Given $f(x)=x^4+2x^3+2x^2-2x-3$, where $x-1$ is a factor of $f(x)$, how is it possible to solve $f(x)$ without the Rational Root Theorem?
Here's my progress: $$f(x)=x^4+2x^3+2x^2-2x-3$$ $$f(x)=(x-1)(x^3+3x^2-5x+3)$$
And there's where I got stuck. I cheated, though, and tried solving this equation with the Rational Root Theorem. That's what I got: $$f(x)=x^4+2x^3+2x^2-2x-3$$ $$f(x)=(x-1)(x^3+3x^2-5x+3)$$ $$f(x)=(x-1)(x+1)(x^2+2x+3)$$ $$f(x)=(x-1)(x+1)(x-(-1-\sqrt2))(x-(-1+\sqrt2))$$
Please, if you're answering this question, I'd really appreciate if you could explain your procedures and what technique you used.
Thank you very much.
$$\begin{aligned} x^4+2x^3+2x^2-2x-3 &=(x^4+2x^2-3)+(2x^3-2x) \\ &=(x^2+3)(x^2-1)+2x(x^2-1) \\ &=(x^2-1)(x^2+3+2x) \\ &=(x^2-1)(x^2+2x+3) \end{aligned}$$