Expand Using the Binomial Theorem (q^2+4/7)^2

${(q^{2} + \frac{4}{7})}^{2}$

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

$\sum_{k = 0}^{2} \frac{2!}{(2 - k)! k!} \cdot {(q^{2})}^{2 - k} \cdot {(\frac{4}{7})}^{k}$

Expand the summation.

$\frac{2!}{(2 - 0)! 0!} \cdot {(q^{2})}^{2 - 0} \cdot {(\frac{4}{7})}^{0} + \frac{2!}{(2 - 1)! 1!} \cdot {(q^{2})}^{2 - 1} \cdot {(\frac{4}{7})}^{1} + \frac{2!}{(2 - 2)! 2!} \cdot {(q^{2})}^{2 - 2} \cdot {(\frac{4}{7})}^{2}$

Simplify the exponents for each term of the expansion.

$1 \cdot {(q^{2})}^{2} \cdot {(\frac{4}{7})}^{0} + 2 \cdot {(q^{2})}^{1} \cdot {(\frac{4}{7})}^{1} + 1 \cdot {(q^{2})}^{0} \cdot {(\frac{4}{7})}^{2}$

Simplify each term.

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

${(q^{2})}^{2} \cdot {(\frac{4}{7})}^{0} + 2 \cdot {(q^{2})}^{1} \cdot {(\frac{4}{7})}^{1} + 1 \cdot {(q^{2})}^{0} \cdot {(\frac{4}{7})}^{2}$

Multiply the exponents in ${(q^{2})}^{2}$ .

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$q^{2 \cdot 2} \cdot {(\frac{4}{7})}^{0} + 2 \cdot {(q^{2})}^{1} \cdot {(\frac{4}{7})}^{1} + 1 \cdot {(q^{2})}^{0} \cdot {(\frac{4}{7})}^{2}$

Multiply $2$ by $2$ .

$q^{4} \cdot {(\frac{4}{7})}^{0} + 2 \cdot {(q^{2})}^{1} \cdot {(\frac{4}{7})}^{1} + 1 \cdot {(q^{2})}^{0} \cdot {(\frac{4}{7})}^{2}$

Apply the product rule to $\frac{4}{7}$ .

$q^{4} \cdot \frac{4^{0}}{7^{0}} + 2 \cdot {(q^{2})}^{1} \cdot {(\frac{4}{7})}^{1} + 1 \cdot {(q^{2})}^{0} \cdot {(\frac{4}{7})}^{2}$

Anything raised to $0$ is $1$ .

$q^{4} \cdot \frac{1}{7^{0}} + 2 \cdot {(q^{2})}^{1} \cdot {(\frac{4}{7})}^{1} + 1 \cdot {(q^{2})}^{0} \cdot {(\frac{4}{7})}^{2}$

Anything raised to $0$ is $1$ .

$q^{4} \cdot \frac{1}{1} + 2 \cdot {(q^{2})}^{1} \cdot {(\frac{4}{7})}^{1} + 1 \cdot {(q^{2})}^{0} \cdot {(\frac{4}{7})}^{2}$

Divide $1$ by $1$ .

$q^{4} \cdot 1 + 2 \cdot {(q^{2})}^{1} \cdot {(\frac{4}{7})}^{1} + 1 \cdot {(q^{2})}^{0} \cdot {(\frac{4}{7})}^{2}$

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

$q^{4} + 2 \cdot {(q^{2})}^{1} \cdot {(\frac{4}{7})}^{1} + 1 \cdot {(q^{2})}^{0} \cdot {(\frac{4}{7})}^{2}$

Multiply the exponents in ${(q^{2})}^{1}$ .

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$q^{4} + 2 \cdot q^{2 \cdot 1} \cdot {(\frac{4}{7})}^{1} + 1 \cdot {(q^{2})}^{0} \cdot {(\frac{4}{7})}^{2}$

Multiply $2$ by $1$ .

$q^{4} + 2 \cdot q^{2} \cdot {(\frac{4}{7})}^{1} + 1 \cdot {(q^{2})}^{0} \cdot {(\frac{4}{7})}^{2}$

Simplify.

$q^{4} + 2 q^{2} \cdot \frac{4}{7} + 1 \cdot {(q^{2})}^{0} \cdot {(\frac{4}{7})}^{2}$

Multiply $2 q^{2} \frac{4}{7}$ .

Combine $2$ and $\frac{4}{7}$ .

$q^{4} + q^{2} \frac{2 \cdot 4}{7} + 1 \cdot {(q^{2})}^{0} \cdot {(\frac{4}{7})}^{2}$

Multiply $2$ by $4$ .

$q^{4} + q^{2} \frac{8}{7} + 1 \cdot {(q^{2})}^{0} \cdot {(\frac{4}{7})}^{2}$

Combine $q^{2}$ and $\frac{8}{7}$ .

$q^{4} + \frac{q^{2} \cdot 8}{7} + 1 \cdot {(q^{2})}^{0} \cdot {(\frac{4}{7})}^{2}$

Move $8$ to the left of $q^{2}$ .

$q^{4} + \frac{8 \cdot q^{2}}{7} + 1 \cdot {(q^{2})}^{0} \cdot {(\frac{4}{7})}^{2}$

Multiply ${(q^{2})}^{0}$ by $1$ .

$q^{4} + \frac{8 q^{2}}{7} + {(q^{2})}^{0} \cdot {(\frac{4}{7})}^{2}$

Multiply the exponents in ${(q^{2})}^{0}$ .

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$q^{4} + \frac{8 q^{2}}{7} + q^{2 \cdot 0} \cdot {(\frac{4}{7})}^{2}$

Multiply $2$ by $0$ .

$q^{4} + \frac{8 q^{2}}{7} + q^{0} \cdot {(\frac{4}{7})}^{2}$

Anything raised to $0$ is $1$ .

$q^{4} + \frac{8 q^{2}}{7} + 1 \cdot {(\frac{4}{7})}^{2}$

Multiply ${(\frac{4}{7})}^{2}$ by $1$ .

$q^{4} + \frac{8 q^{2}}{7} + {(\frac{4}{7})}^{2}$

Apply the product rule to $\frac{4}{7}$ .

$q^{4} + \frac{8 q^{2}}{7} + \frac{4^{2}}{7^{2}}$

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

$q^{4} + \frac{8 q^{2}}{7} + \frac{16}{7^{2}}$

Raise $7$ to the power of $2$ .

$q^{4} + \frac{8 q^{2}}{7} + \frac{16}{49}$

Expand Using the Binomial Theorem (q^2+4/7)^2

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