Find the Product (4-b)(5b^2+5b-4)

$(4 - b) (5 b^{2} + 5 b - 4)$

Expand $(4 - b) (5 b^{2} + 5 b - 4)$ by multiplying each term in the first expression by each term in the second expression.

$4 (5 b^{2}) + 4 (5 b) + 4 \cdot - 4 - b (5 b^{2}) - b (5 b) - b \cdot - 4$

Simplify terms.

Simplify each term.

Multiply $5$ by $4$ .

$20 b^{2} + 4 (5 b) + 4 \cdot - 4 - b (5 b^{2}) - b (5 b) - b \cdot - 4$

Multiply $5$ by $4$ .

$20 b^{2} + 20 b + 4 \cdot - 4 - b (5 b^{2}) - b (5 b) - b \cdot - 4$

Multiply $4$ by $- 4$ .

$20 b^{2} + 20 b - 16 - b (5 b^{2}) - b (5 b) - b \cdot - 4$

Rewrite using the commutative property of multiplication.

$20 b^{2} + 20 b - 16 - 1 \cdot 5 b \cdot b^{2} - b (5 b) - b \cdot - 4$

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

Move $b^{2}$ .

$20 b^{2} + 20 b - 16 - 1 \cdot 5 (b^{2} b) - b (5 b) - b \cdot - 4$

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

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

$20 b^{2} + 20 b - 16 - 1 \cdot 5 (b^{2} b^{1}) - b (5 b) - b \cdot - 4$

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

$20 b^{2} + 20 b - 16 - 1 \cdot 5 b^{2 + 1} - b (5 b) - b \cdot - 4$

Add $2$ and $1$ .

$20 b^{2} + 20 b - 16 - 1 \cdot 5 b^{3} - b (5 b) - b \cdot - 4$

Multiply $- 1$ by $5$ .

$20 b^{2} + 20 b - 16 - 5 b^{3} - b (5 b) - b \cdot - 4$

Rewrite using the commutative property of multiplication.

$20 b^{2} + 20 b - 16 - 5 b^{3} - 1 \cdot 5 b \cdot b - b \cdot - 4$

Multiply $b$ by $b$ by adding the exponents.

Move $b$ .

$20 b^{2} + 20 b - 16 - 5 b^{3} - 1 \cdot 5 (b \cdot b) - b \cdot - 4$

Multiply $b$ by $b$ .

$20 b^{2} + 20 b - 16 - 5 b^{3} - 1 \cdot 5 b^{2} - b \cdot - 4$

Multiply $- 1$ by $5$ .

$20 b^{2} + 20 b - 16 - 5 b^{3} - 5 b^{2} - b \cdot - 4$

Multiply $- 4$ by $- 1$ .

$20 b^{2} + 20 b - 16 - 5 b^{3} - 5 b^{2} + 4 b$

Simplify by adding terms.

Subtract $5 b^{2}$ from $20 b^{2}$ .

$15 b^{2} + 20 b - 16 - 5 b^{3} + 4 b$

Add $20 b$ and $4 b$ .

$15 b^{2} + 24 b - 16 - 5 b^{3}$