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I am aware that for any polynomial with real coefficients, the imaginary roots (if there are any) must come in complex pairs. However, I always believed that you could still have an imaginary number - without a conjugate - as a root. However, I came upon the question:

Which of the following is a polynomial with roots $0$, $4$, and $i$.
a. $x^3 - 4x^2 + x - 4$
b. $x^3 - ( 4+i )x^2 + 4ix$
c. $x^4 - 4x^3 + x^2 - 4x$
d. $x^2 - 4x$
3. $x^4 - 4x^3 + x^2 + 4x$

Using the roots $0, 4, $and $i$, I found the factors to be $(x)(x-4)(x-i)$, which simplifies to choice b, $x^3 - ( 4+i )x^2 + 4ix$. However, the answer in the book states:

C Complex roots occur in conjugate pairs. If $i$ is a root of the polynomial, then $-i$ is also a root. Use the four roots to determine the factors of the polynomial. Then multiply to get the polynomial.
$(x)(x-4)(x-i)(x+i)$
...
$x^4 - 4x^3 + x^2 - 4x$

I'm not sure how they can assume that their are complex conjugate pairs, since there are no constraints that the coefficients be real.

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    $\begingroup$ b) doesn't have real coefficients so the law doesn't apply to it. All the others will come in conjugate pairs. $\endgroup$
    – Ian Miller
    Jan 15 '17 at 3:59
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    $\begingroup$ The question must be implicitly assuming real coefficients, in which case the answer given is correct. $\endgroup$
    – ziggurism
    Jan 15 '17 at 3:59
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    $\begingroup$ What book is it? They seem to have redefined "polynomial" to mean "real polynomial", for which their statement is true. But it's a very strange definition that will get you into a lot of trouble when you move on to ring theory later. $\endgroup$
    – Paul Z
    Jan 15 '17 at 5:05
  • $\begingroup$ @PaulZ McGraw Hill's SAT Subject Test Math Level 2 $\endgroup$
    – Eric Wiener
    Jan 15 '17 at 5:06
  • $\begingroup$ My earlier answer here is useful here too. $\endgroup$
    – JDługosz
    Jan 15 '17 at 9:55
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The book's answer is nonsense. Your calculation gives a polynomial with roots at $0,\,4,\,i$ as desired and is a very direct way to find the minimal polynomial that will have a root at all of those places. More simply, one can see this phenomenon since $x-i=0$ is a polynomial with a single root at $i$. Complex roots only necessarily come in conjugate pairs for polynomials with real coefficients. As you note, the book's answer would be correct if you were asked to find a real polynomial with those roots.

(Not to mention that this question is terribly ambiguous, since it doesn't seem clear whether the polynomial should have only those three roots, or those three should be among the roots, in which case both (b) and (c) would work)

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    $\begingroup$ Thanks. I just wanted to make sure I wasn't missing something. $\endgroup$
    – Eric Wiener
    Jan 15 '17 at 3:58
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    $\begingroup$ Although this answer is certainly correct mathematically, if @EricWiener is actually studying for the SAT he should keep their redefinition in mind and understand why they reject answer B. Arguing mathematics is not much use against standardized-test writers. $\endgroup$
    – Paul Z
    Jan 15 '17 at 15:57
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    $\begingroup$ @PaulZ In all the other practice tests I've taken (and the two other books I've finished) answer B would have been accepted. That was the reason for my confusion. $\endgroup$
    – Eric Wiener
    Jan 15 '17 at 15:59
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While Milo Brandt's answer is perfect, I want to give a quick demonstration that the non-real roots of a real polynomial must come in conjugate pairs. Given any complex $x$ I shall use "$x^*$" to denote the complex conjugate of $x$. Also, given any complex polynomial $f$ I shall use "$f^*$" to denote $f$ with each coefficient replaced by its conjugate.

Take any real polynomial $f$ and any complex root $r$ of $f$.

Then by long division we can find a complex polynomial $g$ and a complex constant $c$ such that for all complex $x$ we have $f(x) = g(x) \cdot (x-r) + c$. Then $0 = f(r) = g(r) \cdot 0 + c = c$.

Now simply take the conjugate on both sides of the identity and use the basic properties of complex conjugation to get $f^*(x^*) = g^*(x^*) \cdot (x^*-r^*)$ for any complex $x$.

But $f^* = f$ since $f$ has only real coefficients, and we can apply the above identity to the case where $x = r$.

Thus $f(r^*) = g^*(r^*) \cdot 0 = 0$.

Therefore $r^*$ is a root of $f$.

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  • $\begingroup$ Your book might be misleadingly using the assumption that polynomials encountered will be real polynomials (perhaps on the SAT they may)... $\endgroup$
    – user21820
    Jan 15 '17 at 13:59
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    $\begingroup$ And note that this same proof generalizes to any automorphism (in this case complex conjugation) on the larger field (in this case the complex numbers) that fixes a smaller field (in this case the reals), and shows that given any polynomial over the smaller field, its roots in the larger field remain roots after applying the automorphism to them. $\endgroup$
    – user21820
    Jan 15 '17 at 14:02

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