Algebra Examples

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

Step 1

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

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

Step 2

Expand the summation.

$\frac{4!}{(4 - 0)! 0!} {(x)}^{4 - 0} \cdot {(2 y)}^{0} + \frac{4!}{(4 - 1)! 1!} {(x)}^{4 - 1} \cdot {(2 y)}^{1} + \frac{4!}{(4 - 2)! 2!} {(x)}^{4 - 2} \cdot {(2 y)}^{2} + \frac{4!}{(4 - 3)! 3!} {(x)}^{4 - 3} \cdot {(2 y)}^{3} + \frac{4!}{(4 - 4)! 4!} {(x)}^{4 - 4} \cdot {(2 y)}^{4}$

Step 3

Simplify the exponents for each term of the expansion.

$1 \cdot {(x)}^{4} \cdot {(2 y)}^{0} + 4 \cdot {(x)}^{3} \cdot {(2 y)}^{1} + 6 \cdot {(x)}^{2} \cdot {(2 y)}^{2} + 4 \cdot {(x)}^{1} \cdot {(2 y)}^{3} + 1 \cdot {(x)}^{0} \cdot {(2 y)}^{4}$

Step 4

Simplify each term.

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Step 4.1

Multiply ${(x)}^{4}$ by $1$ .

${(x)}^{4} \cdot {(2 y)}^{0} + 4 \cdot {(x)}^{3} \cdot {(2 y)}^{1} + 6 \cdot {(x)}^{2} \cdot {(2 y)}^{2} + 4 \cdot {(x)}^{1} \cdot {(2 y)}^{3} + 1 \cdot {(x)}^{0} \cdot {(2 y)}^{4}$

Step 4.2

Apply the product rule to $2 y$ .

$x^{4} \cdot (2^{0} y^{0}) + 4 \cdot {(x)}^{3} \cdot {(2 y)}^{1} + 6 \cdot {(x)}^{2} \cdot {(2 y)}^{2} + 4 \cdot {(x)}^{1} \cdot {(2 y)}^{3} + 1 \cdot {(x)}^{0} \cdot {(2 y)}^{4}$

Step 4.3

Rewrite using the commutative property of multiplication.

$2^{0} x^{4} y^{0} + 4 \cdot {(x)}^{3} \cdot {(2 y)}^{1} + 6 \cdot {(x)}^{2} \cdot {(2 y)}^{2} + 4 \cdot {(x)}^{1} \cdot {(2 y)}^{3} + 1 \cdot {(x)}^{0} \cdot {(2 y)}^{4}$

Step 4.4

Anything raised to $0$ is $1$ .

$1 x^{4} y^{0} + 4 \cdot {(x)}^{3} \cdot {(2 y)}^{1} + 6 \cdot {(x)}^{2} \cdot {(2 y)}^{2} + 4 \cdot {(x)}^{1} \cdot {(2 y)}^{3} + 1 \cdot {(x)}^{0} \cdot {(2 y)}^{4}$

Step 4.5

Multiply $x^{4}$ by $1$ .

$x^{4} y^{0} + 4 \cdot {(x)}^{3} \cdot {(2 y)}^{1} + 6 \cdot {(x)}^{2} \cdot {(2 y)}^{2} + 4 \cdot {(x)}^{1} \cdot {(2 y)}^{3} + 1 \cdot {(x)}^{0} \cdot {(2 y)}^{4}$

Step 4.6

Anything raised to $0$ is $1$ .

$x^{4} \cdot 1 + 4 \cdot {(x)}^{3} \cdot {(2 y)}^{1} + 6 \cdot {(x)}^{2} \cdot {(2 y)}^{2} + 4 \cdot {(x)}^{1} \cdot {(2 y)}^{3} + 1 \cdot {(x)}^{0} \cdot {(2 y)}^{4}$

Step 4.7

Multiply $x^{4}$ by $1$ .

$x^{4} + 4 \cdot {(x)}^{3} \cdot {(2 y)}^{1} + 6 \cdot {(x)}^{2} \cdot {(2 y)}^{2} + 4 \cdot {(x)}^{1} \cdot {(2 y)}^{3} + 1 \cdot {(x)}^{0} \cdot {(2 y)}^{4}$

Step 4.8

Simplify.

$x^{4} + 4 x^{3} \cdot (2 y) + 6 \cdot {(x)}^{2} \cdot {(2 y)}^{2} + 4 \cdot {(x)}^{1} \cdot {(2 y)}^{3} + 1 \cdot {(x)}^{0} \cdot {(2 y)}^{4}$

Step 4.9

Rewrite using the commutative property of multiplication.

$x^{4} + 4 \cdot 2 x^{3} y + 6 \cdot {(x)}^{2} \cdot {(2 y)}^{2} + 4 \cdot {(x)}^{1} \cdot {(2 y)}^{3} + 1 \cdot {(x)}^{0} \cdot {(2 y)}^{4}$

Step 4.10

Multiply $4$ by $2$ .

$x^{4} + 8 x^{3} y + 6 \cdot {(x)}^{2} \cdot {(2 y)}^{2} + 4 \cdot {(x)}^{1} \cdot {(2 y)}^{3} + 1 \cdot {(x)}^{0} \cdot {(2 y)}^{4}$

Step 4.11

Apply the product rule to $2 y$ .

$x^{4} + 8 x^{3} y + 6 x^{2} \cdot (2^{2} y^{2}) + 4 \cdot {(x)}^{1} \cdot {(2 y)}^{3} + 1 \cdot {(x)}^{0} \cdot {(2 y)}^{4}$

Step 4.12

Rewrite using the commutative property of multiplication.

$x^{4} + 8 x^{3} y + 6 \cdot 2^{2} x^{2} y^{2} + 4 \cdot {(x)}^{1} \cdot {(2 y)}^{3} + 1 \cdot {(x)}^{0} \cdot {(2 y)}^{4}$

Step 4.13

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

$x^{4} + 8 x^{3} y + 6 \cdot 4 x^{2} y^{2} + 4 \cdot {(x)}^{1} \cdot {(2 y)}^{3} + 1 \cdot {(x)}^{0} \cdot {(2 y)}^{4}$

Step 4.14

Multiply $6$ by $4$ .

$x^{4} + 8 x^{3} y + 24 x^{2} y^{2} + 4 \cdot {(x)}^{1} \cdot {(2 y)}^{3} + 1 \cdot {(x)}^{0} \cdot {(2 y)}^{4}$

Step 4.15

Simplify.

$x^{4} + 8 x^{3} y + 24 x^{2} y^{2} + 4 \cdot x \cdot {(2 y)}^{3} + 1 \cdot {(x)}^{0} \cdot {(2 y)}^{4}$

Step 4.16

Apply the product rule to $2 y$ .

$x^{4} + 8 x^{3} y + 24 x^{2} y^{2} + 4 x \cdot (2^{3} y^{3}) + 1 \cdot {(x)}^{0} \cdot {(2 y)}^{4}$

Step 4.17

Rewrite using the commutative property of multiplication.

$x^{4} + 8 x^{3} y + 24 x^{2} y^{2} + 4 \cdot 2^{3} x y^{3} + 1 \cdot {(x)}^{0} \cdot {(2 y)}^{4}$

Step 4.18

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

$x^{4} + 8 x^{3} y + 24 x^{2} y^{2} + 4 \cdot 8 x y^{3} + 1 \cdot {(x)}^{0} \cdot {(2 y)}^{4}$

Step 4.19

Multiply $4$ by $8$ .

$x^{4} + 8 x^{3} y + 24 x^{2} y^{2} + 32 x y^{3} + 1 \cdot {(x)}^{0} \cdot {(2 y)}^{4}$

Step 4.20

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

$x^{4} + 8 x^{3} y + 24 x^{2} y^{2} + 32 x y^{3} + {(x)}^{0} \cdot {(2 y)}^{4}$

Step 4.21

Anything raised to $0$ is $1$ .

$x^{4} + 8 x^{3} y + 24 x^{2} y^{2} + 32 x y^{3} + 1 \cdot {(2 y)}^{4}$

Step 4.22

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

$x^{4} + 8 x^{3} y + 24 x^{2} y^{2} + 32 x y^{3} + {(2 y)}^{4}$

Step 4.23

Apply the product rule to $2 y$ .

$x^{4} + 8 x^{3} y + 24 x^{2} y^{2} + 32 x y^{3} + 2^{4} y^{4}$

Step 4.24

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

$x^{4} + 8 x^{3} y + 24 x^{2} y^{2} + 32 x y^{3} + 16 y^{4}$