Algebra Examples

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

Step 1

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

$\sum_{k = 0}^{2} \frac{2!}{(2 - k)! k!} \cdot {(2 x)}^{2 - k} \cdot {(4 y)}^{k}$

Step 2

Expand the summation.

$\frac{2!}{(2 - 0)! 0!} \cdot {(2 x)}^{2 - 0} \cdot {(4 y)}^{0} + \frac{2!}{(2 - 1)! 1!} \cdot {(2 x)}^{2 - 1} \cdot {(4 y)}^{1} + \frac{2!}{(2 - 2)! 2!} \cdot {(2 x)}^{2 - 2} \cdot {(4 y)}^{2}$

Step 3

Simplify the exponents for each term of the expansion.

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

Step 4

Simplify each term.

Tap for more steps...

Step 4.1

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

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

Step 4.2

Apply the product rule to $2 x$ .

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

Step 4.3

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

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

Step 4.4

Apply the product rule to $4 y$ .

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

Step 4.5

Rewrite using the commutative property of multiplication.

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

Step 4.6

Multiply $4$ by $4^{0}$ by adding the exponents.

Tap for more steps...

Step 4.6.1

Multiply $4$ by $4^{0}$ .

Tap for more steps...

Step 4.6.1.1

Raise $4$ to the power of $1$ .

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

Step 4.6.1.2

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

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

Step 4.6.2

Add $1$ and $0$ .

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

Step 4.7

Simplify $4^{1} x^{2} y^{0}$ .

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

Step 4.8

Simplify.

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

Step 4.9

Multiply $2$ by $2$ .

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

Step 4.10

Simplify.

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

Step 4.11

Rewrite using the commutative property of multiplication.

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

Step 4.12

Multiply $4$ by $4$ .

$4 x^{2} + 16 x y + 1 \cdot {(2 x)}^{0} \cdot {(4 y)}^{2}$

Step 4.13

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

$4 x^{2} + 16 x y + {(2 x)}^{0} \cdot {(4 y)}^{2}$

Step 4.14

Apply the product rule to $2 x$ .

$4 x^{2} + 16 x y + 2^{0} x^{0} \cdot {(4 y)}^{2}$

Step 4.15

Anything raised to $0$ is $1$ .

$4 x^{2} + 16 x y + 1 x^{0} \cdot {(4 y)}^{2}$

Step 4.16

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

$4 x^{2} + 16 x y + x^{0} \cdot {(4 y)}^{2}$

Step 4.17

Anything raised to $0$ is $1$ .

$4 x^{2} + 16 x y + 1 \cdot {(4 y)}^{2}$

Step 4.18

Multiply ${(4 y)}^{2}$ by $1$ .

$4 x^{2} + 16 x y + {(4 y)}^{2}$

Step 4.19

Apply the product rule to $4 y$ .

$4 x^{2} + 16 x y + 4^{2} y^{2}$

Step 4.20

Raise $4$ to the power of $2$ .

$4 x^{2} + 16 x y + 16 y^{2}$