# How do you factor 2a^2+3a+1?

$2 {a}^{2} + 3 a + 1 = 2 \left(a + \frac{1}{2}\right) \left(a + 1\right)$
We evaluate the discriminant: $\Delta = {B}^{2} - 4 A C = 9 - 8 = 1$.
The roots are given by $a = \frac{- B \pm \sqrt{\Delta}}{2 A}$.
Here we have ${a}_{1} = \frac{- 3 + 1}{4} = - \frac{1}{2}$ and ${a}_{2} = \frac{- 3 - 1}{4} = - 1$.
The factored form is $A \left(a - {a}_{1}\right) \left(a - {a}_{2}\right) = 2 \left(a + \frac{1}{2}\right) \left(a + 1\right)$.