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

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

Step 1

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

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

Step 2

Expand the summation.

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

Step 3

Simplify the exponents for each term of the expansion.

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

Step 4

Simplify each term.

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

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

Anything raised to $0$ is $1$ .

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

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

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

Simplify.

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

Evaluate the exponent.

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

Multiply $6$ by $2$ .

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

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

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

Anything raised to $0$ is $1$ .

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

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

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

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

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

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

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