Find all the zeroes of the polynomial 2𝑥4 10𝑥3 + 5𝑥2 + 15𝑥 12, if it is given that two of its zeroes are √3 /2 and √3/2

As per my interpretation the given quadratic polynomial is $2x^4-10x^3+5x^2+15x-12$ and the given two roots are $\sqrt{\frac{3}{2}} \mathrm{and} -\sqrt{\frac{3}{2}}$.
Since $\sqrt{\frac{3}{2}} \mathrm{and} -\sqrt{\frac{3}{2}}$ are the zeroes of the given polynomial, so the factors are $\left(x-\sqrt{\frac{3}{2}}\right) \mathrm{and} \left(x+\sqrt{\frac{3}{2}}\right)$
So we have;
$$\begin{gathered} \left(x-\sqrt{\frac{3}{2}}\right)\left(x+\sqrt{\frac{3}{2}}\right) = 0 \\ \Rightarrow x^2-{\left(\sqrt{\frac{3}{2}}\right)}^2 = 0 \\ \Rightarrow x^2-\frac{3}{2} = 0 \\ \Rightarrow 2x^2-3 = 0 \end{gathered}$$
Now dividing the given polynomial by $2x^2-3$; we have;
$$\begin{gathered} x^2-5x+4 \\ 2x^2-3)\overline{2x^4-10x^3+5x^2+15x-12} \\ 2x^4 -3x^2 \\ \overline{ -10x^3+8x^2+15x} \\ -10x^3 +15x \\ \overline{ 8x^2 -12} \\ 8x^2 -12 \\ \overline{ 0 } \end{gathered}$$
Now factorising the quotient we get;
$$\begin{gathered} x^2-5x+4 = 0 \\ \Rightarrow x^2-4x-x+4 = 0 \\ \Rightarrow x\left(x-4\right)-1\left(x-4\right) = 0 \\ \Rightarrow \left(x-1\right)\left(x-4\right) = 0 \\ \Rightarrow x = 1 \mathrm{and} x = 4 \end{gathered}$$
Therefore other two zeroes of the given polynomial are 1 and 4.