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

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

Step 1

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

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

Step 2

Expand the summation.

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

Step 3

Simplify the exponents for each term of the expansion.

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

Step 4

Simplify each term.

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

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

Apply the product rule to $4 y$ .

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

Rewrite using the commutative property of multiplication.

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

Anything raised to $0$ is $1$ .

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

Multiply $x^{4}$ by $1$ .

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

Anything raised to $0$ is $1$ .

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

Multiply $x^{4}$ by $1$ .

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

Simplify.

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

Rewrite using the commutative property of multiplication.

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

Multiply $4$ by $4$ .

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

Apply the product rule to $4 y$ .

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

Rewrite using the commutative property of multiplication.

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

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

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

Multiply $6$ by $16$ .

$x^{4} + 16 x^{3} y + 96 x^{2} y^{2} + 4 \cdot {(x)}^{1} \cdot {(4 y)}^{3} + 1 \cdot {(x)}^{0} \cdot {(4 y)}^{4}$

Simplify.

$x^{4} + 16 x^{3} y + 96 x^{2} y^{2} + 4 \cdot x \cdot {(4 y)}^{3} + 1 \cdot {(x)}^{0} \cdot {(4 y)}^{4}$

Apply the product rule to $4 y$ .

$x^{4} + 16 x^{3} y + 96 x^{2} y^{2} + 4 x \cdot (4^{3} y^{3}) + 1 \cdot {(x)}^{0} \cdot {(4 y)}^{4}$

Rewrite using the commutative property of multiplication.

$x^{4} + 16 x^{3} y + 96 x^{2} y^{2} + 4 \cdot 4^{3} x y^{3} + 1 \cdot {(x)}^{0} \cdot {(4 y)}^{4}$

Multiply $4$ by $4^{3}$ by adding the exponents.

Multiply $4$ by $4^{3}$ .

Raise $4$ to the power of $1$ .

$x^{4} + 16 x^{3} y + 96 x^{2} y^{2} + 4^{1} \cdot 4^{3} x y^{3} + 1 \cdot {(x)}^{0} \cdot {(4 y)}^{4}$

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$x^{4} + 16 x^{3} y + 96 x^{2} y^{2} + 4^{1 + 3} x y^{3} + 1 \cdot {(x)}^{0} \cdot {(4 y)}^{4}$

Add $1$ and $3$ .

$x^{4} + 16 x^{3} y + 96 x^{2} y^{2} + 4^{4} x y^{3} + 1 \cdot {(x)}^{0} \cdot {(4 y)}^{4}$

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

$x^{4} + 16 x^{3} y + 96 x^{2} y^{2} + 256 x y^{3} + 1 \cdot {(x)}^{0} \cdot {(4 y)}^{4}$

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

$x^{4} + 16 x^{3} y + 96 x^{2} y^{2} + 256 x y^{3} + {(x)}^{0} \cdot {(4 y)}^{4}$

Anything raised to $0$ is $1$ .

$x^{4} + 16 x^{3} y + 96 x^{2} y^{2} + 256 x y^{3} + 1 \cdot {(4 y)}^{4}$

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

$x^{4} + 16 x^{3} y + 96 x^{2} y^{2} + 256 x y^{3} + {(4 y)}^{4}$

Apply the product rule to $4 y$ .

$x^{4} + 16 x^{3} y + 96 x^{2} y^{2} + 256 x y^{3} + 4^{4} y^{4}$

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

$x^{4} + 16 x^{3} y + 96 x^{2} y^{2} + 256 x y^{3} + 256 y^{4}$

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

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