Use factor theorem to find all zeros of polynomial $2x^3+3x^2+x+6$ with one known factor $x+2$ I am to find the factors of  $2x^3+3x^2+x+6$ where I am told that$x+2$ one of the factors.
Using synthetic division to divide $2x^3+3x^2+x+6$ by $x+2$ I confirm there is no remainder so it is a zero and the new quotient is $2x^2-x+3$
So I have: $(x+2)(2x^2-x+3)$
I would now like to factor $(2x^2-x+3)$ but am having a hard time. Since my leading coefficient is not 1, I know that in order to factor by grouping I must find two numbers whose sum is -1 and whose product is 6 (leading coefficient 2 * constant term 3).
I cannot find any so do not know how to proceed with factoring $(2x^2-x+3)$.
I considered:
-1 & 6: product = -6, sum 5
1 & -6: product = -6, sum -5
2 & -3: product = -6, sum -1 # close
-2 & 3: product = -6, sum 1  # also close
-2 & -3: product = 6, sum 5
How can I factor $(2x^2-x+3)$?

$2x^2-x+3=2(x^2-\frac{1}{2}x+\frac{3}{2})$ where the discriminant is $\Delta=(\frac{-1}{2})^2-4\times 1\times \frac{3}{2}=\frac{1}{4}-6<0$. Therefore $x^2-\frac{1}{2}x+\frac{3}{2}$ has no real roots and so it is an irreducible polynomial over $\mathbb{R}$.
