Expand Using the Binomial Theorem (-3m+2n)^4

${(- 3 m + 2 n)}^{4}$

Step 1

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

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

Step 2

Expand the summation.

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

Step 3

Simplify the exponents for each term of the expansion.

$1 \cdot {(- 3 m)}^{4} \cdot {(2 n)}^{0} + 4 \cdot {(- 3 m)}^{3} \cdot {(2 n)}^{1} + 6 \cdot {(- 3 m)}^{2} \cdot {(2 n)}^{2} + 4 \cdot {(- 3 m)}^{1} \cdot {(2 n)}^{3} + 1 \cdot {(- 3 m)}^{0} \cdot {(2 n)}^{4}$

Step 4

Simplify each term.

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

${(- 3 m)}^{4} \cdot {(2 n)}^{0} + 4 \cdot {(- 3 m)}^{3} \cdot {(2 n)}^{1} + 6 \cdot {(- 3 m)}^{2} \cdot {(2 n)}^{2} + 4 \cdot {(- 3 m)}^{1} \cdot {(2 n)}^{3} + 1 \cdot {(- 3 m)}^{0} \cdot {(2 n)}^{4}$

Apply the product rule to $- 3 m$ .

${(- 3)}^{4} m^{4} \cdot {(2 n)}^{0} + 4 \cdot {(- 3 m)}^{3} \cdot {(2 n)}^{1} + 6 \cdot {(- 3 m)}^{2} \cdot {(2 n)}^{2} + 4 \cdot {(- 3 m)}^{1} \cdot {(2 n)}^{3} + 1 \cdot {(- 3 m)}^{0} \cdot {(2 n)}^{4}$

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

$81 m^{4} \cdot {(2 n)}^{0} + 4 \cdot {(- 3 m)}^{3} \cdot {(2 n)}^{1} + 6 \cdot {(- 3 m)}^{2} \cdot {(2 n)}^{2} + 4 \cdot {(- 3 m)}^{1} \cdot {(2 n)}^{3} + 1 \cdot {(- 3 m)}^{0} \cdot {(2 n)}^{4}$

Apply the product rule to $2 n$ .

$81 m^{4} \cdot (2^{0} n^{0}) + 4 \cdot {(- 3 m)}^{3} \cdot {(2 n)}^{1} + 6 \cdot {(- 3 m)}^{2} \cdot {(2 n)}^{2} + 4 \cdot {(- 3 m)}^{1} \cdot {(2 n)}^{3} + 1 \cdot {(- 3 m)}^{0} \cdot {(2 n)}^{4}$

Rewrite using the commutative property of multiplication.

$81 \cdot 2^{0} m^{4} n^{0} + 4 \cdot {(- 3 m)}^{3} \cdot {(2 n)}^{1} + 6 \cdot {(- 3 m)}^{2} \cdot {(2 n)}^{2} + 4 \cdot {(- 3 m)}^{1} \cdot {(2 n)}^{3} + 1 \cdot {(- 3 m)}^{0} \cdot {(2 n)}^{4}$

Anything raised to $0$ is $1$ .

$81 \cdot 1 m^{4} n^{0} + 4 \cdot {(- 3 m)}^{3} \cdot {(2 n)}^{1} + 6 \cdot {(- 3 m)}^{2} \cdot {(2 n)}^{2} + 4 \cdot {(- 3 m)}^{1} \cdot {(2 n)}^{3} + 1 \cdot {(- 3 m)}^{0} \cdot {(2 n)}^{4}$

Multiply $81$ by $1$ .

$81 m^{4} n^{0} + 4 \cdot {(- 3 m)}^{3} \cdot {(2 n)}^{1} + 6 \cdot {(- 3 m)}^{2} \cdot {(2 n)}^{2} + 4 \cdot {(- 3 m)}^{1} \cdot {(2 n)}^{3} + 1 \cdot {(- 3 m)}^{0} \cdot {(2 n)}^{4}$

Anything raised to $0$ is $1$ .

$81 m^{4} \cdot 1 + 4 \cdot {(- 3 m)}^{3} \cdot {(2 n)}^{1} + 6 \cdot {(- 3 m)}^{2} \cdot {(2 n)}^{2} + 4 \cdot {(- 3 m)}^{1} \cdot {(2 n)}^{3} + 1 \cdot {(- 3 m)}^{0} \cdot {(2 n)}^{4}$

Multiply $81$ by $1$ .

$81 m^{4} + 4 \cdot {(- 3 m)}^{3} \cdot {(2 n)}^{1} + 6 \cdot {(- 3 m)}^{2} \cdot {(2 n)}^{2} + 4 \cdot {(- 3 m)}^{1} \cdot {(2 n)}^{3} + 1 \cdot {(- 3 m)}^{0} \cdot {(2 n)}^{4}$

Apply the product rule to $- 3 m$ .

$81 m^{4} + 4 \cdot ({(- 3)}^{3} m^{3}) \cdot {(2 n)}^{1} + 6 \cdot {(- 3 m)}^{2} \cdot {(2 n)}^{2} + 4 \cdot {(- 3 m)}^{1} \cdot {(2 n)}^{3} + 1 \cdot {(- 3 m)}^{0} \cdot {(2 n)}^{4}$

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

$81 m^{4} + 4 \cdot (- 27 m^{3}) \cdot {(2 n)}^{1} + 6 \cdot {(- 3 m)}^{2} \cdot {(2 n)}^{2} + 4 \cdot {(- 3 m)}^{1} \cdot {(2 n)}^{3} + 1 \cdot {(- 3 m)}^{0} \cdot {(2 n)}^{4}$

Multiply $- 27$ by $4$ .

$81 m^{4} - 108 m^{3} \cdot {(2 n)}^{1} + 6 \cdot {(- 3 m)}^{2} \cdot {(2 n)}^{2} + 4 \cdot {(- 3 m)}^{1} \cdot {(2 n)}^{3} + 1 \cdot {(- 3 m)}^{0} \cdot {(2 n)}^{4}$

Simplify.

$81 m^{4} - 108 m^{3} \cdot (2 n) + 6 \cdot {(- 3 m)}^{2} \cdot {(2 n)}^{2} + 4 \cdot {(- 3 m)}^{1} \cdot {(2 n)}^{3} + 1 \cdot {(- 3 m)}^{0} \cdot {(2 n)}^{4}$

Rewrite using the commutative property of multiplication.

$81 m^{4} - 108 \cdot 2 m^{3} n + 6 \cdot {(- 3 m)}^{2} \cdot {(2 n)}^{2} + 4 \cdot {(- 3 m)}^{1} \cdot {(2 n)}^{3} + 1 \cdot {(- 3 m)}^{0} \cdot {(2 n)}^{4}$

Multiply $- 108$ by $2$ .

$81 m^{4} - 216 m^{3} n + 6 \cdot {(- 3 m)}^{2} \cdot {(2 n)}^{2} + 4 \cdot {(- 3 m)}^{1} \cdot {(2 n)}^{3} + 1 \cdot {(- 3 m)}^{0} \cdot {(2 n)}^{4}$

Apply the product rule to $- 3 m$ .

$81 m^{4} - 216 m^{3} n + 6 \cdot ({(- 3)}^{2} m^{2}) \cdot {(2 n)}^{2} + 4 \cdot {(- 3 m)}^{1} \cdot {(2 n)}^{3} + 1 \cdot {(- 3 m)}^{0} \cdot {(2 n)}^{4}$

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

$81 m^{4} - 216 m^{3} n + 6 \cdot (9 m^{2}) \cdot {(2 n)}^{2} + 4 \cdot {(- 3 m)}^{1} \cdot {(2 n)}^{3} + 1 \cdot {(- 3 m)}^{0} \cdot {(2 n)}^{4}$

Multiply $9$ by $6$ .

$81 m^{4} - 216 m^{3} n + 54 m^{2} \cdot {(2 n)}^{2} + 4 \cdot {(- 3 m)}^{1} \cdot {(2 n)}^{3} + 1 \cdot {(- 3 m)}^{0} \cdot {(2 n)}^{4}$

Apply the product rule to $2 n$ .

$81 m^{4} - 216 m^{3} n + 54 m^{2} \cdot (2^{2} n^{2}) + 4 \cdot {(- 3 m)}^{1} \cdot {(2 n)}^{3} + 1 \cdot {(- 3 m)}^{0} \cdot {(2 n)}^{4}$

Rewrite using the commutative property of multiplication.

$81 m^{4} - 216 m^{3} n + 54 \cdot 2^{2} m^{2} n^{2} + 4 \cdot {(- 3 m)}^{1} \cdot {(2 n)}^{3} + 1 \cdot {(- 3 m)}^{0} \cdot {(2 n)}^{4}$

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

$81 m^{4} - 216 m^{3} n + 54 \cdot 4 m^{2} n^{2} + 4 \cdot {(- 3 m)}^{1} \cdot {(2 n)}^{3} + 1 \cdot {(- 3 m)}^{0} \cdot {(2 n)}^{4}$

Multiply $54$ by $4$ .

$81 m^{4} - 216 m^{3} n + 216 m^{2} n^{2} + 4 \cdot {(- 3 m)}^{1} \cdot {(2 n)}^{3} + 1 \cdot {(- 3 m)}^{0} \cdot {(2 n)}^{4}$

Simplify.

$81 m^{4} - 216 m^{3} n + 216 m^{2} n^{2} + 4 \cdot (- 3 m) \cdot {(2 n)}^{3} + 1 \cdot {(- 3 m)}^{0} \cdot {(2 n)}^{4}$

Multiply $- 3$ by $4$ .

$81 m^{4} - 216 m^{3} n + 216 m^{2} n^{2} - 12 m \cdot {(2 n)}^{3} + 1 \cdot {(- 3 m)}^{0} \cdot {(2 n)}^{4}$

Apply the product rule to $2 n$ .

$81 m^{4} - 216 m^{3} n + 216 m^{2} n^{2} - 12 m \cdot (2^{3} n^{3}) + 1 \cdot {(- 3 m)}^{0} \cdot {(2 n)}^{4}$

Rewrite using the commutative property of multiplication.

$81 m^{4} - 216 m^{3} n + 216 m^{2} n^{2} - 12 \cdot 2^{3} m n^{3} + 1 \cdot {(- 3 m)}^{0} \cdot {(2 n)}^{4}$

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

$81 m^{4} - 216 m^{3} n + 216 m^{2} n^{2} - 12 \cdot 8 m n^{3} + 1 \cdot {(- 3 m)}^{0} \cdot {(2 n)}^{4}$

Multiply $- 12$ by $8$ .

$81 m^{4} - 216 m^{3} n + 216 m^{2} n^{2} - 96 m n^{3} + 1 \cdot {(- 3 m)}^{0} \cdot {(2 n)}^{4}$

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

$81 m^{4} - 216 m^{3} n + 216 m^{2} n^{2} - 96 m n^{3} + {(- 3 m)}^{0} \cdot {(2 n)}^{4}$

Apply the product rule to $- 3 m$ .

$81 m^{4} - 216 m^{3} n + 216 m^{2} n^{2} - 96 m n^{3} + {(- 3)}^{0} m^{0} \cdot {(2 n)}^{4}$

Anything raised to $0$ is $1$ .

$81 m^{4} - 216 m^{3} n + 216 m^{2} n^{2} - 96 m n^{3} + 1 m^{0} \cdot {(2 n)}^{4}$

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

$81 m^{4} - 216 m^{3} n + 216 m^{2} n^{2} - 96 m n^{3} + m^{0} \cdot {(2 n)}^{4}$

Anything raised to $0$ is $1$ .

$81 m^{4} - 216 m^{3} n + 216 m^{2} n^{2} - 96 m n^{3} + 1 \cdot {(2 n)}^{4}$

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

$81 m^{4} - 216 m^{3} n + 216 m^{2} n^{2} - 96 m n^{3} + {(2 n)}^{4}$

Apply the product rule to $2 n$ .

$81 m^{4} - 216 m^{3} n + 216 m^{2} n^{2} - 96 m n^{3} + 2^{4} n^{4}$

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

$81 m^{4} - 216 m^{3} n + 216 m^{2} n^{2} - 96 m n^{3} + 16 n^{4}$

Expand Using the Binomial Theorem (-3m+2n)^4

Related Questions

Support LCMGCF.com by disabling your adblocker.