Expand Using the Binomial Theorem (6x+5y)^2

${(6 x + 5 y)}^{2}$

Step 1

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

$\sum_{k = 0}^{2} \frac{2!}{(2 - k)! k!} \cdot {(6 x)}^{2 - k} \cdot {(5 y)}^{k}$

Step 2

Expand the summation.

$\frac{2!}{(2 - 0)! 0!} \cdot {(6 x)}^{2 - 0} \cdot {(5 y)}^{0} + \frac{2!}{(2 - 1)! 1!} \cdot {(6 x)}^{2 - 1} \cdot {(5 y)}^{1} + \frac{2!}{(2 - 2)! 2!} \cdot {(6 x)}^{2 - 2} \cdot {(5 y)}^{2}$

Step 3

Simplify the exponents for each term of the expansion.

$1 \cdot {(6 x)}^{2} \cdot {(5 y)}^{0} + 2 \cdot {(6 x)}^{1} \cdot {(5 y)}^{1} + 1 \cdot {(6 x)}^{0} \cdot {(5 y)}^{2}$

Step 4

Simplify each term.

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

${(6 x)}^{2} \cdot {(5 y)}^{0} + 2 \cdot {(6 x)}^{1} \cdot {(5 y)}^{1} + 1 \cdot {(6 x)}^{0} \cdot {(5 y)}^{2}$

Apply the product rule to $6 x$ .

$6^{2} x^{2} \cdot {(5 y)}^{0} + 2 \cdot {(6 x)}^{1} \cdot {(5 y)}^{1} + 1 \cdot {(6 x)}^{0} \cdot {(5 y)}^{2}$

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

$36 x^{2} \cdot {(5 y)}^{0} + 2 \cdot {(6 x)}^{1} \cdot {(5 y)}^{1} + 1 \cdot {(6 x)}^{0} \cdot {(5 y)}^{2}$

Apply the product rule to $5 y$ .

$36 x^{2} \cdot (5^{0} y^{0}) + 2 \cdot {(6 x)}^{1} \cdot {(5 y)}^{1} + 1 \cdot {(6 x)}^{0} \cdot {(5 y)}^{2}$

Rewrite using the commutative property of multiplication.

$36 \cdot 5^{0} x^{2} y^{0} + 2 \cdot {(6 x)}^{1} \cdot {(5 y)}^{1} + 1 \cdot {(6 x)}^{0} \cdot {(5 y)}^{2}$

Anything raised to $0$ is $1$ .

$36 \cdot 1 x^{2} y^{0} + 2 \cdot {(6 x)}^{1} \cdot {(5 y)}^{1} + 1 \cdot {(6 x)}^{0} \cdot {(5 y)}^{2}$

Multiply $36$ by $1$ .

$36 x^{2} y^{0} + 2 \cdot {(6 x)}^{1} \cdot {(5 y)}^{1} + 1 \cdot {(6 x)}^{0} \cdot {(5 y)}^{2}$

Anything raised to $0$ is $1$ .

$36 x^{2} \cdot 1 + 2 \cdot {(6 x)}^{1} \cdot {(5 y)}^{1} + 1 \cdot {(6 x)}^{0} \cdot {(5 y)}^{2}$

Multiply $36$ by $1$ .

$36 x^{2} + 2 \cdot {(6 x)}^{1} \cdot {(5 y)}^{1} + 1 \cdot {(6 x)}^{0} \cdot {(5 y)}^{2}$

Simplify.

$36 x^{2} + 2 \cdot (6 x) \cdot {(5 y)}^{1} + 1 \cdot {(6 x)}^{0} \cdot {(5 y)}^{2}$

Multiply $6$ by $2$ .

$36 x^{2} + 12 x \cdot {(5 y)}^{1} + 1 \cdot {(6 x)}^{0} \cdot {(5 y)}^{2}$

Simplify.

$36 x^{2} + 12 x \cdot (5 y) + 1 \cdot {(6 x)}^{0} \cdot {(5 y)}^{2}$

Rewrite using the commutative property of multiplication.

$36 x^{2} + 12 \cdot 5 x y + 1 \cdot {(6 x)}^{0} \cdot {(5 y)}^{2}$

Multiply $12$ by $5$ .

$36 x^{2} + 60 x y + 1 \cdot {(6 x)}^{0} \cdot {(5 y)}^{2}$

Multiply ${(6 x)}^{0}$ by $1$ .

$36 x^{2} + 60 x y + {(6 x)}^{0} \cdot {(5 y)}^{2}$

Apply the product rule to $6 x$ .

$36 x^{2} + 60 x y + 6^{0} x^{0} \cdot {(5 y)}^{2}$

Anything raised to $0$ is $1$ .

$36 x^{2} + 60 x y + 1 x^{0} \cdot {(5 y)}^{2}$

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

$36 x^{2} + 60 x y + x^{0} \cdot {(5 y)}^{2}$

Anything raised to $0$ is $1$ .

$36 x^{2} + 60 x y + 1 \cdot {(5 y)}^{2}$

Multiply ${(5 y)}^{2}$ by $1$ .

$36 x^{2} + 60 x y + {(5 y)}^{2}$

Apply the product rule to $5 y$ .

$36 x^{2} + 60 x y + 5^{2} y^{2}$

Raise $5$ to the power of $2$ .

$36 x^{2} + 60 x y + 25 y^{2}$

Expand Using the Binomial Theorem (6x+5y)^2

Related Questions

Support LCMGCF.com by disabling your adblocker.