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

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

Step 1

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

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

Step 2

Expand the summation.

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

Step 3

Simplify the exponents for each term of the expansion.

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

Step 4

Simplify the polynomial result.

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

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

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