# How do you find the product -2d(d^3c^2-4dc^2+2d^2c)+c^2(dc^2-3d^4)?

Jul 24, 2017

See a solution process below:

#### Explanation:

First, expand both terms in parenthesis by multiplying each term within the parenthesis by the term outside the parenthesis:

$\textcolor{red}{- 2 d} \left({d}^{3} {c}^{2} - 4 {\mathrm{dc}}^{2} + 2 {d}^{2} c\right) + \textcolor{b l u e}{{c}^{2}} \left({\mathrm{dc}}^{2} - 3 {d}^{4}\right) \implies$

$\left(\textcolor{red}{- 2 d} \cdot {d}^{3} {c}^{2}\right) - \left(\textcolor{red}{- 2 d} \cdot 4 {\mathrm{dc}}^{2}\right) + \left(\textcolor{red}{- 2 d} \cdot 2 {d}^{2} c\right) + \left(\textcolor{b l u e}{{c}^{2}} \cdot {\mathrm{dc}}^{2}\right) - \left(\textcolor{b l u e}{{c}^{2}} \cdot 3 {d}^{4}\right) \implies$

$- 2 {d}^{4} {c}^{2} - \left(- 8 {d}^{2} {c}^{2}\right) + \left(- 4 {d}^{3} c\right) + {\mathrm{dc}}^{4} - 3 {d}^{4} {c}^{2} \implies$

$- 2 {d}^{4} {c}^{2} + 8 {d}^{2} {c}^{2} - 4 {d}^{3} c + {\mathrm{dc}}^{4} - 3 {d}^{4} {c}^{2}$

Now, group and combine like terms:

$- 2 {d}^{4} {c}^{2} - 3 {d}^{4} {c}^{2} + 8 {d}^{2} {c}^{2} - 4 {d}^{3} c + {\mathrm{dc}}^{4} \implies$

$\left(- 2 - 3\right) {d}^{4} {c}^{2} + 8 {d}^{2} {c}^{2} - 4 {d}^{3} c + {\mathrm{dc}}^{4} \implies$

$- 5 {d}^{4} {c}^{2} + 8 {d}^{2} {c}^{2} - 4 {d}^{3} c + {\mathrm{dc}}^{4} \implies$

$- 5 {d}^{4} {c}^{2} - 4 {d}^{3} c + 8 {d}^{2} {c}^{2} + {\mathrm{dc}}^{4}$