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

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

Step 1

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

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

Step 2

Expand the summation.

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

Step 3

Simplify the exponents for each term of the expansion.

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

Step 4

Simplify each term.

Multiply ${(m)}^{2}$ by $1$ .

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

Apply the product rule to $4 n$ .

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

Rewrite using the commutative property of multiplication.

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

Anything raised to $0$ is $1$ .

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

Multiply $m^{2}$ by $1$ .

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

Anything raised to $0$ is $1$ .

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

Multiply $m^{2}$ by $1$ .

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

Simplify.

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

Simplify.

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

Rewrite using the commutative property of multiplication.

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

Multiply $2$ by $4$ .

$m^{2} + 8 m n + 1 \cdot {(m)}^{0} \cdot {(4 n)}^{2}$

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

$m^{2} + 8 m n + {(m)}^{0} \cdot {(4 n)}^{2}$

Anything raised to $0$ is $1$ .

$m^{2} + 8 m n + 1 \cdot {(4 n)}^{2}$

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

$m^{2} + 8 m n + {(4 n)}^{2}$

Apply the product rule to $4 n$ .

$m^{2} + 8 m n + 4^{2} n^{2}$

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

$m^{2} + 8 m n + 16 n^{2}$

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

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