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

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

Step 1

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

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

Step 2

Expand the summation.

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

Step 3

Simplify the exponents for each term of the expansion.

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

Step 4

Simplify each term.

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

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

Apply the product rule to $2 y$ .

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

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

$16 y^{4} \cdot {(4 x^{3})}^{0} + 4 \cdot {(2 y)}^{3} \cdot {(4 x^{3})}^{1} + 6 \cdot {(2 y)}^{2} \cdot {(4 x^{3})}^{2} + 4 \cdot {(2 y)}^{1} \cdot {(4 x^{3})}^{3} + 1 \cdot {(2 y)}^{0} \cdot {(4 x^{3})}^{4}$

Apply the product rule to $4 x^{3}$ .

$16 y^{4} \cdot (4^{0} {(x^{3})}^{0}) + 4 \cdot {(2 y)}^{3} \cdot {(4 x^{3})}^{1} + 6 \cdot {(2 y)}^{2} \cdot {(4 x^{3})}^{2} + 4 \cdot {(2 y)}^{1} \cdot {(4 x^{3})}^{3} + 1 \cdot {(2 y)}^{0} \cdot {(4 x^{3})}^{4}$

Rewrite using the commutative property of multiplication.

$16 \cdot 4^{0} y^{4} {(x^{3})}^{0} + 4 \cdot {(2 y)}^{3} \cdot {(4 x^{3})}^{1} + 6 \cdot {(2 y)}^{2} \cdot {(4 x^{3})}^{2} + 4 \cdot {(2 y)}^{1} \cdot {(4 x^{3})}^{3} + 1 \cdot {(2 y)}^{0} \cdot {(4 x^{3})}^{4}$

Anything raised to $0$ is $1$ .

$16 \cdot 1 y^{4} {(x^{3})}^{0} + 4 \cdot {(2 y)}^{3} \cdot {(4 x^{3})}^{1} + 6 \cdot {(2 y)}^{2} \cdot {(4 x^{3})}^{2} + 4 \cdot {(2 y)}^{1} \cdot {(4 x^{3})}^{3} + 1 \cdot {(2 y)}^{0} \cdot {(4 x^{3})}^{4}$

Multiply $16$ by $1$ .

$16 y^{4} {(x^{3})}^{0} + 4 \cdot {(2 y)}^{3} \cdot {(4 x^{3})}^{1} + 6 \cdot {(2 y)}^{2} \cdot {(4 x^{3})}^{2} + 4 \cdot {(2 y)}^{1} \cdot {(4 x^{3})}^{3} + 1 \cdot {(2 y)}^{0} \cdot {(4 x^{3})}^{4}$

Multiply the exponents in ${(x^{3})}^{0}$ .

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$16 y^{4} x^{3 \cdot 0} + 4 \cdot {(2 y)}^{3} \cdot {(4 x^{3})}^{1} + 6 \cdot {(2 y)}^{2} \cdot {(4 x^{3})}^{2} + 4 \cdot {(2 y)}^{1} \cdot {(4 x^{3})}^{3} + 1 \cdot {(2 y)}^{0} \cdot {(4 x^{3})}^{4}$

Multiply $3$ by $0$ .

$16 y^{4} x^{0} + 4 \cdot {(2 y)}^{3} \cdot {(4 x^{3})}^{1} + 6 \cdot {(2 y)}^{2} \cdot {(4 x^{3})}^{2} + 4 \cdot {(2 y)}^{1} \cdot {(4 x^{3})}^{3} + 1 \cdot {(2 y)}^{0} \cdot {(4 x^{3})}^{4}$

Anything raised to $0$ is $1$ .

$16 y^{4} \cdot 1 + 4 \cdot {(2 y)}^{3} \cdot {(4 x^{3})}^{1} + 6 \cdot {(2 y)}^{2} \cdot {(4 x^{3})}^{2} + 4 \cdot {(2 y)}^{1} \cdot {(4 x^{3})}^{3} + 1 \cdot {(2 y)}^{0} \cdot {(4 x^{3})}^{4}$

Multiply $16$ by $1$ .

$16 y^{4} + 4 \cdot {(2 y)}^{3} \cdot {(4 x^{3})}^{1} + 6 \cdot {(2 y)}^{2} \cdot {(4 x^{3})}^{2} + 4 \cdot {(2 y)}^{1} \cdot {(4 x^{3})}^{3} + 1 \cdot {(2 y)}^{0} \cdot {(4 x^{3})}^{4}$

Apply the product rule to $2 y$ .

$16 y^{4} + 4 \cdot (2^{3} y^{3}) \cdot {(4 x^{3})}^{1} + 6 \cdot {(2 y)}^{2} \cdot {(4 x^{3})}^{2} + 4 \cdot {(2 y)}^{1} \cdot {(4 x^{3})}^{3} + 1 \cdot {(2 y)}^{0} \cdot {(4 x^{3})}^{4}$

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

$16 y^{4} + 4 \cdot (8 y^{3}) \cdot {(4 x^{3})}^{1} + 6 \cdot {(2 y)}^{2} \cdot {(4 x^{3})}^{2} + 4 \cdot {(2 y)}^{1} \cdot {(4 x^{3})}^{3} + 1 \cdot {(2 y)}^{0} \cdot {(4 x^{3})}^{4}$

Multiply $8$ by $4$ .

$16 y^{4} + 32 y^{3} \cdot {(4 x^{3})}^{1} + 6 \cdot {(2 y)}^{2} \cdot {(4 x^{3})}^{2} + 4 \cdot {(2 y)}^{1} \cdot {(4 x^{3})}^{3} + 1 \cdot {(2 y)}^{0} \cdot {(4 x^{3})}^{4}$

Simplify.

$16 y^{4} + 32 y^{3} \cdot (4 x^{3}) + 6 \cdot {(2 y)}^{2} \cdot {(4 x^{3})}^{2} + 4 \cdot {(2 y)}^{1} \cdot {(4 x^{3})}^{3} + 1 \cdot {(2 y)}^{0} \cdot {(4 x^{3})}^{4}$

Rewrite using the commutative property of multiplication.

$16 y^{4} + 32 \cdot 4 y^{3} x^{3} + 6 \cdot {(2 y)}^{2} \cdot {(4 x^{3})}^{2} + 4 \cdot {(2 y)}^{1} \cdot {(4 x^{3})}^{3} + 1 \cdot {(2 y)}^{0} \cdot {(4 x^{3})}^{4}$

Multiply $32$ by $4$ .

$16 y^{4} + 128 y^{3} x^{3} + 6 \cdot {(2 y)}^{2} \cdot {(4 x^{3})}^{2} + 4 \cdot {(2 y)}^{1} \cdot {(4 x^{3})}^{3} + 1 \cdot {(2 y)}^{0} \cdot {(4 x^{3})}^{4}$

Apply the product rule to $2 y$ .

$16 y^{4} + 128 y^{3} x^{3} + 6 \cdot (2^{2} y^{2}) \cdot {(4 x^{3})}^{2} + 4 \cdot {(2 y)}^{1} \cdot {(4 x^{3})}^{3} + 1 \cdot {(2 y)}^{0} \cdot {(4 x^{3})}^{4}$

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

$16 y^{4} + 128 y^{3} x^{3} + 6 \cdot (4 y^{2}) \cdot {(4 x^{3})}^{2} + 4 \cdot {(2 y)}^{1} \cdot {(4 x^{3})}^{3} + 1 \cdot {(2 y)}^{0} \cdot {(4 x^{3})}^{4}$

Multiply $4$ by $6$ .

$16 y^{4} + 128 y^{3} x^{3} + 24 y^{2} \cdot {(4 x^{3})}^{2} + 4 \cdot {(2 y)}^{1} \cdot {(4 x^{3})}^{3} + 1 \cdot {(2 y)}^{0} \cdot {(4 x^{3})}^{4}$

Apply the product rule to $4 x^{3}$ .

$16 y^{4} + 128 y^{3} x^{3} + 24 y^{2} \cdot (4^{2} {(x^{3})}^{2}) + 4 \cdot {(2 y)}^{1} \cdot {(4 x^{3})}^{3} + 1 \cdot {(2 y)}^{0} \cdot {(4 x^{3})}^{4}$

Rewrite using the commutative property of multiplication.

$16 y^{4} + 128 y^{3} x^{3} + 24 \cdot 4^{2} y^{2} {(x^{3})}^{2} + 4 \cdot {(2 y)}^{1} \cdot {(4 x^{3})}^{3} + 1 \cdot {(2 y)}^{0} \cdot {(4 x^{3})}^{4}$

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

$16 y^{4} + 128 y^{3} x^{3} + 24 \cdot 16 y^{2} {(x^{3})}^{2} + 4 \cdot {(2 y)}^{1} \cdot {(4 x^{3})}^{3} + 1 \cdot {(2 y)}^{0} \cdot {(4 x^{3})}^{4}$

Multiply $24$ by $16$ .

$16 y^{4} + 128 y^{3} x^{3} + 384 y^{2} {(x^{3})}^{2} + 4 \cdot {(2 y)}^{1} \cdot {(4 x^{3})}^{3} + 1 \cdot {(2 y)}^{0} \cdot {(4 x^{3})}^{4}$

Multiply the exponents in ${(x^{3})}^{2}$ .

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$16 y^{4} + 128 y^{3} x^{3} + 384 y^{2} x^{3 \cdot 2} + 4 \cdot {(2 y)}^{1} \cdot {(4 x^{3})}^{3} + 1 \cdot {(2 y)}^{0} \cdot {(4 x^{3})}^{4}$

Multiply $3$ by $2$ .

$16 y^{4} + 128 y^{3} x^{3} + 384 y^{2} x^{6} + 4 \cdot {(2 y)}^{1} \cdot {(4 x^{3})}^{3} + 1 \cdot {(2 y)}^{0} \cdot {(4 x^{3})}^{4}$

Simplify.

$16 y^{4} + 128 y^{3} x^{3} + 384 y^{2} x^{6} + 4 \cdot (2 y) \cdot {(4 x^{3})}^{3} + 1 \cdot {(2 y)}^{0} \cdot {(4 x^{3})}^{4}$

Multiply $2$ by $4$ .

$16 y^{4} + 128 y^{3} x^{3} + 384 y^{2} x^{6} + 8 y \cdot {(4 x^{3})}^{3} + 1 \cdot {(2 y)}^{0} \cdot {(4 x^{3})}^{4}$

Apply the product rule to $4 x^{3}$ .

$16 y^{4} + 128 y^{3} x^{3} + 384 y^{2} x^{6} + 8 y \cdot (4^{3} {(x^{3})}^{3}) + 1 \cdot {(2 y)}^{0} \cdot {(4 x^{3})}^{4}$

Rewrite using the commutative property of multiplication.

$16 y^{4} + 128 y^{3} x^{3} + 384 y^{2} x^{6} + 8 \cdot 4^{3} y {(x^{3})}^{3} + 1 \cdot {(2 y)}^{0} \cdot {(4 x^{3})}^{4}$

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

$16 y^{4} + 128 y^{3} x^{3} + 384 y^{2} x^{6} + 8 \cdot 64 y {(x^{3})}^{3} + 1 \cdot {(2 y)}^{0} \cdot {(4 x^{3})}^{4}$

Multiply $8$ by $64$ .

$16 y^{4} + 128 y^{3} x^{3} + 384 y^{2} x^{6} + 512 y {(x^{3})}^{3} + 1 \cdot {(2 y)}^{0} \cdot {(4 x^{3})}^{4}$

Multiply the exponents in ${(x^{3})}^{3}$ .

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$16 y^{4} + 128 y^{3} x^{3} + 384 y^{2} x^{6} + 512 y x^{3 \cdot 3} + 1 \cdot {(2 y)}^{0} \cdot {(4 x^{3})}^{4}$

Multiply $3$ by $3$ .

$16 y^{4} + 128 y^{3} x^{3} + 384 y^{2} x^{6} + 512 y x^{9} + 1 \cdot {(2 y)}^{0} \cdot {(4 x^{3})}^{4}$

Multiply ${(2 y)}^{0}$ by $1$ .

$16 y^{4} + 128 y^{3} x^{3} + 384 y^{2} x^{6} + 512 y x^{9} + {(2 y)}^{0} \cdot {(4 x^{3})}^{4}$

Apply the product rule to $2 y$ .

$16 y^{4} + 128 y^{3} x^{3} + 384 y^{2} x^{6} + 512 y x^{9} + 2^{0} y^{0} \cdot {(4 x^{3})}^{4}$

Anything raised to $0$ is $1$ .

$16 y^{4} + 128 y^{3} x^{3} + 384 y^{2} x^{6} + 512 y x^{9} + 1 y^{0} \cdot {(4 x^{3})}^{4}$

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

$16 y^{4} + 128 y^{3} x^{3} + 384 y^{2} x^{6} + 512 y x^{9} + y^{0} \cdot {(4 x^{3})}^{4}$

Anything raised to $0$ is $1$ .

$16 y^{4} + 128 y^{3} x^{3} + 384 y^{2} x^{6} + 512 y x^{9} + 1 \cdot {(4 x^{3})}^{4}$

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

$16 y^{4} + 128 y^{3} x^{3} + 384 y^{2} x^{6} + 512 y x^{9} + {(4 x^{3})}^{4}$

Apply the product rule to $4 x^{3}$ .

$16 y^{4} + 128 y^{3} x^{3} + 384 y^{2} x^{6} + 512 y x^{9} + 4^{4} {(x^{3})}^{4}$

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

$16 y^{4} + 128 y^{3} x^{3} + 384 y^{2} x^{6} + 512 y x^{9} + 256 {(x^{3})}^{4}$

Multiply the exponents in ${(x^{3})}^{4}$ .

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$16 y^{4} + 128 y^{3} x^{3} + 384 y^{2} x^{6} + 512 y x^{9} + 256 x^{3 \cdot 4}$

Multiply $3$ by $4$ .

$16 y^{4} + 128 y^{3} x^{3} + 384 y^{2} x^{6} + 512 y x^{9} + 256 x^{12}$

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

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