Expand Using the Binomial Theorem (x-14)^2

${(x - 14)}^{2}$

Step 1

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

$\sum_{k = 0}^{2} \frac{2!}{(2 - k)! k!} \cdot {(x)}^{2 - k} \cdot {(- 14)}^{k}$

Step 2

Expand the summation.

$\frac{2!}{(2 - 0)! 0!} {(x)}^{2 - 0} \cdot {(- 14)}^{0} + \frac{2!}{(2 - 1)! 1!} {(x)}^{2 - 1} \cdot {(- 14)}^{1} + \frac{2!}{(2 - 2)! 2!} {(x)}^{2 - 2} \cdot {(- 14)}^{2}$

Step 3

Simplify the exponents for each term of the expansion.

$1 \cdot {(x)}^{2} \cdot {(- 14)}^{0} + 2 \cdot {(x)}^{1} \cdot {(- 14)}^{1} + 1 \cdot {(x)}^{0} \cdot {(- 14)}^{2}$

Step 4

Simplify each term.

Multiply ${(x)}^{2}$ by $1$ .

${(x)}^{2} \cdot {(- 14)}^{0} + 2 \cdot {(x)}^{1} \cdot {(- 14)}^{1} + 1 \cdot {(x)}^{0} \cdot {(- 14)}^{2}$

Anything raised to $0$ is $1$ .

$x^{2} \cdot 1 + 2 \cdot {(x)}^{1} \cdot {(- 14)}^{1} + 1 \cdot {(x)}^{0} \cdot {(- 14)}^{2}$

Multiply $x^{2}$ by $1$ .

$x^{2} + 2 \cdot {(x)}^{1} \cdot {(- 14)}^{1} + 1 \cdot {(x)}^{0} \cdot {(- 14)}^{2}$

Simplify.

$x^{2} + 2 \cdot x \cdot {(- 14)}^{1} + 1 \cdot {(x)}^{0} \cdot {(- 14)}^{2}$

Evaluate the exponent.

$x^{2} + 2 x \cdot - 14 + 1 \cdot {(x)}^{0} \cdot {(- 14)}^{2}$

Multiply $- 14$ by $2$ .

$x^{2} - 28 x + 1 \cdot {(x)}^{0} \cdot {(- 14)}^{2}$

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

$x^{2} - 28 x + {(x)}^{0} \cdot {(- 14)}^{2}$

Anything raised to $0$ is $1$ .

$x^{2} - 28 x + 1 \cdot {(- 14)}^{2}$

Multiply ${(- 14)}^{2}$ by $1$ .

$x^{2} - 28 x + {(- 14)}^{2}$

Raise $- 14$ to the power of $2$ .

$x^{2} - 28 x + 196$

Expand Using the Binomial Theorem (x-14)^2

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