# How do you multiply (7n^2+8n+7)(7n^2+n-5)?

Feb 28, 2017

$49 {n}^{4} + 63 {n}^{3} + 22 {n}^{2} = 33 n - 35$

#### Explanation:

Distribution-
$\left(7 {n}^{2} + 8 n + 7\right) \left(7 {n}^{2} + n - 5\right)$

Multiply $7 {n}^{2}$, $8 n$, and $7$ EACH by $7 {n}^{2}$, $n$, and $- 5$.

For $7 {n}^{2}$ :
$7 {n}^{2} \cdot 7 {n}^{2} = 49 {n}^{4}$
$7 {n}^{2} \cdot n = 7 {n}^{3}$
$7 {n}^{2} \cdot - 5 = - 35 {n}^{2}$

do the same for $8 n$ and $7$
You should get:
$\left(49 {n}^{4} + 7 {n}^{3} - 35 {n}^{2}\right) + \left(56 {n}^{3} + 8 {n}^{2} - 40 n\right) + \left(49 {n}^{2} + 7 n - 35\right)$

Now combine like terms-

$49 {n}^{4} + 63 {n}^{3} + 22 {n}^{2} - 33 n - 35$