Expand Using the Binomial Theorem (3+4d)^2

${(3 + 4 d)}^{2}$

Step 1

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

$\sum_{k = 0}^{2} \frac{2!}{(2 - k)! k!} \cdot {(3)}^{2 - k} \cdot {(4 d)}^{k}$

Step 2

Expand the summation.

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

Step 3

Simplify the exponents for each term of the expansion.

$1 \cdot {(3)}^{2} \cdot {(4 d)}^{0} + 2 \cdot {(3)}^{1} \cdot {(4 d)}^{1} + 1 \cdot {(3)}^{0} \cdot {(4 d)}^{2}$

Step 4

Simplify each term.

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

${(3)}^{2} \cdot {(4 d)}^{0} + 2 \cdot {(3)}^{1} \cdot {(4 d)}^{1} + 1 \cdot {(3)}^{0} \cdot {(4 d)}^{2}$

Raise $3$ to the power of $2$ .

$9 \cdot {(4 d)}^{0} + 2 \cdot {(3)}^{1} \cdot {(4 d)}^{1} + 1 \cdot {(3)}^{0} \cdot {(4 d)}^{2}$

Apply the product rule to $4 d$ .

$9 \cdot (4^{0} d^{0}) + 2 \cdot {(3)}^{1} \cdot {(4 d)}^{1} + 1 \cdot {(3)}^{0} \cdot {(4 d)}^{2}$

Anything raised to $0$ is $1$ .

$9 \cdot (1 d^{0}) + 2 \cdot {(3)}^{1} \cdot {(4 d)}^{1} + 1 \cdot {(3)}^{0} \cdot {(4 d)}^{2}$

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

$9 \cdot d^{0} + 2 \cdot {(3)}^{1} \cdot {(4 d)}^{1} + 1 \cdot {(3)}^{0} \cdot {(4 d)}^{2}$

Anything raised to $0$ is $1$ .

$9 \cdot 1 + 2 \cdot {(3)}^{1} \cdot {(4 d)}^{1} + 1 \cdot {(3)}^{0} \cdot {(4 d)}^{2}$

Multiply $9$ by $1$ .

$9 + 2 \cdot {(3)}^{1} \cdot {(4 d)}^{1} + 1 \cdot {(3)}^{0} \cdot {(4 d)}^{2}$

Evaluate the exponent.

$9 + 2 \cdot 3 \cdot {(4 d)}^{1} + 1 \cdot {(3)}^{0} \cdot {(4 d)}^{2}$

Multiply $2$ by $3$ .

$9 + 6 \cdot {(4 d)}^{1} + 1 \cdot {(3)}^{0} \cdot {(4 d)}^{2}$

Simplify.

$9 + 6 \cdot (4 d) + 1 \cdot {(3)}^{0} \cdot {(4 d)}^{2}$

Multiply $4$ by $6$ .

$9 + 24 d + 1 \cdot {(3)}^{0} \cdot {(4 d)}^{2}$

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

$9 + 24 d + {(3)}^{0} \cdot {(4 d)}^{2}$

Anything raised to $0$ is $1$ .

$9 + 24 d + 1 \cdot {(4 d)}^{2}$

Multiply ${(4 d)}^{2}$ by $1$ .

$9 + 24 d + {(4 d)}^{2}$

Apply the product rule to $4 d$ .

$9 + 24 d + 4^{2} d^{2}$

Raise $4$ to the power of $2$ .

$9 + 24 d + 16 d^{2}$

Expand Using the Binomial Theorem (3+4d)^2

Related Questions

Support LCMGCF.com by disabling your adblocker.