Expand Using the Binomial Theorem (3x+2y)^4

${(3 x + 2 y)}^{4}$

Step 1

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

$\sum_{k = 0}^{4} \frac{4!}{(4 - k)! k!} \cdot {(3 x)}^{4 - k} \cdot {(2 y)}^{k}$

Step 2

Expand the summation.

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

Step 3

Simplify the exponents for each term of the expansion.

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

Step 4

Simplify each term.

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

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

Apply the product rule to $3 x$ .

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

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

$81 x^{4} \cdot {(2 y)}^{0} + 4 \cdot {(3 x)}^{3} \cdot {(2 y)}^{1} + 6 \cdot {(3 x)}^{2} \cdot {(2 y)}^{2} + 4 \cdot {(3 x)}^{1} \cdot {(2 y)}^{3} + 1 \cdot {(3 x)}^{0} \cdot {(2 y)}^{4}$

Apply the product rule to $2 y$ .

$81 x^{4} \cdot (2^{0} y^{0}) + 4 \cdot {(3 x)}^{3} \cdot {(2 y)}^{1} + 6 \cdot {(3 x)}^{2} \cdot {(2 y)}^{2} + 4 \cdot {(3 x)}^{1} \cdot {(2 y)}^{3} + 1 \cdot {(3 x)}^{0} \cdot {(2 y)}^{4}$

Rewrite using the commutative property of multiplication.

$81 \cdot 2^{0} x^{4} y^{0} + 4 \cdot {(3 x)}^{3} \cdot {(2 y)}^{1} + 6 \cdot {(3 x)}^{2} \cdot {(2 y)}^{2} + 4 \cdot {(3 x)}^{1} \cdot {(2 y)}^{3} + 1 \cdot {(3 x)}^{0} \cdot {(2 y)}^{4}$

Anything raised to $0$ is $1$ .

$81 \cdot 1 x^{4} y^{0} + 4 \cdot {(3 x)}^{3} \cdot {(2 y)}^{1} + 6 \cdot {(3 x)}^{2} \cdot {(2 y)}^{2} + 4 \cdot {(3 x)}^{1} \cdot {(2 y)}^{3} + 1 \cdot {(3 x)}^{0} \cdot {(2 y)}^{4}$

Multiply $81$ by $1$ .

$81 x^{4} y^{0} + 4 \cdot {(3 x)}^{3} \cdot {(2 y)}^{1} + 6 \cdot {(3 x)}^{2} \cdot {(2 y)}^{2} + 4 \cdot {(3 x)}^{1} \cdot {(2 y)}^{3} + 1 \cdot {(3 x)}^{0} \cdot {(2 y)}^{4}$

Anything raised to $0$ is $1$ .

$81 x^{4} \cdot 1 + 4 \cdot {(3 x)}^{3} \cdot {(2 y)}^{1} + 6 \cdot {(3 x)}^{2} \cdot {(2 y)}^{2} + 4 \cdot {(3 x)}^{1} \cdot {(2 y)}^{3} + 1 \cdot {(3 x)}^{0} \cdot {(2 y)}^{4}$

Multiply $81$ by $1$ .

$81 x^{4} + 4 \cdot {(3 x)}^{3} \cdot {(2 y)}^{1} + 6 \cdot {(3 x)}^{2} \cdot {(2 y)}^{2} + 4 \cdot {(3 x)}^{1} \cdot {(2 y)}^{3} + 1 \cdot {(3 x)}^{0} \cdot {(2 y)}^{4}$

Apply the product rule to $3 x$ .

$81 x^{4} + 4 \cdot (3^{3} x^{3}) \cdot {(2 y)}^{1} + 6 \cdot {(3 x)}^{2} \cdot {(2 y)}^{2} + 4 \cdot {(3 x)}^{1} \cdot {(2 y)}^{3} + 1 \cdot {(3 x)}^{0} \cdot {(2 y)}^{4}$

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

$81 x^{4} + 4 \cdot (27 x^{3}) \cdot {(2 y)}^{1} + 6 \cdot {(3 x)}^{2} \cdot {(2 y)}^{2} + 4 \cdot {(3 x)}^{1} \cdot {(2 y)}^{3} + 1 \cdot {(3 x)}^{0} \cdot {(2 y)}^{4}$

Multiply $27$ by $4$ .

$81 x^{4} + 108 x^{3} \cdot {(2 y)}^{1} + 6 \cdot {(3 x)}^{2} \cdot {(2 y)}^{2} + 4 \cdot {(3 x)}^{1} \cdot {(2 y)}^{3} + 1 \cdot {(3 x)}^{0} \cdot {(2 y)}^{4}$

Simplify.

$81 x^{4} + 108 x^{3} \cdot (2 y) + 6 \cdot {(3 x)}^{2} \cdot {(2 y)}^{2} + 4 \cdot {(3 x)}^{1} \cdot {(2 y)}^{3} + 1 \cdot {(3 x)}^{0} \cdot {(2 y)}^{4}$

Rewrite using the commutative property of multiplication.

$81 x^{4} + 108 \cdot 2 x^{3} y + 6 \cdot {(3 x)}^{2} \cdot {(2 y)}^{2} + 4 \cdot {(3 x)}^{1} \cdot {(2 y)}^{3} + 1 \cdot {(3 x)}^{0} \cdot {(2 y)}^{4}$

Multiply $108$ by $2$ .

$81 x^{4} + 216 x^{3} y + 6 \cdot {(3 x)}^{2} \cdot {(2 y)}^{2} + 4 \cdot {(3 x)}^{1} \cdot {(2 y)}^{3} + 1 \cdot {(3 x)}^{0} \cdot {(2 y)}^{4}$

Apply the product rule to $3 x$ .

$81 x^{4} + 216 x^{3} y + 6 \cdot (3^{2} x^{2}) \cdot {(2 y)}^{2} + 4 \cdot {(3 x)}^{1} \cdot {(2 y)}^{3} + 1 \cdot {(3 x)}^{0} \cdot {(2 y)}^{4}$

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

$81 x^{4} + 216 x^{3} y + 6 \cdot (9 x^{2}) \cdot {(2 y)}^{2} + 4 \cdot {(3 x)}^{1} \cdot {(2 y)}^{3} + 1 \cdot {(3 x)}^{0} \cdot {(2 y)}^{4}$

Multiply $9$ by $6$ .

$81 x^{4} + 216 x^{3} y + 54 x^{2} \cdot {(2 y)}^{2} + 4 \cdot {(3 x)}^{1} \cdot {(2 y)}^{3} + 1 \cdot {(3 x)}^{0} \cdot {(2 y)}^{4}$

Apply the product rule to $2 y$ .

$81 x^{4} + 216 x^{3} y + 54 x^{2} \cdot (2^{2} y^{2}) + 4 \cdot {(3 x)}^{1} \cdot {(2 y)}^{3} + 1 \cdot {(3 x)}^{0} \cdot {(2 y)}^{4}$

Rewrite using the commutative property of multiplication.

$81 x^{4} + 216 x^{3} y + 54 \cdot 2^{2} x^{2} y^{2} + 4 \cdot {(3 x)}^{1} \cdot {(2 y)}^{3} + 1 \cdot {(3 x)}^{0} \cdot {(2 y)}^{4}$

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

$81 x^{4} + 216 x^{3} y + 54 \cdot 4 x^{2} y^{2} + 4 \cdot {(3 x)}^{1} \cdot {(2 y)}^{3} + 1 \cdot {(3 x)}^{0} \cdot {(2 y)}^{4}$

Multiply $54$ by $4$ .

$81 x^{4} + 216 x^{3} y + 216 x^{2} y^{2} + 4 \cdot {(3 x)}^{1} \cdot {(2 y)}^{3} + 1 \cdot {(3 x)}^{0} \cdot {(2 y)}^{4}$

Simplify.

$81 x^{4} + 216 x^{3} y + 216 x^{2} y^{2} + 4 \cdot (3 x) \cdot {(2 y)}^{3} + 1 \cdot {(3 x)}^{0} \cdot {(2 y)}^{4}$

Multiply $3$ by $4$ .

$81 x^{4} + 216 x^{3} y + 216 x^{2} y^{2} + 12 x \cdot {(2 y)}^{3} + 1 \cdot {(3 x)}^{0} \cdot {(2 y)}^{4}$

Apply the product rule to $2 y$ .

$81 x^{4} + 216 x^{3} y + 216 x^{2} y^{2} + 12 x \cdot (2^{3} y^{3}) + 1 \cdot {(3 x)}^{0} \cdot {(2 y)}^{4}$

Rewrite using the commutative property of multiplication.

$81 x^{4} + 216 x^{3} y + 216 x^{2} y^{2} + 12 \cdot 2^{3} x y^{3} + 1 \cdot {(3 x)}^{0} \cdot {(2 y)}^{4}$

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

$81 x^{4} + 216 x^{3} y + 216 x^{2} y^{2} + 12 \cdot 8 x y^{3} + 1 \cdot {(3 x)}^{0} \cdot {(2 y)}^{4}$

Multiply $12$ by $8$ .

$81 x^{4} + 216 x^{3} y + 216 x^{2} y^{2} + 96 x y^{3} + 1 \cdot {(3 x)}^{0} \cdot {(2 y)}^{4}$

Multiply ${(3 x)}^{0}$ by $1$ .

$81 x^{4} + 216 x^{3} y + 216 x^{2} y^{2} + 96 x y^{3} + {(3 x)}^{0} \cdot {(2 y)}^{4}$

Apply the product rule to $3 x$ .

$81 x^{4} + 216 x^{3} y + 216 x^{2} y^{2} + 96 x y^{3} + 3^{0} x^{0} \cdot {(2 y)}^{4}$

Anything raised to $0$ is $1$ .

$81 x^{4} + 216 x^{3} y + 216 x^{2} y^{2} + 96 x y^{3} + 1 x^{0} \cdot {(2 y)}^{4}$

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

$81 x^{4} + 216 x^{3} y + 216 x^{2} y^{2} + 96 x y^{3} + x^{0} \cdot {(2 y)}^{4}$

Anything raised to $0$ is $1$ .

$81 x^{4} + 216 x^{3} y + 216 x^{2} y^{2} + 96 x y^{3} + 1 \cdot {(2 y)}^{4}$

Multiply ${(2 y)}^{4}$ by $1$ .

$81 x^{4} + 216 x^{3} y + 216 x^{2} y^{2} + 96 x y^{3} + {(2 y)}^{4}$

Apply the product rule to $2 y$ .

$81 x^{4} + 216 x^{3} y + 216 x^{2} y^{2} + 96 x y^{3} + 2^{4} y^{4}$

Raise $2$ to the power of $4$ .

$81 x^{4} + 216 x^{3} y + 216 x^{2} y^{2} + 96 x y^{3} + 16 y^{4}$

Expand Using the Binomial Theorem (3x+2y)^4

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