Expand Using the Binomial Theorem (2x+9y)^2

${(2 x + 9 y)}^{2}$

Step 1

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

$\sum_{k = 0}^{2} \frac{2!}{(2 - k)! k!} \cdot {(2 x)}^{2 - k} \cdot {(9 y)}^{k}$

Step 2

Expand the summation.

$\frac{2!}{(2 - 0)! 0!} \cdot {(2 x)}^{2 - 0} \cdot {(9 y)}^{0} + \frac{2!}{(2 - 1)! 1!} \cdot {(2 x)}^{2 - 1} \cdot {(9 y)}^{1} + \frac{2!}{(2 - 2)! 2!} \cdot {(2 x)}^{2 - 2} \cdot {(9 y)}^{2}$

Step 3

Simplify the exponents for each term of the expansion.

$1 \cdot {(2 x)}^{2} \cdot {(9 y)}^{0} + 2 \cdot {(2 x)}^{1} \cdot {(9 y)}^{1} + 1 \cdot {(2 x)}^{0} \cdot {(9 y)}^{2}$

Step 4

Simplify each term.

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

${(2 x)}^{2} \cdot {(9 y)}^{0} + 2 \cdot {(2 x)}^{1} \cdot {(9 y)}^{1} + 1 \cdot {(2 x)}^{0} \cdot {(9 y)}^{2}$

Apply the product rule to $2 x$ .

$2^{2} x^{2} \cdot {(9 y)}^{0} + 2 \cdot {(2 x)}^{1} \cdot {(9 y)}^{1} + 1 \cdot {(2 x)}^{0} \cdot {(9 y)}^{2}$

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

$4 x^{2} \cdot {(9 y)}^{0} + 2 \cdot {(2 x)}^{1} \cdot {(9 y)}^{1} + 1 \cdot {(2 x)}^{0} \cdot {(9 y)}^{2}$

Apply the product rule to $9 y$ .

$4 x^{2} \cdot (9^{0} y^{0}) + 2 \cdot {(2 x)}^{1} \cdot {(9 y)}^{1} + 1 \cdot {(2 x)}^{0} \cdot {(9 y)}^{2}$

Rewrite using the commutative property of multiplication.

$4 \cdot 9^{0} x^{2} y^{0} + 2 \cdot {(2 x)}^{1} \cdot {(9 y)}^{1} + 1 \cdot {(2 x)}^{0} \cdot {(9 y)}^{2}$

Anything raised to $0$ is $1$ .

$4 \cdot 1 x^{2} y^{0} + 2 \cdot {(2 x)}^{1} \cdot {(9 y)}^{1} + 1 \cdot {(2 x)}^{0} \cdot {(9 y)}^{2}$

Multiply $4$ by $1$ .

$4 x^{2} y^{0} + 2 \cdot {(2 x)}^{1} \cdot {(9 y)}^{1} + 1 \cdot {(2 x)}^{0} \cdot {(9 y)}^{2}$

Anything raised to $0$ is $1$ .

$4 x^{2} \cdot 1 + 2 \cdot {(2 x)}^{1} \cdot {(9 y)}^{1} + 1 \cdot {(2 x)}^{0} \cdot {(9 y)}^{2}$

Multiply $4$ by $1$ .

$4 x^{2} + 2 \cdot {(2 x)}^{1} \cdot {(9 y)}^{1} + 1 \cdot {(2 x)}^{0} \cdot {(9 y)}^{2}$

Simplify.

$4 x^{2} + 2 \cdot (2 x) \cdot {(9 y)}^{1} + 1 \cdot {(2 x)}^{0} \cdot {(9 y)}^{2}$

Multiply $2$ by $2$ .

$4 x^{2} + 4 x \cdot {(9 y)}^{1} + 1 \cdot {(2 x)}^{0} \cdot {(9 y)}^{2}$

Simplify.

$4 x^{2} + 4 x \cdot (9 y) + 1 \cdot {(2 x)}^{0} \cdot {(9 y)}^{2}$

Rewrite using the commutative property of multiplication.

$4 x^{2} + 4 \cdot 9 x y + 1 \cdot {(2 x)}^{0} \cdot {(9 y)}^{2}$

Multiply $4$ by $9$ .

$4 x^{2} + 36 x y + 1 \cdot {(2 x)}^{0} \cdot {(9 y)}^{2}$

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

$4 x^{2} + 36 x y + {(2 x)}^{0} \cdot {(9 y)}^{2}$

Apply the product rule to $2 x$ .

$4 x^{2} + 36 x y + 2^{0} x^{0} \cdot {(9 y)}^{2}$

Anything raised to $0$ is $1$ .

$4 x^{2} + 36 x y + 1 x^{0} \cdot {(9 y)}^{2}$

Multiply $x^{0}$ by $1$ .

$4 x^{2} + 36 x y + x^{0} \cdot {(9 y)}^{2}$

Anything raised to $0$ is $1$ .

$4 x^{2} + 36 x y + 1 \cdot {(9 y)}^{2}$

Multiply ${(9 y)}^{2}$ by $1$ .

$4 x^{2} + 36 x y + {(9 y)}^{2}$

Apply the product rule to $9 y$ .

$4 x^{2} + 36 x y + 9^{2} y^{2}$

Raise $9$ to the power of $2$ .

$4 x^{2} + 36 x y + 81 y^{2}$

Expand Using the Binomial Theorem (2x+9y)^2

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