Expand Using the Binomial Theorem (4x-1)^5

${(4 x - 1)}^{5}$

Step 1

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

$\sum_{k = 0}^{5} \frac{5!}{(5 - k)! k!} \cdot {(4 x)}^{5 - k} \cdot {(- 1)}^{k}$

Step 2

Expand the summation.

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

Step 3

Simplify the exponents for each term of the expansion.

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

Step 4

Simplify the polynomial result.

$1024 x^{5} - 1280 x^{4} + 640 x^{3} - 160 x^{2} + 20 x - 1$

Expand Using the Binomial Theorem (4x-1)^5

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