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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.

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    $\begingroup$ 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
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    $\begingroup$ 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
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    $\begingroup$ 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
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    $\begingroup$ For a polynomial with real coefficients, the complex solutions come in conjugate pairs. $\endgroup$ – user124862 Jun 28 '14 at 1:27
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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 $$p(x)=\frac{1}{3}(3x^4-17x^3+25x^2+23x-22).$$

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  • $\begingroup$ 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
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    $\begingroup$ 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

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