Expand Using the Binomial Theorem (9p+2q)^4

${(9 p + 2 q)}^{4}$

Step 1

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

$\sum_{k = 0}^{4} \frac{4!}{(4 - k)! k!} \cdot {(9 p)}^{4 - k} \cdot {(2 q)}^{k}$

Step 2

Expand the summation.

$\frac{4!}{(4 - 0)! 0!} \cdot {(9 p)}^{4 - 0} \cdot {(2 q)}^{0} + \frac{4!}{(4 - 1)! 1!} \cdot {(9 p)}^{4 - 1} \cdot {(2 q)}^{1} + \frac{4!}{(4 - 2)! 2!} \cdot {(9 p)}^{4 - 2} \cdot {(2 q)}^{2} + \frac{4!}{(4 - 3)! 3!} \cdot {(9 p)}^{4 - 3} \cdot {(2 q)}^{3} + \frac{4!}{(4 - 4)! 4!} \cdot {(9 p)}^{4 - 4} \cdot {(2 q)}^{4}$

Step 3

Simplify the exponents for each term of the expansion.

$1 \cdot {(9 p)}^{4} \cdot {(2 q)}^{0} + 4 \cdot {(9 p)}^{3} \cdot {(2 q)}^{1} + 6 \cdot {(9 p)}^{2} \cdot {(2 q)}^{2} + 4 \cdot {(9 p)}^{1} \cdot {(2 q)}^{3} + 1 \cdot {(9 p)}^{0} \cdot {(2 q)}^{4}$

Step 4

Simplify each term.

Multiply ${(9 p)}^{4}$ by $1$ .

${(9 p)}^{4} \cdot {(2 q)}^{0} + 4 \cdot {(9 p)}^{3} \cdot {(2 q)}^{1} + 6 \cdot {(9 p)}^{2} \cdot {(2 q)}^{2} + 4 \cdot {(9 p)}^{1} \cdot {(2 q)}^{3} + 1 \cdot {(9 p)}^{0} \cdot {(2 q)}^{4}$

Apply the product rule to $9 p$ .

$9^{4} p^{4} \cdot {(2 q)}^{0} + 4 \cdot {(9 p)}^{3} \cdot {(2 q)}^{1} + 6 \cdot {(9 p)}^{2} \cdot {(2 q)}^{2} + 4 \cdot {(9 p)}^{1} \cdot {(2 q)}^{3} + 1 \cdot {(9 p)}^{0} \cdot {(2 q)}^{4}$

Raise $9$ to the power of $4$ .

$6561 p^{4} \cdot {(2 q)}^{0} + 4 \cdot {(9 p)}^{3} \cdot {(2 q)}^{1} + 6 \cdot {(9 p)}^{2} \cdot {(2 q)}^{2} + 4 \cdot {(9 p)}^{1} \cdot {(2 q)}^{3} + 1 \cdot {(9 p)}^{0} \cdot {(2 q)}^{4}$

Apply the product rule to $2 q$ .

$6561 p^{4} \cdot (2^{0} q^{0}) + 4 \cdot {(9 p)}^{3} \cdot {(2 q)}^{1} + 6 \cdot {(9 p)}^{2} \cdot {(2 q)}^{2} + 4 \cdot {(9 p)}^{1} \cdot {(2 q)}^{3} + 1 \cdot {(9 p)}^{0} \cdot {(2 q)}^{4}$

Rewrite using the commutative property of multiplication.

$6561 \cdot 2^{0} p^{4} q^{0} + 4 \cdot {(9 p)}^{3} \cdot {(2 q)}^{1} + 6 \cdot {(9 p)}^{2} \cdot {(2 q)}^{2} + 4 \cdot {(9 p)}^{1} \cdot {(2 q)}^{3} + 1 \cdot {(9 p)}^{0} \cdot {(2 q)}^{4}$

Anything raised to $0$ is $1$ .

$6561 \cdot 1 p^{4} q^{0} + 4 \cdot {(9 p)}^{3} \cdot {(2 q)}^{1} + 6 \cdot {(9 p)}^{2} \cdot {(2 q)}^{2} + 4 \cdot {(9 p)}^{1} \cdot {(2 q)}^{3} + 1 \cdot {(9 p)}^{0} \cdot {(2 q)}^{4}$

Multiply $6561$ by $1$ .

$6561 p^{4} q^{0} + 4 \cdot {(9 p)}^{3} \cdot {(2 q)}^{1} + 6 \cdot {(9 p)}^{2} \cdot {(2 q)}^{2} + 4 \cdot {(9 p)}^{1} \cdot {(2 q)}^{3} + 1 \cdot {(9 p)}^{0} \cdot {(2 q)}^{4}$

Anything raised to $0$ is $1$ .

$6561 p^{4} \cdot 1 + 4 \cdot {(9 p)}^{3} \cdot {(2 q)}^{1} + 6 \cdot {(9 p)}^{2} \cdot {(2 q)}^{2} + 4 \cdot {(9 p)}^{1} \cdot {(2 q)}^{3} + 1 \cdot {(9 p)}^{0} \cdot {(2 q)}^{4}$

Multiply $6561$ by $1$ .

$6561 p^{4} + 4 \cdot {(9 p)}^{3} \cdot {(2 q)}^{1} + 6 \cdot {(9 p)}^{2} \cdot {(2 q)}^{2} + 4 \cdot {(9 p)}^{1} \cdot {(2 q)}^{3} + 1 \cdot {(9 p)}^{0} \cdot {(2 q)}^{4}$

Apply the product rule to $9 p$ .

$6561 p^{4} + 4 \cdot (9^{3} p^{3}) \cdot {(2 q)}^{1} + 6 \cdot {(9 p)}^{2} \cdot {(2 q)}^{2} + 4 \cdot {(9 p)}^{1} \cdot {(2 q)}^{3} + 1 \cdot {(9 p)}^{0} \cdot {(2 q)}^{4}$

Raise $9$ to the power of $3$ .

$6561 p^{4} + 4 \cdot (729 p^{3}) \cdot {(2 q)}^{1} + 6 \cdot {(9 p)}^{2} \cdot {(2 q)}^{2} + 4 \cdot {(9 p)}^{1} \cdot {(2 q)}^{3} + 1 \cdot {(9 p)}^{0} \cdot {(2 q)}^{4}$

Multiply $729$ by $4$ .

$6561 p^{4} + 2916 p^{3} \cdot {(2 q)}^{1} + 6 \cdot {(9 p)}^{2} \cdot {(2 q)}^{2} + 4 \cdot {(9 p)}^{1} \cdot {(2 q)}^{3} + 1 \cdot {(9 p)}^{0} \cdot {(2 q)}^{4}$

Simplify.

$6561 p^{4} + 2916 p^{3} \cdot (2 q) + 6 \cdot {(9 p)}^{2} \cdot {(2 q)}^{2} + 4 \cdot {(9 p)}^{1} \cdot {(2 q)}^{3} + 1 \cdot {(9 p)}^{0} \cdot {(2 q)}^{4}$

Rewrite using the commutative property of multiplication.

$6561 p^{4} + 2916 \cdot 2 p^{3} q + 6 \cdot {(9 p)}^{2} \cdot {(2 q)}^{2} + 4 \cdot {(9 p)}^{1} \cdot {(2 q)}^{3} + 1 \cdot {(9 p)}^{0} \cdot {(2 q)}^{4}$

Multiply $2916$ by $2$ .

$6561 p^{4} + 5832 p^{3} q + 6 \cdot {(9 p)}^{2} \cdot {(2 q)}^{2} + 4 \cdot {(9 p)}^{1} \cdot {(2 q)}^{3} + 1 \cdot {(9 p)}^{0} \cdot {(2 q)}^{4}$

Apply the product rule to $9 p$ .

$6561 p^{4} + 5832 p^{3} q + 6 \cdot (9^{2} p^{2}) \cdot {(2 q)}^{2} + 4 \cdot {(9 p)}^{1} \cdot {(2 q)}^{3} + 1 \cdot {(9 p)}^{0} \cdot {(2 q)}^{4}$

Raise $9$ to the power of $2$ .

$6561 p^{4} + 5832 p^{3} q + 6 \cdot (81 p^{2}) \cdot {(2 q)}^{2} + 4 \cdot {(9 p)}^{1} \cdot {(2 q)}^{3} + 1 \cdot {(9 p)}^{0} \cdot {(2 q)}^{4}$

Multiply $81$ by $6$ .

$6561 p^{4} + 5832 p^{3} q + 486 p^{2} \cdot {(2 q)}^{2} + 4 \cdot {(9 p)}^{1} \cdot {(2 q)}^{3} + 1 \cdot {(9 p)}^{0} \cdot {(2 q)}^{4}$

Apply the product rule to $2 q$ .

$6561 p^{4} + 5832 p^{3} q + 486 p^{2} \cdot (2^{2} q^{2}) + 4 \cdot {(9 p)}^{1} \cdot {(2 q)}^{3} + 1 \cdot {(9 p)}^{0} \cdot {(2 q)}^{4}$

Rewrite using the commutative property of multiplication.

$6561 p^{4} + 5832 p^{3} q + 486 \cdot 2^{2} p^{2} q^{2} + 4 \cdot {(9 p)}^{1} \cdot {(2 q)}^{3} + 1 \cdot {(9 p)}^{0} \cdot {(2 q)}^{4}$

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

$6561 p^{4} + 5832 p^{3} q + 486 \cdot 4 p^{2} q^{2} + 4 \cdot {(9 p)}^{1} \cdot {(2 q)}^{3} + 1 \cdot {(9 p)}^{0} \cdot {(2 q)}^{4}$

Multiply $486$ by $4$ .

$6561 p^{4} + 5832 p^{3} q + 1944 p^{2} q^{2} + 4 \cdot {(9 p)}^{1} \cdot {(2 q)}^{3} + 1 \cdot {(9 p)}^{0} \cdot {(2 q)}^{4}$

Simplify.

$6561 p^{4} + 5832 p^{3} q + 1944 p^{2} q^{2} + 4 \cdot (9 p) \cdot {(2 q)}^{3} + 1 \cdot {(9 p)}^{0} \cdot {(2 q)}^{4}$

Multiply $9$ by $4$ .

$6561 p^{4} + 5832 p^{3} q + 1944 p^{2} q^{2} + 36 p \cdot {(2 q)}^{3} + 1 \cdot {(9 p)}^{0} \cdot {(2 q)}^{4}$

Apply the product rule to $2 q$ .

$6561 p^{4} + 5832 p^{3} q + 1944 p^{2} q^{2} + 36 p \cdot (2^{3} q^{3}) + 1 \cdot {(9 p)}^{0} \cdot {(2 q)}^{4}$

Rewrite using the commutative property of multiplication.

$6561 p^{4} + 5832 p^{3} q + 1944 p^{2} q^{2} + 36 \cdot 2^{3} p q^{3} + 1 \cdot {(9 p)}^{0} \cdot {(2 q)}^{4}$

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

$6561 p^{4} + 5832 p^{3} q + 1944 p^{2} q^{2} + 36 \cdot 8 p q^{3} + 1 \cdot {(9 p)}^{0} \cdot {(2 q)}^{4}$

Multiply $36$ by $8$ .

$6561 p^{4} + 5832 p^{3} q + 1944 p^{2} q^{2} + 288 p q^{3} + 1 \cdot {(9 p)}^{0} \cdot {(2 q)}^{4}$

Multiply ${(9 p)}^{0}$ by $1$ .

$6561 p^{4} + 5832 p^{3} q + 1944 p^{2} q^{2} + 288 p q^{3} + {(9 p)}^{0} \cdot {(2 q)}^{4}$

Apply the product rule to $9 p$ .

$6561 p^{4} + 5832 p^{3} q + 1944 p^{2} q^{2} + 288 p q^{3} + 9^{0} p^{0} \cdot {(2 q)}^{4}$

Anything raised to $0$ is $1$ .

$6561 p^{4} + 5832 p^{3} q + 1944 p^{2} q^{2} + 288 p q^{3} + 1 p^{0} \cdot {(2 q)}^{4}$

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

$6561 p^{4} + 5832 p^{3} q + 1944 p^{2} q^{2} + 288 p q^{3} + p^{0} \cdot {(2 q)}^{4}$

Anything raised to $0$ is $1$ .

$6561 p^{4} + 5832 p^{3} q + 1944 p^{2} q^{2} + 288 p q^{3} + 1 \cdot {(2 q)}^{4}$

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

$6561 p^{4} + 5832 p^{3} q + 1944 p^{2} q^{2} + 288 p q^{3} + {(2 q)}^{4}$

Apply the product rule to $2 q$ .

$6561 p^{4} + 5832 p^{3} q + 1944 p^{2} q^{2} + 288 p q^{3} + 2^{4} q^{4}$

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

$6561 p^{4} + 5832 p^{3} q + 1944 p^{2} q^{2} + 288 p q^{3} + 16 q^{4}$

Expand Using the Binomial Theorem (9p+2q)^4

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