Expand Using the Binomial Theorem (4a^2+b)^2

${(4 a^{2} + b)}^{2}$

Step 1

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

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

Step 2

Expand the summation.

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

Step 3

Simplify the exponents for each term of the expansion.

$1 \cdot {(4 a^{2})}^{2} \cdot {(b)}^{0} + 2 \cdot {(4 a^{2})}^{1} \cdot {(b)}^{1} + 1 \cdot {(4 a^{2})}^{0} \cdot {(b)}^{2}$

Step 4

Simplify each term.

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

${(4 a^{2})}^{2} \cdot {(b)}^{0} + 2 \cdot {(4 a^{2})}^{1} \cdot {(b)}^{1} + 1 \cdot {(4 a^{2})}^{0} \cdot {(b)}^{2}$

Apply the product rule to $4 a^{2}$ .

$4^{2} {(a^{2})}^{2} \cdot {(b)}^{0} + 2 \cdot {(4 a^{2})}^{1} \cdot {(b)}^{1} + 1 \cdot {(4 a^{2})}^{0} \cdot {(b)}^{2}$

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

$16 {(a^{2})}^{2} \cdot {(b)}^{0} + 2 \cdot {(4 a^{2})}^{1} \cdot {(b)}^{1} + 1 \cdot {(4 a^{2})}^{0} \cdot {(b)}^{2}$

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

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

$16 a^{2 \cdot 2} \cdot {(b)}^{0} + 2 \cdot {(4 a^{2})}^{1} \cdot {(b)}^{1} + 1 \cdot {(4 a^{2})}^{0} \cdot {(b)}^{2}$

Multiply $2$ by $2$ .

$16 a^{4} \cdot {(b)}^{0} + 2 \cdot {(4 a^{2})}^{1} \cdot {(b)}^{1} + 1 \cdot {(4 a^{2})}^{0} \cdot {(b)}^{2}$

Anything raised to $0$ is $1$ .

$16 a^{4} \cdot 1 + 2 \cdot {(4 a^{2})}^{1} \cdot {(b)}^{1} + 1 \cdot {(4 a^{2})}^{0} \cdot {(b)}^{2}$

Multiply $16$ by $1$ .

$16 a^{4} + 2 \cdot {(4 a^{2})}^{1} \cdot {(b)}^{1} + 1 \cdot {(4 a^{2})}^{0} \cdot {(b)}^{2}$

Simplify.

$16 a^{4} + 2 \cdot (4 a^{2}) \cdot {(b)}^{1} + 1 \cdot {(4 a^{2})}^{0} \cdot {(b)}^{2}$

Multiply $4$ by $2$ .

$16 a^{4} + 8 a^{2} \cdot {(b)}^{1} + 1 \cdot {(4 a^{2})}^{0} \cdot {(b)}^{2}$

Simplify.

$16 a^{4} + 8 a^{2} \cdot b + 1 \cdot {(4 a^{2})}^{0} \cdot {(b)}^{2}$

Multiply ${(4 a^{2})}^{0}$ by $1$ .

$16 a^{4} + 8 a^{2} b + {(4 a^{2})}^{0} \cdot {(b)}^{2}$

Apply the product rule to $4 a^{2}$ .

$16 a^{4} + 8 a^{2} b + 4^{0} {(a^{2})}^{0} \cdot {(b)}^{2}$

Anything raised to $0$ is $1$ .

$16 a^{4} + 8 a^{2} b + 1 {(a^{2})}^{0} \cdot {(b)}^{2}$

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

$16 a^{4} + 8 a^{2} b + {(a^{2})}^{0} \cdot {(b)}^{2}$

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

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

$16 a^{4} + 8 a^{2} b + a^{2 \cdot 0} \cdot {(b)}^{2}$

Multiply $2$ by $0$ .

$16 a^{4} + 8 a^{2} b + a^{0} \cdot {(b)}^{2}$

Anything raised to $0$ is $1$ .

$16 a^{4} + 8 a^{2} b + 1 \cdot {(b)}^{2}$

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

$16 a^{4} + 8 a^{2} b + b^{2}$

Expand Using the Binomial Theorem (4a^2+b)^2

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