Expand Using the Binomial Theorem (x+y)^4

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

Step 1

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

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

Step 2

Expand the summation.

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

Step 3

Simplify the exponents for each term of the expansion.

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

Step 4

Simplify each term.

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

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

Anything raised to $0$ is $1$ .

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

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

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

Simplify.

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

Simplify.

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

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

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

Anything raised to $0$ is $1$ .

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

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

$x^{4} + 4 x^{3} y + 6 x^{2} y^{2} + 4 x y^{3} + y^{4}$

Expand Using the Binomial Theorem (x+y)^4

Related Questions

Support LCMGCF.com by disabling your adblocker.