Simplify (n^2+6n-4)(2n-4)

$(n^{2} + 6 n - 4) (2 n - 4)$

Step 1

Expand $(n^{2} + 6 n - 4) (2 n - 4)$ by multiplying each term in the first expression by each term in the second expression.

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

Step 2

Simplify terms.

Simplify each term.

Rewrite using the commutative property of multiplication.

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

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

Move $n$ .

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

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

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

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

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

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

Add $1$ and $2$ .

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

Move $- 4$ to the left of $n^{2}$ .

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

Rewrite using the commutative property of multiplication.

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

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

Move $n$ .

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

Multiply $n$ by $n$ .

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

Multiply $6$ by $2$ .

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

Multiply $- 4$ by $6$ .

$2 n^{3} - 4 n^{2} + 12 n^{2} - 24 n - 4 (2 n) - 4 \cdot - 4$

Multiply $2$ by $- 4$ .

$2 n^{3} - 4 n^{2} + 12 n^{2} - 24 n - 8 n - 4 \cdot - 4$

Multiply $- 4$ by $- 4$ .

$2 n^{3} - 4 n^{2} + 12 n^{2} - 24 n - 8 n + 16$

Simplify by adding terms.

Add $- 4 n^{2}$ and $12 n^{2}$ .

$2 n^{3} + 8 n^{2} - 24 n - 8 n + 16$

Subtract $8 n$ from $- 24 n$ .

$2 n^{3} + 8 n^{2} - 32 n + 16$