Solve $3x^3=({x^2+\sqrt{18}x+\sqrt{32}})(x^2-\sqrt{18}x-\sqrt{32})-4x^2$

Question

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:

$$x^4-3x^3-22x^2-48x-32=0$$

Answer 1

Expand $$x^4-3 x^3-22 x^2-48 x-32=0$$ Try to factorize $$x^4-3 x^3-22 x^2-48 x-32=\left(a x+b+x^2\right) \left(c x+d+x^2\right)$$ RHS gives $$x^4+x^3 (a+c)+x^2 (a c+b+d)+x (a d+b c)+b d$$ thus we must have

$$ \begin{cases} a+c=-3\\ ac+b+d=-22\\ ad+bc=-48\\ bd=-32\\ \end{cases} $$

Final factorization should be $$\left(x^2-6 x-8\right) \left(x^2+3 x+4\right)=0$$

Answer 2

(Fill in the gaps as needed. If you're stuck, show what you've tried.)

1. Expanding everything, the equation is
   $$ x^4 - 3x^3 - 22x^2 - 48x - 32 = 0. $$

2. Factorize this.

3. Apply the quadratic formula to find the roots.

Answer 3

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

$$3x^3=x^4-(\sqrt{18}x+\sqrt{32})^2 -4x^2$$ $$x^4-3x^3-4x^2-(\sqrt{18}x+\sqrt{32})^2 =0$$ $$x^4-(3x+4)x^2-2(3x+4)^2 =0$$ $$(x^2+(3x+4))(x^2-2(3x+4))=0$$