Simplify (3n-1)(2n^2+4n+4)

$(3 n - 1) (2 n^{2} + 4 n + 4)$

Step 1

Expand $(3 n - 1) (2 n^{2} + 4 n + 4)$ by multiplying each term in the first expression by each term in the second expression.

$3 n (2 n^{2}) + 3 n (4 n) + 3 n \cdot 4 - 1 (2 n^{2}) - 1 (4 n) - 1 \cdot 4$

Step 2

Simplify terms.

Simplify each term.

Rewrite using the commutative property of multiplication.

$3 \cdot 2 n \cdot n^{2} + 3 n (4 n) + 3 n \cdot 4 - 1 (2 n^{2}) - 1 (4 n) - 1 \cdot 4$

Multiply $n$ by $n^{2}$ by adding the exponents.

Move $n^{2}$ .

$3 \cdot 2 (n^{2} n) + 3 n (4 n) + 3 n \cdot 4 - 1 (2 n^{2}) - 1 (4 n) - 1 \cdot 4$

Multiply $n^{2}$ by $n$ .

Raise $n$ to the power of $1$ .

$3 \cdot 2 (n^{2} n^{1}) + 3 n (4 n) + 3 n \cdot 4 - 1 (2 n^{2}) - 1 (4 n) - 1 \cdot 4$

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$3 \cdot 2 n^{2 + 1} + 3 n (4 n) + 3 n \cdot 4 - 1 (2 n^{2}) - 1 (4 n) - 1 \cdot 4$

Add $2$ and $1$ .

$3 \cdot 2 n^{3} + 3 n (4 n) + 3 n \cdot 4 - 1 (2 n^{2}) - 1 (4 n) - 1 \cdot 4$

Multiply $3$ by $2$ .

$6 n^{3} + 3 n (4 n) + 3 n \cdot 4 - 1 (2 n^{2}) - 1 (4 n) - 1 \cdot 4$

Rewrite using the commutative property of multiplication.

$6 n^{3} + 3 \cdot 4 n \cdot n + 3 n \cdot 4 - 1 (2 n^{2}) - 1 (4 n) - 1 \cdot 4$

Multiply $n$ by $n$ by adding the exponents.

Move $n$ .

$6 n^{3} + 3 \cdot 4 (n \cdot n) + 3 n \cdot 4 - 1 (2 n^{2}) - 1 (4 n) - 1 \cdot 4$

Multiply $n$ by $n$ .

$6 n^{3} + 3 \cdot 4 n^{2} + 3 n \cdot 4 - 1 (2 n^{2}) - 1 (4 n) - 1 \cdot 4$

Multiply $3$ by $4$ .

$6 n^{3} + 12 n^{2} + 3 n \cdot 4 - 1 (2 n^{2}) - 1 (4 n) - 1 \cdot 4$

Multiply $4$ by $3$ .

$6 n^{3} + 12 n^{2} + 12 n - 1 (2 n^{2}) - 1 (4 n) - 1 \cdot 4$

Multiply $2$ by $- 1$ .

$6 n^{3} + 12 n^{2} + 12 n - 2 n^{2} - 1 (4 n) - 1 \cdot 4$

Multiply $4$ by $- 1$ .

$6 n^{3} + 12 n^{2} + 12 n - 2 n^{2} - 4 n - 1 \cdot 4$

Multiply $- 1$ by $4$ .

$6 n^{3} + 12 n^{2} + 12 n - 2 n^{2} - 4 n - 4$

Simplify by adding terms.

Subtract $2 n^{2}$ from $12 n^{2}$ .

$6 n^{3} + 10 n^{2} + 12 n - 4 n - 4$

Subtract $4 n$ from $12 n$ .

$6 n^{3} + 10 n^{2} + 8 n - 4$