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

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

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

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

Expand the summation.

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

Simplify the exponents for each term of the expansion.

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

Simplify each term.

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

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

Anything raised to $0$ is $1$ .

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

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

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

Simplify.

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

Simplify.

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

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

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

Anything raised to $0$ is $1$ .

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

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

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

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

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