# How do you solve 3x^2+5x+2=0 ?

Mar 13, 2016

$\left(3 x + 2\right) \left(x + 1\right)$

#### Explanation:

3 and 2 are both prime. Thus the only factors of them is 1 and 3 for 3; 1 and 2 for 2.

Assuming there are integer solutions for $x$ then it has to be true that one of the following is true. If not, then it is down to the formula.

$\left(3 x \pm 2\right) \left(x \pm 1\right)$ as the first option
$\left(3 x \pm 1\right) \left(x \pm 2\right)$ as the second option

In the given question:
The constant 2 is positive so the signs in the brackets are the same.
The $5 x$ is positive so both signs are +

$\textcolor{b l u e}{\text{Test 1}}$

$\left(3 x + 2\right) \left(x + 1\right)$

$= \text{ } 3 {x}^{2} + 3 x + 2 x + 2$

$\textcolor{b l u e}{= \text{ } 3 {x}^{2} + 5 x + 2}$