Algebra Examples

${(3 x + 2 y)}^{2}$

Step 1

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

$\sum_{k = 0}^{2} \frac{2!}{(2 - k)! k!} \cdot {(3 x)}^{2 - k} \cdot {(2 y)}^{k}$

Step 2

Expand the summation.

$\frac{2!}{(2 - 0)! 0!} \cdot {(3 x)}^{2 - 0} \cdot {(2 y)}^{0} + \frac{2!}{(2 - 1)! 1!} \cdot {(3 x)}^{2 - 1} \cdot {(2 y)}^{1} + \frac{2!}{(2 - 2)! 2!} \cdot {(3 x)}^{2 - 2} \cdot {(2 y)}^{2}$

Step 3

Simplify the exponents for each term of the expansion.

$1 \cdot {(3 x)}^{2} \cdot {(2 y)}^{0} + 2 \cdot {(3 x)}^{1} \cdot {(2 y)}^{1} + 1 \cdot {(3 x)}^{0} \cdot {(2 y)}^{2}$

Step 4

Simplify each term.

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Step 4.1

Multiply ${(3 x)}^{2}$ by $1$ .

${(3 x)}^{2} \cdot {(2 y)}^{0} + 2 \cdot {(3 x)}^{1} \cdot {(2 y)}^{1} + 1 \cdot {(3 x)}^{0} \cdot {(2 y)}^{2}$

Step 4.2

Apply the product rule to $3 x$ .

$3^{2} x^{2} \cdot {(2 y)}^{0} + 2 \cdot {(3 x)}^{1} \cdot {(2 y)}^{1} + 1 \cdot {(3 x)}^{0} \cdot {(2 y)}^{2}$

Step 4.3

Raise $3$ to the power of $2$ .

$9 x^{2} \cdot {(2 y)}^{0} + 2 \cdot {(3 x)}^{1} \cdot {(2 y)}^{1} + 1 \cdot {(3 x)}^{0} \cdot {(2 y)}^{2}$

Step 4.4

Apply the product rule to $2 y$ .

$9 x^{2} \cdot (2^{0} y^{0}) + 2 \cdot {(3 x)}^{1} \cdot {(2 y)}^{1} + 1 \cdot {(3 x)}^{0} \cdot {(2 y)}^{2}$

Step 4.5

Rewrite using the commutative property of multiplication.

$9 \cdot 2^{0} x^{2} y^{0} + 2 \cdot {(3 x)}^{1} \cdot {(2 y)}^{1} + 1 \cdot {(3 x)}^{0} \cdot {(2 y)}^{2}$

Step 4.6

Anything raised to $0$ is $1$ .

$9 \cdot 1 x^{2} y^{0} + 2 \cdot {(3 x)}^{1} \cdot {(2 y)}^{1} + 1 \cdot {(3 x)}^{0} \cdot {(2 y)}^{2}$

Step 4.7

Multiply $9$ by $1$ .

$9 x^{2} y^{0} + 2 \cdot {(3 x)}^{1} \cdot {(2 y)}^{1} + 1 \cdot {(3 x)}^{0} \cdot {(2 y)}^{2}$

Step 4.8

Anything raised to $0$ is $1$ .

$9 x^{2} \cdot 1 + 2 \cdot {(3 x)}^{1} \cdot {(2 y)}^{1} + 1 \cdot {(3 x)}^{0} \cdot {(2 y)}^{2}$

Step 4.9

Multiply $9$ by $1$ .

$9 x^{2} + 2 \cdot {(3 x)}^{1} \cdot {(2 y)}^{1} + 1 \cdot {(3 x)}^{0} \cdot {(2 y)}^{2}$

Step 4.10

Simplify.

$9 x^{2} + 2 \cdot (3 x) \cdot {(2 y)}^{1} + 1 \cdot {(3 x)}^{0} \cdot {(2 y)}^{2}$

Step 4.11

Multiply $3$ by $2$ .

$9 x^{2} + 6 x \cdot {(2 y)}^{1} + 1 \cdot {(3 x)}^{0} \cdot {(2 y)}^{2}$

Step 4.12

Simplify.

$9 x^{2} + 6 x \cdot (2 y) + 1 \cdot {(3 x)}^{0} \cdot {(2 y)}^{2}$

Step 4.13

Rewrite using the commutative property of multiplication.

$9 x^{2} + 6 \cdot 2 x y + 1 \cdot {(3 x)}^{0} \cdot {(2 y)}^{2}$

Step 4.14

Multiply $6$ by $2$ .

$9 x^{2} + 12 x y + 1 \cdot {(3 x)}^{0} \cdot {(2 y)}^{2}$

Step 4.15

Multiply ${(3 x)}^{0}$ by $1$ .

$9 x^{2} + 12 x y + {(3 x)}^{0} \cdot {(2 y)}^{2}$

Step 4.16

Apply the product rule to $3 x$ .

$9 x^{2} + 12 x y + 3^{0} x^{0} \cdot {(2 y)}^{2}$

Step 4.17

Anything raised to $0$ is $1$ .

$9 x^{2} + 12 x y + 1 x^{0} \cdot {(2 y)}^{2}$

Step 4.18

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

$9 x^{2} + 12 x y + x^{0} \cdot {(2 y)}^{2}$

Step 4.19

Anything raised to $0$ is $1$ .

$9 x^{2} + 12 x y + 1 \cdot {(2 y)}^{2}$

Step 4.20

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

$9 x^{2} + 12 x y + {(2 y)}^{2}$

Step 4.21

Apply the product rule to $2 y$ .

$9 x^{2} + 12 x y + 2^{2} y^{2}$

Step 4.22

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

$9 x^{2} + 12 x y + 4 y^{2}$