Solve (b+5)(b-4)(b+2)=

$$( b + 5 ) ( b - 4 ) ( b + 2 ) =$$

Evaluate

$$\left(b-4\right)\left(b+2\right)\left(b+5\right)$$

Solution Steps

Apply the distributive property by multiplying each term of b+5 by each term of b-4.

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

Combine -4b and 5b to get b.

$$\left(b^{2}+b-20\right)\left(b+2\right)$$

Apply the distributive property by multiplying each term of b^{2}+b-20 by each term of b+2.

$$b^{3}+2b^{2}+b^{2}+2b-20b-40$$

Combine 2b^{2} and b^{2} to get 3b^{2}.

$$b^{3}+3b^{2}+2b-20b-40$$

Combine 2b and -20b to get -18b.

$$b^{3}+3b^{2}-18b-40$$

Expand

$$b^{3}+3b^{2}-18b-40$$

Solution Steps

Apply the distributive property by multiplying each term of b+5 by each term of b-4.

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

Combine -4b and 5b to get b.

$$\left(b^{2}+b-20\right)\left(b+2\right)$$

Apply the distributive property by multiplying each term of b^{2}+b-20 by each term of b+2.

$$b^{3}+2b^{2}+b^{2}+2b-20b-40$$

Combine 2b^{2} and b^{2} to get 3b^{2}.

$$b^{3}+3b^{2}+2b-20b-40$$

Combine 2b and -20b to get -18b.

$$b^{3}+3b^{2}-18b-40$$