Expand Using the Binomial Theorem (5x+3)^4

${(5 x + 3)}^{4}$

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

$\sum_{k = 0}^{4} \frac{4!}{(4 - k)! k!} \cdot {(5 x)}^{4 - k} \cdot {(3)}^{k}$

Expand the summation.

$\frac{4!}{(4 - 0)! 0!} \cdot {(5 x)}^{4 - 0} \cdot {(3)}^{0} + \frac{4!}{(4 - 1)! 1!} \cdot {(5 x)}^{4 - 1} \cdot {(3)}^{1} + \frac{4!}{(4 - 2)! 2!} \cdot {(5 x)}^{4 - 2} \cdot {(3)}^{2} + \frac{4!}{(4 - 3)! 3!} \cdot {(5 x)}^{4 - 3} \cdot {(3)}^{3} + \frac{4!}{(4 - 4)! 4!} \cdot {(5 x)}^{4 - 4} \cdot {(3)}^{4}$

Simplify the exponents for each term of the expansion.

$1 \cdot {(5 x)}^{4} \cdot {(3)}^{0} + 4 \cdot {(5 x)}^{3} \cdot {(3)}^{1} + 6 \cdot {(5 x)}^{2} \cdot {(3)}^{2} + 4 \cdot {(5 x)}^{1} \cdot {(3)}^{3} + 1 \cdot {(5 x)}^{0} \cdot {(3)}^{4}$

Simplify each term.

Multiply ${(5 x)}^{4}$ by $1$ .

${(5 x)}^{4} \cdot {(3)}^{0} + 4 \cdot {(5 x)}^{3} \cdot {(3)}^{1} + 6 \cdot {(5 x)}^{2} \cdot {(3)}^{2} + 4 \cdot {(5 x)}^{1} \cdot {(3)}^{3} + 1 \cdot {(5 x)}^{0} \cdot {(3)}^{4}$

Apply the product rule to $5 x$ .

$5^{4} x^{4} \cdot {(3)}^{0} + 4 \cdot {(5 x)}^{3} \cdot {(3)}^{1} + 6 \cdot {(5 x)}^{2} \cdot {(3)}^{2} + 4 \cdot {(5 x)}^{1} \cdot {(3)}^{3} + 1 \cdot {(5 x)}^{0} \cdot {(3)}^{4}$

Raise $5$ to the power of $4$ .

$625 x^{4} \cdot {(3)}^{0} + 4 \cdot {(5 x)}^{3} \cdot {(3)}^{1} + 6 \cdot {(5 x)}^{2} \cdot {(3)}^{2} + 4 \cdot {(5 x)}^{1} \cdot {(3)}^{3} + 1 \cdot {(5 x)}^{0} \cdot {(3)}^{4}$

Anything raised to $0$ is $1$ .

$625 x^{4} \cdot 1 + 4 \cdot {(5 x)}^{3} \cdot {(3)}^{1} + 6 \cdot {(5 x)}^{2} \cdot {(3)}^{2} + 4 \cdot {(5 x)}^{1} \cdot {(3)}^{3} + 1 \cdot {(5 x)}^{0} \cdot {(3)}^{4}$

Multiply $625$ by $1$ .

$625 x^{4} + 4 \cdot {(5 x)}^{3} \cdot {(3)}^{1} + 6 \cdot {(5 x)}^{2} \cdot {(3)}^{2} + 4 \cdot {(5 x)}^{1} \cdot {(3)}^{3} + 1 \cdot {(5 x)}^{0} \cdot {(3)}^{4}$

Apply the product rule to $5 x$ .

$625 x^{4} + 4 \cdot (5^{3} x^{3}) \cdot {(3)}^{1} + 6 \cdot {(5 x)}^{2} \cdot {(3)}^{2} + 4 \cdot {(5 x)}^{1} \cdot {(3)}^{3} + 1 \cdot {(5 x)}^{0} \cdot {(3)}^{4}$

Raise $5$ to the power of $3$ .

$625 x^{4} + 4 \cdot (125 x^{3}) \cdot {(3)}^{1} + 6 \cdot {(5 x)}^{2} \cdot {(3)}^{2} + 4 \cdot {(5 x)}^{1} \cdot {(3)}^{3} + 1 \cdot {(5 x)}^{0} \cdot {(3)}^{4}$

Multiply $125$ by $4$ .

$625 x^{4} + 500 x^{3} \cdot {(3)}^{1} + 6 \cdot {(5 x)}^{2} \cdot {(3)}^{2} + 4 \cdot {(5 x)}^{1} \cdot {(3)}^{3} + 1 \cdot {(5 x)}^{0} \cdot {(3)}^{4}$

Evaluate the exponent.

$625 x^{4} + 500 x^{3} \cdot 3 + 6 \cdot {(5 x)}^{2} \cdot {(3)}^{2} + 4 \cdot {(5 x)}^{1} \cdot {(3)}^{3} + 1 \cdot {(5 x)}^{0} \cdot {(3)}^{4}$

Multiply $3$ by $500$ .

$625 x^{4} + 1500 x^{3} + 6 \cdot {(5 x)}^{2} \cdot {(3)}^{2} + 4 \cdot {(5 x)}^{1} \cdot {(3)}^{3} + 1 \cdot {(5 x)}^{0} \cdot {(3)}^{4}$

Apply the product rule to $5 x$ .

$625 x^{4} + 1500 x^{3} + 6 \cdot (5^{2} x^{2}) \cdot {(3)}^{2} + 4 \cdot {(5 x)}^{1} \cdot {(3)}^{3} + 1 \cdot {(5 x)}^{0} \cdot {(3)}^{4}$

Raise $5$ to the power of $2$ .

$625 x^{4} + 1500 x^{3} + 6 \cdot (25 x^{2}) \cdot {(3)}^{2} + 4 \cdot {(5 x)}^{1} \cdot {(3)}^{3} + 1 \cdot {(5 x)}^{0} \cdot {(3)}^{4}$

Multiply $25$ by $6$ .

$625 x^{4} + 1500 x^{3} + 150 x^{2} \cdot {(3)}^{2} + 4 \cdot {(5 x)}^{1} \cdot {(3)}^{3} + 1 \cdot {(5 x)}^{0} \cdot {(3)}^{4}$

Raise $3$ to the power of $2$ .

$625 x^{4} + 1500 x^{3} + 150 x^{2} \cdot 9 + 4 \cdot {(5 x)}^{1} \cdot {(3)}^{3} + 1 \cdot {(5 x)}^{0} \cdot {(3)}^{4}$

Multiply $9$ by $150$ .

$625 x^{4} + 1500 x^{3} + 1350 x^{2} + 4 \cdot {(5 x)}^{1} \cdot {(3)}^{3} + 1 \cdot {(5 x)}^{0} \cdot {(3)}^{4}$

Simplify.

$625 x^{4} + 1500 x^{3} + 1350 x^{2} + 4 \cdot (5 x) \cdot {(3)}^{3} + 1 \cdot {(5 x)}^{0} \cdot {(3)}^{4}$

Multiply $5$ by $4$ .

$625 x^{4} + 1500 x^{3} + 1350 x^{2} + 20 x \cdot {(3)}^{3} + 1 \cdot {(5 x)}^{0} \cdot {(3)}^{4}$

Raise $3$ to the power of $3$ .

$625 x^{4} + 1500 x^{3} + 1350 x^{2} + 20 x \cdot 27 + 1 \cdot {(5 x)}^{0} \cdot {(3)}^{4}$

Multiply $27$ by $20$ .

$625 x^{4} + 1500 x^{3} + 1350 x^{2} + 540 x + 1 \cdot {(5 x)}^{0} \cdot {(3)}^{4}$

Multiply ${(5 x)}^{0}$ by $1$ .

$625 x^{4} + 1500 x^{3} + 1350 x^{2} + 540 x + {(5 x)}^{0} \cdot {(3)}^{4}$

Apply the product rule to $5 x$ .

$625 x^{4} + 1500 x^{3} + 1350 x^{2} + 540 x + 5^{0} x^{0} \cdot {(3)}^{4}$

Anything raised to $0$ is $1$ .

$625 x^{4} + 1500 x^{3} + 1350 x^{2} + 540 x + 1 x^{0} \cdot {(3)}^{4}$

Multiply $x^{0}$ by $1$ .

$625 x^{4} + 1500 x^{3} + 1350 x^{2} + 540 x + x^{0} \cdot {(3)}^{4}$

Anything raised to $0$ is $1$ .

$625 x^{4} + 1500 x^{3} + 1350 x^{2} + 540 x + 1 \cdot {(3)}^{4}$

Multiply ${(3)}^{4}$ by $1$ .

$625 x^{4} + 1500 x^{3} + 1350 x^{2} + 540 x + {(3)}^{4}$

Raise $3$ to the power of $4$ .

$625 x^{4} + 1500 x^{3} + 1350 x^{2} + 540 x + 81$

Expand Using the Binomial Theorem (5x+3)^4

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