# How do you find the zeroes for y=x^2+3x+2?

Jul 2, 2015

You may notice that $2 = 1 \cdot 2$ and $3 = 1 + 2$
$y = {x}^{2} + 3 x + 2 = \left(x + 1\right) \cdot \left(x + 2\right)$
For $y = 0$ either one of the factors need to be $= 0$
This leads to $x = - 1 \mathmr{and} x = - 2$