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

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

Step 1

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

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

Step 2

Expand the summation.

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

Step 3

Simplify the exponents for each term of the expansion.

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

Step 4

Simplify each term.

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

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

Apply the product rule to $9 x$ .

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

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

$81 x^{2} \cdot {(- y)}^{0} + 2 \cdot {(9 x)}^{1} \cdot {(- y)}^{1} + 1 \cdot {(9 x)}^{0} \cdot {(- y)}^{2}$

Apply the product rule to $- y$ .

$81 x^{2} \cdot ({(- 1)}^{0} y^{0}) + 2 \cdot {(9 x)}^{1} \cdot {(- y)}^{1} + 1 \cdot {(9 x)}^{0} \cdot {(- y)}^{2}$

Rewrite using the commutative property of multiplication.

$81 \cdot {(- 1)}^{0} x^{2} y^{0} + 2 \cdot {(9 x)}^{1} \cdot {(- y)}^{1} + 1 \cdot {(9 x)}^{0} \cdot {(- y)}^{2}$

Anything raised to $0$ is $1$ .

$81 \cdot 1 x^{2} y^{0} + 2 \cdot {(9 x)}^{1} \cdot {(- y)}^{1} + 1 \cdot {(9 x)}^{0} \cdot {(- y)}^{2}$

Multiply $81$ by $1$ .

$81 x^{2} y^{0} + 2 \cdot {(9 x)}^{1} \cdot {(- y)}^{1} + 1 \cdot {(9 x)}^{0} \cdot {(- y)}^{2}$

Anything raised to $0$ is $1$ .

$81 x^{2} \cdot 1 + 2 \cdot {(9 x)}^{1} \cdot {(- y)}^{1} + 1 \cdot {(9 x)}^{0} \cdot {(- y)}^{2}$

Multiply $81$ by $1$ .

$81 x^{2} + 2 \cdot {(9 x)}^{1} \cdot {(- y)}^{1} + 1 \cdot {(9 x)}^{0} \cdot {(- y)}^{2}$

Simplify.

$81 x^{2} + 2 \cdot (9 x) \cdot {(- y)}^{1} + 1 \cdot {(9 x)}^{0} \cdot {(- y)}^{2}$

Multiply $9$ by $2$ .

$81 x^{2} + 18 x \cdot {(- y)}^{1} + 1 \cdot {(9 x)}^{0} \cdot {(- y)}^{2}$

Simplify.

$81 x^{2} + 18 x \cdot (- y) + 1 \cdot {(9 x)}^{0} \cdot {(- y)}^{2}$

Rewrite using the commutative property of multiplication.

$81 x^{2} + 18 \cdot - 1 x y + 1 \cdot {(9 x)}^{0} \cdot {(- y)}^{2}$

Multiply $18$ by $- 1$ .

$81 x^{2} - 18 x y + 1 \cdot {(9 x)}^{0} \cdot {(- y)}^{2}$

Multiply ${(9 x)}^{0}$ by $1$ .

$81 x^{2} - 18 x y + {(9 x)}^{0} \cdot {(- y)}^{2}$

Apply the product rule to $9 x$ .

$81 x^{2} - 18 x y + 9^{0} x^{0} \cdot {(- y)}^{2}$

Anything raised to $0$ is $1$ .

$81 x^{2} - 18 x y + 1 x^{0} \cdot {(- y)}^{2}$

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

$81 x^{2} - 18 x y + x^{0} \cdot {(- y)}^{2}$

Anything raised to $0$ is $1$ .

$81 x^{2} - 18 x y + 1 \cdot {(- y)}^{2}$

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

$81 x^{2} - 18 x y + {(- y)}^{2}$

Apply the product rule to $- y$ .

$81 x^{2} - 18 x y + {(- 1)}^{2} y^{2}$

Raise $- 1$ to the power of $2$ .

$81 x^{2} - 18 x y + 1 y^{2}$

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

$81 x^{2} - 18 x y + y^{2}$

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

Related Questions

Support LCMGCF.com by disabling your adblocker.