# How do you subtract (w ^ { 2} + 4w + 4) - ( w + 2)?

May 31, 2018 May 31, 2018

$\omega = - 2$ and $- 1$

#### Explanation:

As per the question, we have

$\left({\omega}^{2} + 4 \omega + 4\right) - \left(\omega + 2\right)$

And I assume that this is a solvable quadratic equation, $f \left(\omega\right) = 0$

$\therefore \left({\omega}^{2} + 4 \omega + 4\right) - \left(\omega + 2\right) = 0$

Opening the brackets, we get

${\omega}^{2} + 4 \omega + 4 - \omega - 2 = 0$

Now as per our Pre-Algebra knowledge

${\omega}^{2} + 3 \omega + 2 = 0$

$\therefore \left(\omega + 1\right) \left(\omega + 2\right) = 0$

$\therefore \omega = - 1 , - 2$

May 31, 2018

${w}^{2} + 3 w + 2$

#### Explanation:

The key here is that we subtract terms with the same degree from each other.

Since there is no other ${w}^{2}$ term, we will just be subtracting

$\textcolor{b l u e}{4 w - w}$

and

$\textcolor{c y a n}{4 - 2}$

Putting it all together, we get

${w}^{2} + \textcolor{b l u e}{4 w - w} + \textcolor{c y a n}{4 - 2}$

which simplifies to

${w}^{2} + 3 w + 2$

Hope this helps!