# How do you find the product (g-4h)(g-4h)?

May 3, 2017

See the solution process below:

#### Explanation:

This is a special form of a quadratic which follows this rule:

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

Substituting $g$ for $a$ and $4 h$ for $b$ gives:

$\left(g - 4 h\right) \left(g - 4 h\right) = {g}^{2} - 2 g 4 h + {\left(4 h\right)}^{2} = {g}^{2} - 8 g h + 16 {h}^{2}$

We can also multiply this out the long way to obtain the same solution. To multiply these two terms you multiply each individual term in the left parenthesis by each individual term in the right parenthesis.

$\left(\textcolor{red}{g} - \textcolor{red}{4 h}\right) \left(\textcolor{b l u e}{g} - \textcolor{b l u e}{4 h}\right)$ becomes:

$\left(\textcolor{red}{g} \times \textcolor{b l u e}{g}\right) - \left(\textcolor{red}{g} \times \textcolor{b l u e}{4 h}\right) - \left(\textcolor{red}{4 h} \times \textcolor{b l u e}{g}\right) + \left(\textcolor{red}{4 h} \times \textcolor{b l u e}{4 h}\right)$

${g}^{2} - 4 g h - 4 g h + 16 {h}^{2}$

We can now combine like terms:

${g}^{2} + \left(- 4 - 4\right) g h + 16 {h}^{2}$

${g}^{2} + \left(- 8\right) g h + 16 {h}^{2}$

${g}^{2} - 8 g h + 16 {h}^{2}$