1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Is Descartes' Rule of Signs also off limits? $\endgroup$ – hardmath Feb 4 '16 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 '16 at 1:20
  • $\begingroup$ Sometimes you can get lucky with carefully doing grouping to factor. $\endgroup$ – randomgirl Feb 4 '16 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) $\endgroup$ – Cameron Williams Feb 4 '16 at 1:39
5
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.