By Rational Root Theorem, all rational roots of a polynomial are in the form $\frac{p}{q}$, where $p$ divides the constant term $3$ and $q$ divides the leading coefficient $4$. One such root is $-\frac{3}{2}$. Factor the polynomial by dividing it by $2x+3$.
$$\left(2x+3\right)\left(2x^{2}-3x+1\right)$$
Consider $2x^{2}-3x+1$. Factor the expression by grouping. First, the expression needs to be rewritten as $2x^{2}+ax+bx+1$. To find $a$ and $b$, set up a system to be solved.
$$a+b=-3$$ $$ab=2\times 1=2$$
Since $ab$ is positive, $a$ and $b$ have the same sign. Since $a+b$ is negative, $a$ and $b$ are both negative. The only such pair is the system solution.
$$a=-2$$ $$b=-1$$
Rewrite $2x^{2}-3x+1$ as $\left(2x^{2}-2x\right)+\left(-x+1\right)$.
$$\left(2x^{2}-2x\right)+\left(-x+1\right)$$
Factor out $2x$ in the first and $-1$ in the second group.
$$2x\left(x-1\right)-\left(x-1\right)$$
Factor out common term $x-1$ by using distributive property.
