Expand Using the Binomial Theorem (ax+by)^4

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

Step 1

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

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

Step 2

Expand the summation.

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

Step 3

Simplify the exponents for each term of the expansion.

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

Step 4

Simplify each term.

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

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

Apply the product rule to $a x$ .

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

Apply the product rule to $b y$ .

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

Anything raised to $0$ is $1$ .

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

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

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

Anything raised to $0$ is $1$ .

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

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

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

Apply the product rule to $a x$ .

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

Simplify.

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

Apply the product rule to $a x$ .

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

Apply the product rule to $b y$ .

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

Simplify.

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

Apply the product rule to $b y$ .

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

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

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

Apply the product rule to $a x$ .

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

Anything raised to $0$ is $1$ .

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

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

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

Anything raised to $0$ is $1$ .

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

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

$a^{4} x^{4} + 4 a^{3} x^{3} b y + 6 a^{2} x^{2} b^{2} y^{2} + 4 a x b^{3} y^{3} + {(b y)}^{4}$

Apply the product rule to $b y$ .

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

Expand Using the Binomial Theorem (ax+by)^4

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