Expand Using the Binomial Theorem (7u-3t)^2

${(7 u - 3 t)}^{2}$

Step 1

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

$\sum_{k = 0}^{2} \frac{2!}{(2 - k)! k!} \cdot {(7 u)}^{2 - k} \cdot {(- 3 t)}^{k}$

Step 2

Expand the summation.

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

Step 3

Simplify the exponents for each term of the expansion.

$1 \cdot {(7 u)}^{2} \cdot {(- 3 t)}^{0} + 2 \cdot {(7 u)}^{1} \cdot {(- 3 t)}^{1} + 1 \cdot {(7 u)}^{0} \cdot {(- 3 t)}^{2}$

Step 4

Simplify each term.

Multiply ${(7 u)}^{2}$ by $1$ .

${(7 u)}^{2} \cdot {(- 3 t)}^{0} + 2 \cdot {(7 u)}^{1} \cdot {(- 3 t)}^{1} + 1 \cdot {(7 u)}^{0} \cdot {(- 3 t)}^{2}$

Apply the product rule to $7 u$ .

$7^{2} u^{2} \cdot {(- 3 t)}^{0} + 2 \cdot {(7 u)}^{1} \cdot {(- 3 t)}^{1} + 1 \cdot {(7 u)}^{0} \cdot {(- 3 t)}^{2}$

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

$49 u^{2} \cdot {(- 3 t)}^{0} + 2 \cdot {(7 u)}^{1} \cdot {(- 3 t)}^{1} + 1 \cdot {(7 u)}^{0} \cdot {(- 3 t)}^{2}$

Apply the product rule to $- 3 t$ .

$49 u^{2} \cdot ({(- 3)}^{0} t^{0}) + 2 \cdot {(7 u)}^{1} \cdot {(- 3 t)}^{1} + 1 \cdot {(7 u)}^{0} \cdot {(- 3 t)}^{2}$

Rewrite using the commutative property of multiplication.

$49 \cdot {(- 3)}^{0} u^{2} t^{0} + 2 \cdot {(7 u)}^{1} \cdot {(- 3 t)}^{1} + 1 \cdot {(7 u)}^{0} \cdot {(- 3 t)}^{2}$

Anything raised to $0$ is $1$ .

$49 \cdot 1 u^{2} t^{0} + 2 \cdot {(7 u)}^{1} \cdot {(- 3 t)}^{1} + 1 \cdot {(7 u)}^{0} \cdot {(- 3 t)}^{2}$

Multiply $49$ by $1$ .

$49 u^{2} t^{0} + 2 \cdot {(7 u)}^{1} \cdot {(- 3 t)}^{1} + 1 \cdot {(7 u)}^{0} \cdot {(- 3 t)}^{2}$

Anything raised to $0$ is $1$ .

$49 u^{2} \cdot 1 + 2 \cdot {(7 u)}^{1} \cdot {(- 3 t)}^{1} + 1 \cdot {(7 u)}^{0} \cdot {(- 3 t)}^{2}$

Multiply $49$ by $1$ .

$49 u^{2} + 2 \cdot {(7 u)}^{1} \cdot {(- 3 t)}^{1} + 1 \cdot {(7 u)}^{0} \cdot {(- 3 t)}^{2}$

Simplify.

$49 u^{2} + 2 \cdot (7 u) \cdot {(- 3 t)}^{1} + 1 \cdot {(7 u)}^{0} \cdot {(- 3 t)}^{2}$

Multiply $7$ by $2$ .

$49 u^{2} + 14 u \cdot {(- 3 t)}^{1} + 1 \cdot {(7 u)}^{0} \cdot {(- 3 t)}^{2}$

Simplify.

$49 u^{2} + 14 u \cdot (- 3 t) + 1 \cdot {(7 u)}^{0} \cdot {(- 3 t)}^{2}$

Rewrite using the commutative property of multiplication.

$49 u^{2} + 14 \cdot - 3 u t + 1 \cdot {(7 u)}^{0} \cdot {(- 3 t)}^{2}$

Multiply $14$ by $- 3$ .

$49 u^{2} - 42 u t + 1 \cdot {(7 u)}^{0} \cdot {(- 3 t)}^{2}$

Multiply ${(7 u)}^{0}$ by $1$ .

$49 u^{2} - 42 u t + {(7 u)}^{0} \cdot {(- 3 t)}^{2}$

Apply the product rule to $7 u$ .

$49 u^{2} - 42 u t + 7^{0} u^{0} \cdot {(- 3 t)}^{2}$

Anything raised to $0$ is $1$ .

$49 u^{2} - 42 u t + 1 u^{0} \cdot {(- 3 t)}^{2}$

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

$49 u^{2} - 42 u t + u^{0} \cdot {(- 3 t)}^{2}$

Anything raised to $0$ is $1$ .

$49 u^{2} - 42 u t + 1 \cdot {(- 3 t)}^{2}$

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

$49 u^{2} - 42 u t + {(- 3 t)}^{2}$

Apply the product rule to $- 3 t$ .

$49 u^{2} - 42 u t + {(- 3)}^{2} t^{2}$

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

$49 u^{2} - 42 u t + 9 t^{2}$

Expand Using the Binomial Theorem (7u-3t)^2

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