Expand Using the Binomial Theorem (4p+3q)^4

${(4 p + 3 q)}^{4}$

Step 1

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

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

Step 2

Expand the summation.

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

Step 3

Simplify the exponents for each term of the expansion.

$1 \cdot {(4 p)}^{4} \cdot {(3 q)}^{0} + 4 \cdot {(4 p)}^{3} \cdot {(3 q)}^{1} + 6 \cdot {(4 p)}^{2} \cdot {(3 q)}^{2} + 4 \cdot {(4 p)}^{1} \cdot {(3 q)}^{3} + 1 \cdot {(4 p)}^{0} \cdot {(3 q)}^{4}$

Step 4

Simplify each term.

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

${(4 p)}^{4} \cdot {(3 q)}^{0} + 4 \cdot {(4 p)}^{3} \cdot {(3 q)}^{1} + 6 \cdot {(4 p)}^{2} \cdot {(3 q)}^{2} + 4 \cdot {(4 p)}^{1} \cdot {(3 q)}^{3} + 1 \cdot {(4 p)}^{0} \cdot {(3 q)}^{4}$

Apply the product rule to $4 p$ .

$4^{4} p^{4} \cdot {(3 q)}^{0} + 4 \cdot {(4 p)}^{3} \cdot {(3 q)}^{1} + 6 \cdot {(4 p)}^{2} \cdot {(3 q)}^{2} + 4 \cdot {(4 p)}^{1} \cdot {(3 q)}^{3} + 1 \cdot {(4 p)}^{0} \cdot {(3 q)}^{4}$

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

$256 p^{4} \cdot {(3 q)}^{0} + 4 \cdot {(4 p)}^{3} \cdot {(3 q)}^{1} + 6 \cdot {(4 p)}^{2} \cdot {(3 q)}^{2} + 4 \cdot {(4 p)}^{1} \cdot {(3 q)}^{3} + 1 \cdot {(4 p)}^{0} \cdot {(3 q)}^{4}$

Apply the product rule to $3 q$ .

$256 p^{4} \cdot (3^{0} q^{0}) + 4 \cdot {(4 p)}^{3} \cdot {(3 q)}^{1} + 6 \cdot {(4 p)}^{2} \cdot {(3 q)}^{2} + 4 \cdot {(4 p)}^{1} \cdot {(3 q)}^{3} + 1 \cdot {(4 p)}^{0} \cdot {(3 q)}^{4}$

Rewrite using the commutative property of multiplication.

$256 \cdot 3^{0} p^{4} q^{0} + 4 \cdot {(4 p)}^{3} \cdot {(3 q)}^{1} + 6 \cdot {(4 p)}^{2} \cdot {(3 q)}^{2} + 4 \cdot {(4 p)}^{1} \cdot {(3 q)}^{3} + 1 \cdot {(4 p)}^{0} \cdot {(3 q)}^{4}$

Anything raised to $0$ is $1$ .

$256 \cdot 1 p^{4} q^{0} + 4 \cdot {(4 p)}^{3} \cdot {(3 q)}^{1} + 6 \cdot {(4 p)}^{2} \cdot {(3 q)}^{2} + 4 \cdot {(4 p)}^{1} \cdot {(3 q)}^{3} + 1 \cdot {(4 p)}^{0} \cdot {(3 q)}^{4}$

Multiply $256$ by $1$ .

$256 p^{4} q^{0} + 4 \cdot {(4 p)}^{3} \cdot {(3 q)}^{1} + 6 \cdot {(4 p)}^{2} \cdot {(3 q)}^{2} + 4 \cdot {(4 p)}^{1} \cdot {(3 q)}^{3} + 1 \cdot {(4 p)}^{0} \cdot {(3 q)}^{4}$

Anything raised to $0$ is $1$ .

$256 p^{4} \cdot 1 + 4 \cdot {(4 p)}^{3} \cdot {(3 q)}^{1} + 6 \cdot {(4 p)}^{2} \cdot {(3 q)}^{2} + 4 \cdot {(4 p)}^{1} \cdot {(3 q)}^{3} + 1 \cdot {(4 p)}^{0} \cdot {(3 q)}^{4}$

Multiply $256$ by $1$ .

$256 p^{4} + 4 \cdot {(4 p)}^{3} \cdot {(3 q)}^{1} + 6 \cdot {(4 p)}^{2} \cdot {(3 q)}^{2} + 4 \cdot {(4 p)}^{1} \cdot {(3 q)}^{3} + 1 \cdot {(4 p)}^{0} \cdot {(3 q)}^{4}$

Apply the product rule to $4 p$ .

$256 p^{4} + 4 \cdot (4^{3} p^{3}) \cdot {(3 q)}^{1} + 6 \cdot {(4 p)}^{2} \cdot {(3 q)}^{2} + 4 \cdot {(4 p)}^{1} \cdot {(3 q)}^{3} + 1 \cdot {(4 p)}^{0} \cdot {(3 q)}^{4}$

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

Move $4^{3}$ .

$256 p^{4} + 4^{3} \cdot 4 \cdot p^{3} \cdot {(3 q)}^{1} + 6 \cdot {(4 p)}^{2} \cdot {(3 q)}^{2} + 4 \cdot {(4 p)}^{1} \cdot {(3 q)}^{3} + 1 \cdot {(4 p)}^{0} \cdot {(3 q)}^{4}$

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

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

$256 p^{4} + 4^{3} \cdot 4^{1} \cdot p^{3} \cdot {(3 q)}^{1} + 6 \cdot {(4 p)}^{2} \cdot {(3 q)}^{2} + 4 \cdot {(4 p)}^{1} \cdot {(3 q)}^{3} + 1 \cdot {(4 p)}^{0} \cdot {(3 q)}^{4}$

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

$256 p^{4} + 4^{3 + 1} \cdot p^{3} \cdot {(3 q)}^{1} + 6 \cdot {(4 p)}^{2} \cdot {(3 q)}^{2} + 4 \cdot {(4 p)}^{1} \cdot {(3 q)}^{3} + 1 \cdot {(4 p)}^{0} \cdot {(3 q)}^{4}$

Add $3$ and $1$ .

$256 p^{4} + 4^{4} \cdot p^{3} \cdot {(3 q)}^{1} + 6 \cdot {(4 p)}^{2} \cdot {(3 q)}^{2} + 4 \cdot {(4 p)}^{1} \cdot {(3 q)}^{3} + 1 \cdot {(4 p)}^{0} \cdot {(3 q)}^{4}$

Simplify $4^{4} \cdot p^{3} \cdot {(3 q)}^{1}$ .

$256 p^{4} + 4^{4} \cdot p^{3} \cdot (3 q) + 6 \cdot {(4 p)}^{2} \cdot {(3 q)}^{2} + 4 \cdot {(4 p)}^{1} \cdot {(3 q)}^{3} + 1 \cdot {(4 p)}^{0} \cdot {(3 q)}^{4}$

Rewrite using the commutative property of multiplication.

$256 p^{4} + 4^{4} \cdot 3 p^{3} q + 6 \cdot {(4 p)}^{2} \cdot {(3 q)}^{2} + 4 \cdot {(4 p)}^{1} \cdot {(3 q)}^{3} + 1 \cdot {(4 p)}^{0} \cdot {(3 q)}^{4}$

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

$256 p^{4} + 256 \cdot 3 p^{3} q + 6 \cdot {(4 p)}^{2} \cdot {(3 q)}^{2} + 4 \cdot {(4 p)}^{1} \cdot {(3 q)}^{3} + 1 \cdot {(4 p)}^{0} \cdot {(3 q)}^{4}$

Multiply $256$ by $3$ .

$256 p^{4} + 768 p^{3} q + 6 \cdot {(4 p)}^{2} \cdot {(3 q)}^{2} + 4 \cdot {(4 p)}^{1} \cdot {(3 q)}^{3} + 1 \cdot {(4 p)}^{0} \cdot {(3 q)}^{4}$

Apply the product rule to $4 p$ .

$256 p^{4} + 768 p^{3} q + 6 \cdot (4^{2} p^{2}) \cdot {(3 q)}^{2} + 4 \cdot {(4 p)}^{1} \cdot {(3 q)}^{3} + 1 \cdot {(4 p)}^{0} \cdot {(3 q)}^{4}$

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

$256 p^{4} + 768 p^{3} q + 6 \cdot (16 p^{2}) \cdot {(3 q)}^{2} + 4 \cdot {(4 p)}^{1} \cdot {(3 q)}^{3} + 1 \cdot {(4 p)}^{0} \cdot {(3 q)}^{4}$

Multiply $16$ by $6$ .

$256 p^{4} + 768 p^{3} q + 96 p^{2} \cdot {(3 q)}^{2} + 4 \cdot {(4 p)}^{1} \cdot {(3 q)}^{3} + 1 \cdot {(4 p)}^{0} \cdot {(3 q)}^{4}$

Apply the product rule to $3 q$ .

$256 p^{4} + 768 p^{3} q + 96 p^{2} \cdot (3^{2} q^{2}) + 4 \cdot {(4 p)}^{1} \cdot {(3 q)}^{3} + 1 \cdot {(4 p)}^{0} \cdot {(3 q)}^{4}$

Rewrite using the commutative property of multiplication.

$256 p^{4} + 768 p^{3} q + 96 \cdot 3^{2} p^{2} q^{2} + 4 \cdot {(4 p)}^{1} \cdot {(3 q)}^{3} + 1 \cdot {(4 p)}^{0} \cdot {(3 q)}^{4}$

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

$256 p^{4} + 768 p^{3} q + 96 \cdot 9 p^{2} q^{2} + 4 \cdot {(4 p)}^{1} \cdot {(3 q)}^{3} + 1 \cdot {(4 p)}^{0} \cdot {(3 q)}^{4}$

Multiply $96$ by $9$ .

$256 p^{4} + 768 p^{3} q + 864 p^{2} q^{2} + 4 \cdot {(4 p)}^{1} \cdot {(3 q)}^{3} + 1 \cdot {(4 p)}^{0} \cdot {(3 q)}^{4}$

Simplify.

$256 p^{4} + 768 p^{3} q + 864 p^{2} q^{2} + 4 \cdot (4 p) \cdot {(3 q)}^{3} + 1 \cdot {(4 p)}^{0} \cdot {(3 q)}^{4}$

Multiply $4$ by $4$ .

$256 p^{4} + 768 p^{3} q + 864 p^{2} q^{2} + 16 p \cdot {(3 q)}^{3} + 1 \cdot {(4 p)}^{0} \cdot {(3 q)}^{4}$

Apply the product rule to $3 q$ .

$256 p^{4} + 768 p^{3} q + 864 p^{2} q^{2} + 16 p \cdot (3^{3} q^{3}) + 1 \cdot {(4 p)}^{0} \cdot {(3 q)}^{4}$

Rewrite using the commutative property of multiplication.

$256 p^{4} + 768 p^{3} q + 864 p^{2} q^{2} + 16 \cdot 3^{3} p q^{3} + 1 \cdot {(4 p)}^{0} \cdot {(3 q)}^{4}$

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

$256 p^{4} + 768 p^{3} q + 864 p^{2} q^{2} + 16 \cdot 27 p q^{3} + 1 \cdot {(4 p)}^{0} \cdot {(3 q)}^{4}$

Multiply $16$ by $27$ .

$256 p^{4} + 768 p^{3} q + 864 p^{2} q^{2} + 432 p q^{3} + 1 \cdot {(4 p)}^{0} \cdot {(3 q)}^{4}$

Multiply ${(4 p)}^{0}$ by $1$ .

$256 p^{4} + 768 p^{3} q + 864 p^{2} q^{2} + 432 p q^{3} + {(4 p)}^{0} \cdot {(3 q)}^{4}$

Apply the product rule to $4 p$ .

$256 p^{4} + 768 p^{3} q + 864 p^{2} q^{2} + 432 p q^{3} + 4^{0} p^{0} \cdot {(3 q)}^{4}$

Anything raised to $0$ is $1$ .

$256 p^{4} + 768 p^{3} q + 864 p^{2} q^{2} + 432 p q^{3} + 1 p^{0} \cdot {(3 q)}^{4}$

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

$256 p^{4} + 768 p^{3} q + 864 p^{2} q^{2} + 432 p q^{3} + p^{0} \cdot {(3 q)}^{4}$

Anything raised to $0$ is $1$ .

$256 p^{4} + 768 p^{3} q + 864 p^{2} q^{2} + 432 p q^{3} + 1 \cdot {(3 q)}^{4}$

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

$256 p^{4} + 768 p^{3} q + 864 p^{2} q^{2} + 432 p q^{3} + {(3 q)}^{4}$

Apply the product rule to $3 q$ .

$256 p^{4} + 768 p^{3} q + 864 p^{2} q^{2} + 432 p q^{3} + 3^{4} q^{4}$

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

$256 p^{4} + 768 p^{3} q + 864 p^{2} q^{2} + 432 p q^{3} + 81 q^{4}$

Expand Using the Binomial Theorem (4p+3q)^4

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