Expand Using the Binomial Theorem (a+6)^2

${(a + 6)}^{2}$

Step 1

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

$\sum_{k = 0}^{2} \frac{2!}{(2 - k)! k!} \cdot {(a)}^{2 - k} \cdot {(6)}^{k}$

Step 2

Expand the summation.

$\frac{2!}{(2 - 0)! 0!} {(a)}^{2 - 0} \cdot {(6)}^{0} + \frac{2!}{(2 - 1)! 1!} {(a)}^{2 - 1} \cdot {(6)}^{1} + \frac{2!}{(2 - 2)! 2!} {(a)}^{2 - 2} \cdot {(6)}^{2}$

Step 3

Simplify the exponents for each term of the expansion.

$1 \cdot {(a)}^{2} \cdot {(6)}^{0} + 2 \cdot {(a)}^{1} \cdot {(6)}^{1} + 1 \cdot {(a)}^{0} \cdot {(6)}^{2}$

Step 4

Simplify each term.

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

${(a)}^{2} \cdot {(6)}^{0} + 2 \cdot {(a)}^{1} \cdot {(6)}^{1} + 1 \cdot {(a)}^{0} \cdot {(6)}^{2}$

Anything raised to $0$ is $1$ .

$a^{2} \cdot 1 + 2 \cdot {(a)}^{1} \cdot {(6)}^{1} + 1 \cdot {(a)}^{0} \cdot {(6)}^{2}$

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

$a^{2} + 2 \cdot {(a)}^{1} \cdot {(6)}^{1} + 1 \cdot {(a)}^{0} \cdot {(6)}^{2}$

Simplify.

$a^{2} + 2 \cdot a \cdot {(6)}^{1} + 1 \cdot {(a)}^{0} \cdot {(6)}^{2}$

Evaluate the exponent.

$a^{2} + 2 a \cdot 6 + 1 \cdot {(a)}^{0} \cdot {(6)}^{2}$

Multiply $6$ by $2$ .

$a^{2} + 12 a + 1 \cdot {(a)}^{0} \cdot {(6)}^{2}$

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

$a^{2} + 12 a + {(a)}^{0} \cdot {(6)}^{2}$

Anything raised to $0$ is $1$ .

$a^{2} + 12 a + 1 \cdot {(6)}^{2}$

Multiply ${(6)}^{2}$ by $1$ .

$a^{2} + 12 a + {(6)}^{2}$

Raise $6$ to the power of $2$ .

$a^{2} + 12 a + 36$

Expand Using the Binomial Theorem (a+6)^2

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