How do you find the cubic polynomial $f(x)$ with integer coefficients, where $1-3i$ and $-2$ are the roots of $f(x)$?
It has been too long since I have worked with $i$ and I have been unable to generate it.
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1Hint: The third root is $1+3i$. – André Nicolas Jun 15 '15 at 21:24
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Hint conjugate roots, e.g. see this answer and this one. – Bill Dubuque Jun 15 '15 at 21:49
4 Answers
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The thing to realize is that roots occur in conjugate pairs. So, if $1-3i$ is one root, then $1+3i$ is another root. Hence, the polynomial can be formed by multiplying out
$$(x-(1-3i))(x-(1+3i))(x-(-2)).$$
Ken
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For real coefficients, we must have either real roots, or roots that are complex conjugate pairs.
Inasmuch as $1-i3$ is a root, then $1+i3$ must also be.
The cubic polynomial has the form
$$a(z+2)(z-1+i3)(z-1-i3)=a(z+2)(z^2-2z+10)=a(z^3+6z+20)$$
Mark Viola
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The conjugate root theorem tells us that for a polynomial f(x) with real coefficients, if a+bi is a root, then a-bi is also a root.
Therefore 1-3i must also be a root. Thus, you may multiply the roots together in the form (x-r1)(x-r2)(x-r3) to obtain the required polynomial.
Cataline
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The other solutions correctly point out that the third root must be $1+3i$, and therefore the polynomial must be $$(x-(1-3i))(x-(1+3i))(x-2)$$ or a multiple of it (notice that multiplying through the entire thing by a constant does not change the zeroes).
I will add to this one tip for simplifying the expression: Notice that you can regroup and multiply out the factors that involve $i$ as follows: $$(x-(1-3i))(x-(1+3i)) = ((x-1) + 3i)((x-1) - 3i) = (x-1)^2 + 9$$ which might make the problem slightly easier to complete.
