# How do you multiply  2z(3z+2)(z-1)-(z+7)(z-9)?

Jun 23, 2018

Use the distributive property and FOIL and combine like terms to get our answer, $6 {z}^{3} - 3 {z}^{2} - 2 z + 63$.

#### Explanation:

Let's split this into two parts. Focusing on the first part, $2 z \left(3 z + 2\right) \left(z - 1\right)$, we can use the distributive property to simplify it:

$2 z \left(3 z + 2\right) \left(z - 1\right)$ = $\left(6 {z}^{2} + 4 z\right) \left(z - 1\right)$

Next, let's use FOIL to get to its simplest form:

$\left(6 {z}^{2}\right) \left(z\right) + \left(6 {z}^{2}\right) \left(- 1\right) + \left(4 z\right) \left(z\right) + \left(4 z\right) \left(- 1\right)$
$6 {z}^{3} - 6 {z}^{2} + 4 {z}^{2} - 4 z$
$6 {z}^{3} - 2 {z}^{2} - 4 z$

Let's also use FOIL to simplify the other half of our expression:

$\left(z + 7\right) \left(z - 9\right)$
${z}^{2} - 9 z + 7 z - 63$
${z}^{2} - 2 z - 63$

Finally, let's combine the two terms and simplify:

$6 {z}^{3} - 2 {z}^{2} - 4 z - \left({z}^{2} - 2 z - 63\right)$
$6 {z}^{3} - 2 {z}^{2} - 4 z - {z}^{2} + 2 z + 63$
$6 {z}^{3} - 2 {z}^{2} - {z}^{2} - 4 z + 2 z + 63$
$6 {z}^{3} - 3 {z}^{2} - 2 z + 63$