Expand Using the Binomial Theorem (3x-2)^9

${(3 x - 2)}^{9}$

Step 1

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

$\sum_{k = 0}^{9} \frac{9!}{(9 - k)! k!} \cdot {(3 x)}^{9 - k} \cdot {(- 2)}^{k}$

Step 2

Expand the summation.

$\frac{9!}{(9 - 0)! 0!} \cdot {(3 x)}^{9 - 0} \cdot {(- 2)}^{0} + \frac{9!}{(9 - 1)! 1!} \cdot {(3 x)}^{9 - 1} \cdot {(- 2)}^{1} + \dots + \frac{9!}{(9 - 9)! 9!} \cdot {(3 x)}^{9 - 9} \cdot {(- 2)}^{9}$

Step 3

Simplify the exponents for each term of the expansion.

$1 \cdot {(3 x)}^{9} \cdot {(- 2)}^{0} + 9 \cdot {(3 x)}^{8} \cdot {(- 2)}^{1} + \dots + 1 \cdot {(3 x)}^{0} \cdot {(- 2)}^{9}$

Step 4

Simplify the polynomial result.

$19683 x^{9} - 118098 x^{8} + 314928 x^{7} - 489888 x^{6} + 489888 x^{5} - 326592 x^{4} + 145152 x^{3} - 41472 x^{2} + 6912 x - 512$

Expand Using the Binomial Theorem (3x-2)^9

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