1
$\begingroup$

This is not a duplicate of theory of equations finding roots from given polynomial.

Given that the roots (both real and complex) of a polynomial are $\frac{2}{3}$, $-1$, $3+\sqrt2i$, and $3+\sqrt2i$, find the polynomial. All coefficients of the polynomial are real integer values.

What I have so far: $$(3x-2)(x+1)(x-\sqrt2\times i)=0$$ If I were solving other similar problems with two complex roots, I would probably be able to cancel them out, but I'm confused about how to do the $(x-\sqrt2i)$ part. Is this actually two complex roots that meet? Also, what degree is this polynomial? I would like an algebraic explanation please.

$\endgroup$
  • 1
    $\begingroup$ Are you sure you're not being asked for a real polynomial? $\endgroup$ – Git Gud Jun 28 '14 at 1:07
  • $\begingroup$ I'll edit the question to be more specific. $\endgroup$ – Jason Chen Jun 28 '14 at 1:07
  • 3
    $\begingroup$ If the coefficients of the polynomial are real, then $3-\sqrt{2}{i}$ is also a solution. $\endgroup$ – André Nicolas Jun 28 '14 at 1:08
  • 2
    $\begingroup$ If the coefficients are real, there is no polynomial whose complete list of roots is as given. $\endgroup$ – André Nicolas Jun 28 '14 at 1:10
  • 1
    $\begingroup$ For a polynomial with real coefficients, the complex solutions come in conjugate pairs. $\endgroup$ – user124862 Jun 28 '14 at 1:27
2
$\begingroup$

The conjugate factor theorem states that for a polynomial $p(x)$ with real coefficients, the complex roots come in conjugate pairs, or if $a+bi$ is a root, then $a-bi$ is also a root. From this we see that your polynomial has the roots $$\frac{2}{3}, -1, 3+\sqrt{2}i, 3-\sqrt{2}i.$$ Therefore, $$(x-\frac{2}{3})(x+1)(x-3-\sqrt{2}i)(x-3+\sqrt{2}i)=0.$$ $$\therefore (x^2+\frac{x}{3}-\frac{2}{3})(x-3-\sqrt{2}i)(x-3+\sqrt{2}i)=0$$ $$\therefore (x^2+\frac{x}{3}-\frac{2}{3})(x^2-6x+11)=0.$$ Continuing the expansion results in the polynomial $$p(x)=\frac{1}{3}(3x^4-17x^3+25x^2+23x-22).$$

$\endgroup$
  • $\begingroup$ We call it the conjugate factor theorem in Australia, everybody else calls it the complex conjugate root theorem I believe. $\endgroup$ – user124862 Jun 28 '14 at 2:17
  • 1
    $\begingroup$ Check it out, here. $\endgroup$ – user124862 Jun 28 '14 at 2:22
  • $\begingroup$ Also. Whack the polynomial into WolframAlpha and it'll give you roots, showing the polynomial to be correct. $\endgroup$ – user124862 Jun 28 '14 at 2:30
  • $\begingroup$ Would it be better to make the first term $(3x-2)$? Then you would be able to remove the coefficient in front of the final answer. $\endgroup$ – Jason Chen Jun 28 '14 at 4:53
  • $\begingroup$ Either way works. $\endgroup$ – user124862 Jun 28 '14 at 4:58

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.