# How do you multiply (p^4+3p^2-8)(p+1)?

Jun 19, 2017

See a solution process below:

#### Explanation:

To multiply these two terms you multiply each individual term in the left parenthesis by each individual term in the right parenthesis.

$\left(\textcolor{red}{{p}^{4}} + \textcolor{red}{3 {p}^{2}} - \textcolor{red}{8}\right) \left(\textcolor{b l u e}{p} + \textcolor{b l u e}{1}\right)$ becomes:

$\left(\textcolor{red}{{p}^{4}} \times \textcolor{b l u e}{p}\right) + \left(\textcolor{red}{{p}^{4}} \times \textcolor{b l u e}{1}\right) + \left(\textcolor{red}{3 {p}^{2}} \times \textcolor{b l u e}{p}\right) + \left(\textcolor{red}{3 {p}^{2}} \times \textcolor{b l u e}{1}\right) - \left(\textcolor{red}{8} \times \textcolor{b l u e}{p}\right) - \left(\textcolor{red}{8} \times \textcolor{b l u e}{1}\right)$

${p}^{5} + {p}^{4} + 3 {p}^{3} + 3 {p}^{2} - 8 p - 8$

Jun 20, 2017

color(green)(p^5+p^4+3p^3+3p^2-8p-8

#### Explanation:

$\left({p}^{4} + 3 {p}^{2} - 8\right) \left(p + 1\right)$

$\textcolor{w h i t e}{a a a a a a a a a a a a a}$${p}^{4} + 3 {p}^{2} - 8$
$\textcolor{w h i t e}{a a a a a a a a a a a}$$\times \underline{p + 1}$
$\textcolor{w h i t e}{a a a a a a a a a a a a a}$${p}^{5} + 3 {p}^{3} - 8 p$
$\textcolor{w h i t e}{a a a a a a a a a a a a a a a}$${p}^{4} + 3 {p}^{2} - 8$
$\textcolor{w h i t e}{a a a a a a a a a a a a a}$$\overline{{p}^{5} + {p}^{4} + 3 {p}^{3} + 3 {p}^{2} - 8 p - 8}$

$\textcolor{w h i t e}{a a a a a a a a a a a a a}$$\textcolor{g r e e n}{{p}^{5} + {p}^{4} + 3 {p}^{3} + 3 {p}^{2} - 8 p - 8}$