Solve for $x$ in
$$3x^3=({x^2+\sqrt{18}x+\sqrt{32}})(x^2-\sqrt{18}x-\sqrt{32})-4x^2$$
I'm really not good with 4th degree equations but since the 1st term in the RHS looked a simple (a+b)(a-b) application, I tried solving that but I'm really not able to reach to the final answer as it only gets more complicated…
can anyone please help me out?
The equation I got was:
$$x^4-3x^3-22x^2-48x-32=0$$
Best Answer
Note that the given equation
$$3x^3=({x^2+\sqrt{18}x+\sqrt{32}})(x^2-\sqrt{18}x-\sqrt{32})-4x^2$$
is purposefully structured to allow convenient factorization as follows
$$3x^3=x^4-(\sqrt{18}x+\sqrt{32})^2 -4x^2$$ $$x^4-3x^3-4x^2-(\sqrt{18}x+\sqrt{32})^2 =0$$ $$x^4-(3x+4)x^2-2(3x+4)^2 =0$$ $$(x^2+(3x+4))(x^2-2(3x+4))=0$$