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.

  • $\begingroup$ Is Descartes' Rule of Signs also off limits? $\endgroup$
    – hardmath
    Feb 4, 2016 at 1:19
  • $\begingroup$ Here, RRT says that the possible rational roots of the equation are $\pm1$ and $\pm3$. When tested, $-1$ and $1$ are found to be roots, while neither $3$ nor $-3$ is a root. Thus, besides $1$ and $-1$, all roots of the equation are either irrational or non-real. $\endgroup$
    – Arcturus
    Feb 4, 2016 at 1:20
  • $\begingroup$ Sometimes you can get lucky with carefully doing grouping to factor. $\endgroup$
    – randomgirl
    Feb 4, 2016 at 1:23
  • $\begingroup$ Often the roots will be "nice enough" to do by hand/inspection. Check the integers between $-3$ and $3$. This is often where some of the roots will be (at least in a quiz/exam situation where getting the roots is part of the purpose of the problem) $\endgroup$ Feb 4, 2016 at 1:39

1 Answer 1


$$\begin{aligned} x^4+2x^3+2x^2-2x-3 &=(x^4+2x^2-3)+(2x^3-2x) \\ &=(x^2+3)(x^2-1)+2x(x^2-1) \\ &=(x^2-1)(x^2+3+2x) \\ &=(x^2-1)(x^2+2x+3) \end{aligned}$$


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