Expand Using the Binomial Theorem ((x+3)-5)^2

${((x + 3) - 5)}^{2}$

Step 1

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

$\sum_{k = 0}^{2} \frac{2!}{(2 - k)! k!} \cdot {(x + 3)}^{2 - k} \cdot {(- 5)}^{k}$

Step 2

Expand the summation.

$\frac{2!}{(2 - 0)! 0!} \cdot {(x + 3)}^{2 - 0} \cdot {(- 5)}^{0} + \frac{2!}{(2 - 1)! 1!} \cdot {(x + 3)}^{2 - 1} \cdot {(- 5)}^{1} + \frac{2!}{(2 - 2)! 2!} \cdot {(x + 3)}^{2 - 2} \cdot {(- 5)}^{2}$

Step 3

Simplify the exponents for each term of the expansion.

$1 \cdot {(x + 3)}^{2} \cdot {(- 5)}^{0} + 2 \cdot {(x + 3)}^{1} \cdot {(- 5)}^{1} + 1 \cdot {(x + 3)}^{0} \cdot {(- 5)}^{2}$

Step 4

Simplify the polynomial result.

Simplify each term.

Multiply ${(x + 3)}^{2}$ by $1$ .

${(x + 3)}^{2} \cdot {(- 5)}^{0} + 2 \cdot {(x + 3)}^{1} \cdot {(- 5)}^{1} + 1 \cdot {(x + 3)}^{0} \cdot {(- 5)}^{2}$

Rewrite ${(x + 3)}^{2}$ as $(x + 3) (x + 3)$ .

$(x + 3) (x + 3) \cdot {(- 5)}^{0} + 2 \cdot {(x + 3)}^{1} \cdot {(- 5)}^{1} + 1 \cdot {(x + 3)}^{0} \cdot {(- 5)}^{2}$

Expand $(x + 3) (x + 3)$ using the FOIL Method.

Apply the distributive property.

$(x (x + 3) + 3 (x + 3)) \cdot {(- 5)}^{0} + 2 \cdot {(x + 3)}^{1} \cdot {(- 5)}^{1} + 1 \cdot {(x + 3)}^{0} \cdot {(- 5)}^{2}$

Apply the distributive property.

$(x \cdot x + x \cdot 3 + 3 (x + 3)) \cdot {(- 5)}^{0} + 2 \cdot {(x + 3)}^{1} \cdot {(- 5)}^{1} + 1 \cdot {(x + 3)}^{0} \cdot {(- 5)}^{2}$

Apply the distributive property.

$(x \cdot x + x \cdot 3 + 3 x + 3 \cdot 3) \cdot {(- 5)}^{0} + 2 \cdot {(x + 3)}^{1} \cdot {(- 5)}^{1} + 1 \cdot {(x + 3)}^{0} \cdot {(- 5)}^{2}$

Simplify and combine like terms.

Simplify each term.

Multiply $x$ by $x$ .

$(x^{2} + x \cdot 3 + 3 x + 3 \cdot 3) \cdot {(- 5)}^{0} + 2 \cdot {(x + 3)}^{1} \cdot {(- 5)}^{1} + 1 \cdot {(x + 3)}^{0} \cdot {(- 5)}^{2}$

Move $3$ to the left of $x$ .

$(x^{2} + 3 \cdot x + 3 x + 3 \cdot 3) \cdot {(- 5)}^{0} + 2 \cdot {(x + 3)}^{1} \cdot {(- 5)}^{1} + 1 \cdot {(x + 3)}^{0} \cdot {(- 5)}^{2}$

Multiply $3$ by $3$ .

$(x^{2} + 3 x + 3 x + 9) \cdot {(- 5)}^{0} + 2 \cdot {(x + 3)}^{1} \cdot {(- 5)}^{1} + 1 \cdot {(x + 3)}^{0} \cdot {(- 5)}^{2}$

Add $3 x$ and $3 x$ .

$(x^{2} + 6 x + 9) \cdot {(- 5)}^{0} + 2 \cdot {(x + 3)}^{1} \cdot {(- 5)}^{1} + 1 \cdot {(x + 3)}^{0} \cdot {(- 5)}^{2}$

Anything raised to $0$ is $1$ .

$(x^{2} + 6 x + 9) \cdot 1 + 2 \cdot {(x + 3)}^{1} \cdot {(- 5)}^{1} + 1 \cdot {(x + 3)}^{0} \cdot {(- 5)}^{2}$

Multiply $x^{2} + 6 x + 9$ by $1$ .

$x^{2} + 6 x + 9 + 2 \cdot {(x + 3)}^{1} \cdot {(- 5)}^{1} + 1 \cdot {(x + 3)}^{0} \cdot {(- 5)}^{2}$

Simplify.

$x^{2} + 6 x + 9 + 2 \cdot (x + 3) \cdot {(- 5)}^{1} + 1 \cdot {(x + 3)}^{0} \cdot {(- 5)}^{2}$

Apply the distributive property.

$x^{2} + 6 x + 9 + (2 x + 2 \cdot 3) \cdot {(- 5)}^{1} + 1 \cdot {(x + 3)}^{0} \cdot {(- 5)}^{2}$

Multiply $2$ by $3$ .

$x^{2} + 6 x + 9 + (2 x + 6) \cdot {(- 5)}^{1} + 1 \cdot {(x + 3)}^{0} \cdot {(- 5)}^{2}$

Evaluate the exponent.

$x^{2} + 6 x + 9 + (2 x + 6) \cdot - 5 + 1 \cdot {(x + 3)}^{0} \cdot {(- 5)}^{2}$

Apply the distributive property.

$x^{2} + 6 x + 9 + 2 x \cdot - 5 + 6 \cdot - 5 + 1 \cdot {(x + 3)}^{0} \cdot {(- 5)}^{2}$

Multiply $- 5$ by $2$ .

$x^{2} + 6 x + 9 - 10 x + 6 \cdot - 5 + 1 \cdot {(x + 3)}^{0} \cdot {(- 5)}^{2}$

Multiply $6$ by $- 5$ .

$x^{2} + 6 x + 9 - 10 x - 30 + 1 \cdot {(x + 3)}^{0} \cdot {(- 5)}^{2}$

Multiply ${(x + 3)}^{0}$ by $1$ .

$x^{2} + 6 x + 9 - 10 x - 30 + {(x + 3)}^{0} \cdot {(- 5)}^{2}$

Anything raised to $0$ is $1$ .

$x^{2} + 6 x + 9 - 10 x - 30 + 1 \cdot {(- 5)}^{2}$

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

$x^{2} + 6 x + 9 - 10 x - 30 + {(- 5)}^{2}$

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

$x^{2} + 6 x + 9 - 10 x - 30 + 25$

Simplify by adding terms.

Subtract $10 x$ from $6 x$ .

$x^{2} - 4 x + 9 - 30 + 25$

Simplify by adding and subtracting.

Subtract $30$ from $9$ .

$x^{2} - 4 x - 21 + 25$

Add $- 21$ and $25$ .

$x^{2} - 4 x + 4$

Expand Using the Binomial Theorem ((x+3)-5)^2

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