 - 0} \cdot {(2)}^{0} + \frac{8!}{(8 - 1)! 1!} \cdot {(3 x)}^{8 - 1} \cdot {(2)}^{1} + \dots + \frac{8!}{(8 - 8)! 8!} \cdot {(3 x)}^{8 - 8} \cdot {(2)}^{8}$

Step 3

Simplify the exponents for each term of the expansion.

$1 \cdot {(3 x)}^{8} \cdot {(2)}^{0} + 8 \cdot {(3 x)}^{7} \cdot {(2)}^{1} + \dots + 1 \cdot {(3 x)}^{0} \cdot {(2)}^{8}$

Step 4

Simplify the polynomial result.

$6561 x^{8} + 34992 x^{7} + 81648 x^{6} + 108864 x^{5} + 90720 x^{4} + 48384 x^{3} + 16128 x^{2} + 3072 x + 256$

Expand Using the Binomial Theorem (3x+2)^8

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