# What is the standard form of y= (x+5)(x-2)^2 ?

Jul 12, 2017

See a solution process below:

#### Explanation:

First, expand the term being squared on the right hand of the equation using this rule:

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

Substituting $x$ for $a$ and $2$ for $b$ gives:

$y = \left(x + 5\right) {\left(x - 2\right)}^{2}$

$y = \left(x + 5\right) \left({x}^{2} - \left(2 \cdot x \cdot 2\right) + {2}^{2}\right)$

$y = \left(x + 5\right) \left({x}^{2} - 4 x + 4\right)$

Next, we can multiply the two remaining terms by multiplying each term in the parenthesis on the left by each term in the parenthesis on the left:

$y = \left(\textcolor{red}{x} + \textcolor{red}{5}\right) \left(\textcolor{b l u e}{{x}^{2}} - \textcolor{b l u e}{4 x} + \textcolor{b l u e}{4}\right)$

Becomes:

$\left(\textcolor{red}{x} \times \textcolor{b l u e}{{x}^{2}}\right) - \left(\textcolor{red}{x} \times \textcolor{b l u e}{4 x}\right) + \left(\textcolor{red}{x} \times \textcolor{b l u e}{4}\right) + \left(\textcolor{red}{5} \times \textcolor{b l u e}{{x}^{2}}\right) - \left(\textcolor{red}{5} \times \textcolor{b l u e}{4 x}\right) + \left(\textcolor{red}{5} \times \textcolor{b l u e}{4}\right)$

$y = {x}^{3} - 4 {x}^{2} + 4 x + 5 {x}^{2} - 20 x + 20$

We can now group and combine like terms in descending order by the power of the exponent for the $x$ variables::

$y = {x}^{3} - 4 {x}^{2} + 5 {x}^{2} + 4 x - 20 x + 20$

$y = {x}^{3} + 1 {x}^{2} + \left(- 16\right) x + 20$

$y = {x}^{3} + {x}^{2} - 16 x + 20$