Expand Using the Binomial Theorem (9-5x)^2

${(9 - 5 x)}^{2}$

Step 1

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

$\sum_{k = 0}^{2} \frac{2!}{(2 - k)! k!} \cdot {(9)}^{2 - k} \cdot {(- 5 x)}^{k}$

Step 2

Expand the summation.

$\frac{2!}{(2 - 0)! 0!} \cdot {(9)}^{2 - 0} \cdot {(- 5 x)}^{0} + \frac{2!}{(2 - 1)! 1!} \cdot {(9)}^{2 - 1} \cdot {(- 5 x)}^{1} + \frac{2!}{(2 - 2)! 2!} \cdot {(9)}^{2 - 2} \cdot {(- 5 x)}^{2}$

Step 3

Simplify the exponents for each term of the expansion.

$1 \cdot {(9)}^{2} \cdot {(- 5 x)}^{0} + 2 \cdot {(9)}^{1} \cdot {(- 5 x)}^{1} + 1 \cdot {(9)}^{0} \cdot {(- 5 x)}^{2}$

Step 4

Simplify each term.

Multiply ${(9)}^{2}$ by $1$ .

${(9)}^{2} \cdot {(- 5 x)}^{0} + 2 \cdot {(9)}^{1} \cdot {(- 5 x)}^{1} + 1 \cdot {(9)}^{0} \cdot {(- 5 x)}^{2}$

Raise $9$ to the power of $2$ .

$81 \cdot {(- 5 x)}^{0} + 2 \cdot {(9)}^{1} \cdot {(- 5 x)}^{1} + 1 \cdot {(9)}^{0} \cdot {(- 5 x)}^{2}$

Apply the product rule to $- 5 x$ .

$81 \cdot ({(- 5)}^{0} x^{0}) + 2 \cdot {(9)}^{1} \cdot {(- 5 x)}^{1} + 1 \cdot {(9)}^{0} \cdot {(- 5 x)}^{2}$

Anything raised to $0$ is $1$ .

$81 \cdot (1 x^{0}) + 2 \cdot {(9)}^{1} \cdot {(- 5 x)}^{1} + 1 \cdot {(9)}^{0} \cdot {(- 5 x)}^{2}$

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

$81 \cdot x^{0} + 2 \cdot {(9)}^{1} \cdot {(- 5 x)}^{1} + 1 \cdot {(9)}^{0} \cdot {(- 5 x)}^{2}$

Anything raised to $0$ is $1$ .

$81 \cdot 1 + 2 \cdot {(9)}^{1} \cdot {(- 5 x)}^{1} + 1 \cdot {(9)}^{0} \cdot {(- 5 x)}^{2}$

Multiply $81$ by $1$ .

$81 + 2 \cdot {(9)}^{1} \cdot {(- 5 x)}^{1} + 1 \cdot {(9)}^{0} \cdot {(- 5 x)}^{2}$

Evaluate the exponent.

$81 + 2 \cdot 9 \cdot {(- 5 x)}^{1} + 1 \cdot {(9)}^{0} \cdot {(- 5 x)}^{2}$

Multiply $2$ by $9$ .

$81 + 18 \cdot {(- 5 x)}^{1} + 1 \cdot {(9)}^{0} \cdot {(- 5 x)}^{2}$

Simplify.

$81 + 18 \cdot (- 5 x) + 1 \cdot {(9)}^{0} \cdot {(- 5 x)}^{2}$

Multiply $- 5$ by $18$ .

$81 - 90 x + 1 \cdot {(9)}^{0} \cdot {(- 5 x)}^{2}$

Multiply ${(9)}^{0}$ by $1$ .

$81 - 90 x + {(9)}^{0} \cdot {(- 5 x)}^{2}$

Anything raised to $0$ is $1$ .

$81 - 90 x + 1 \cdot {(- 5 x)}^{2}$

Multiply ${(- 5 x)}^{2}$ by $1$ .

$81 - 90 x + {(- 5 x)}^{2}$

Apply the product rule to $- 5 x$ .

$81 - 90 x + {(- 5)}^{2} x^{2}$

Raise $- 5$ to the power of $2$ .

$81 - 90 x + 25 x^{2}$

Expand Using the Binomial Theorem (9-5x)^2

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