Expand Using the Binomial Theorem (a-7)^4

${(a - 7)}^{4}$

Step 1

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

$\sum_{k = 0}^{4} \frac{4!}{(4 - k)! k!} \cdot {(a)}^{4 - k} \cdot {(- 7)}^{k}$

Step 2

Expand the summation.

$\frac{4!}{(4 - 0)! 0!} {(a)}^{4 - 0} \cdot {(- 7)}^{0} + \frac{4!}{(4 - 1)! 1!} {(a)}^{4 - 1} \cdot {(- 7)}^{1} + \frac{4!}{(4 - 2)! 2!} {(a)}^{4 - 2} \cdot {(- 7)}^{2} + \frac{4!}{(4 - 3)! 3!} {(a)}^{4 - 3} \cdot {(- 7)}^{3} + \frac{4!}{(4 - 4)! 4!} {(a)}^{4 - 4} \cdot {(- 7)}^{4}$

Step 3

Simplify the exponents for each term of the expansion.

$1 \cdot {(a)}^{4} \cdot {(- 7)}^{0} + 4 \cdot {(a)}^{3} \cdot {(- 7)}^{1} + 6 \cdot {(a)}^{2} \cdot {(- 7)}^{2} + 4 \cdot {(a)}^{1} \cdot {(- 7)}^{3} + 1 \cdot {(a)}^{0} \cdot {(- 7)}^{4}$

Step 4

Simplify each term.

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

${(a)}^{4} \cdot {(- 7)}^{0} + 4 \cdot {(a)}^{3} \cdot {(- 7)}^{1} + 6 \cdot {(a)}^{2} \cdot {(- 7)}^{2} + 4 \cdot {(a)}^{1} \cdot {(- 7)}^{3} + 1 \cdot {(a)}^{0} \cdot {(- 7)}^{4}$

Anything raised to $0$ is $1$ .

$a^{4} \cdot 1 + 4 \cdot {(a)}^{3} \cdot {(- 7)}^{1} + 6 \cdot {(a)}^{2} \cdot {(- 7)}^{2} + 4 \cdot {(a)}^{1} \cdot {(- 7)}^{3} + 1 \cdot {(a)}^{0} \cdot {(- 7)}^{4}$

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

$a^{4} + 4 \cdot {(a)}^{3} \cdot {(- 7)}^{1} + 6 \cdot {(a)}^{2} \cdot {(- 7)}^{2} + 4 \cdot {(a)}^{1} \cdot {(- 7)}^{3} + 1 \cdot {(a)}^{0} \cdot {(- 7)}^{4}$

Evaluate the exponent.

$a^{4} + 4 a^{3} \cdot - 7 + 6 \cdot {(a)}^{2} \cdot {(- 7)}^{2} + 4 \cdot {(a)}^{1} \cdot {(- 7)}^{3} + 1 \cdot {(a)}^{0} \cdot {(- 7)}^{4}$

Multiply $- 7$ by $4$ .

$a^{4} - 28 a^{3} + 6 \cdot {(a)}^{2} \cdot {(- 7)}^{2} + 4 \cdot {(a)}^{1} \cdot {(- 7)}^{3} + 1 \cdot {(a)}^{0} \cdot {(- 7)}^{4}$

Raise $- 7$ to the power of $2$ .

$a^{4} - 28 a^{3} + 6 a^{2} \cdot 49 + 4 \cdot {(a)}^{1} \cdot {(- 7)}^{3} + 1 \cdot {(a)}^{0} \cdot {(- 7)}^{4}$

Multiply $49$ by $6$ .

$a^{4} - 28 a^{3} + 294 a^{2} + 4 \cdot {(a)}^{1} \cdot {(- 7)}^{3} + 1 \cdot {(a)}^{0} \cdot {(- 7)}^{4}$

Simplify.

$a^{4} - 28 a^{3} + 294 a^{2} + 4 \cdot a \cdot {(- 7)}^{3} + 1 \cdot {(a)}^{0} \cdot {(- 7)}^{4}$

Raise $- 7$ to the power of $3$ .

$a^{4} - 28 a^{3} + 294 a^{2} + 4 a \cdot - 343 + 1 \cdot {(a)}^{0} \cdot {(- 7)}^{4}$

Multiply $- 343$ by $4$ .

$a^{4} - 28 a^{3} + 294 a^{2} - 1372 a + 1 \cdot {(a)}^{0} \cdot {(- 7)}^{4}$

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

$a^{4} - 28 a^{3} + 294 a^{2} - 1372 a + {(a)}^{0} \cdot {(- 7)}^{4}$

Anything raised to $0$ is $1$ .

$a^{4} - 28 a^{3} + 294 a^{2} - 1372 a + 1 \cdot {(- 7)}^{4}$

Multiply ${(- 7)}^{4}$ by $1$ .

$a^{4} - 28 a^{3} + 294 a^{2} - 1372 a + {(- 7)}^{4}$

Raise $- 7$ to the power of $4$ .

$a^{4} - 28 a^{3} + 294 a^{2} - 1372 a + 2401$

Expand Using the Binomial Theorem (a-7)^4

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