I am a student learning competition math, and in a past test I found, you need to find the sum of the reciprocals of the roots of the polynomial $x^4-7x^3+4x^2+7x-4 = 0$. I have watched some videos on how to solve higher order polynomials, and I found that I like grouping. However, when I try to solve this with grouping, I end up with $x^4+4(x^2-1)-7x(x^2-1)$ which can be simplified to $x^4 + (x^2-1)(4-7x)$. I am not sure what to do now, and am having trouble figuring it out from here. Any help would be appreciated!
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I think you lost an $x$ in your last formula -- it should end with $(x^2 - 1) (4 - 7x)$. (You can click "edit" below your question to fix this.) – John Hughes Jan 29 '19 at 01:00
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Also: do you mean the sum of the reciprocals of the roots of the polynomial? – John Hughes Jan 29 '19 at 01:01
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@John Hughes Yes, I just mis-typed on those. – SpencerLS Jan 29 '19 at 01:02
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If $r$ is a root of the polynomial $p(x) = x^4-7x^3+4x^2+7x-4 $, then $1/r$ is a root of $p(1/x) = x^{-4}-7x^{-3}+4x^{-2}+7x^{-1}-4$, and vice-versa, because $x = 0$ is not a root of either one.
That means that $1/r$ is also a root of $x^4 p(1/x)$, which is a polynomial, namely, $$ x^4 p(1/x) = 1-7x+4x^2+7x^3-4x^4 = -4(-\frac{1}{4}+\frac{7}{4}x-x^2-\frac{7}{4}x^3+x^4). $$
Now if you recall, when you have a monic polynomial (coefficient of highest power is $1$), the coefficient of the NEXT highest power is the negative of the sum of the roots. So $\frac{7}{4}$ is the sum of the roots of the polynomial in parens, which has the same roots as $x^4 p(1/x)$, i.e., the number you're looking for.
Notice that I never had to actually find any of the roots to know what their sum was (or the sum of their reciprocals).
John Hughes
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Thanks! I would still like to learn how to factor polynomials like this. I like your solution though! I never knew that about monic polynomials! Isn't the first polynomial a monic polynomial though? If it is wouldn't the sum of the roots be -7? – SpencerLS Jan 29 '19 at 01:21
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The sum of the roots of the first polynomial is indeed $-7$...but you need the sum of the reciprocals of the roots, which is why I had to hoke up the second polynomial. BTW, the proof of that monic thing is easy: just multiply out $(x-a)(x-b) \ldots (x-k)$ where $a, b, \ldots k$ are the $n$ roots, obviously. The coefficient of $x^{n-1}$ ends up being $a + b + \ldots + k$. (You have to do a little dancing for repeated roots, but not much!) – John Hughes Jan 29 '19 at 02:32
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Just a quick question: Is $1/r$ always a root of $x^n p(1/x)$? (where n is the largest power of $p(x)$) – SpencerLS Jan 29 '19 at 02:57
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Yes...almost. If $r$ is a nonzero root of $p(x)$, then it's true. Proof: evaluate by plugging in $1/r$ for $x$! You get $(1/r)^n p(1/(1/r)) = r^{-n} p(r) = r^{-n} \cdot 0 = 0$. Of course, if $r = 0$, then the question doesn't even make sense. – John Hughes Jan 29 '19 at 03:15
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@JohnHughes: First of all, I think it should be $-4( \frac{-1} 4$ – J. W. Tanner Jan 30 '19 at 04:15
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@JohnHughes: Secondly, and more importantly for the solution to the problem posed, I think the reciprocals are roots of the monic polynomial in parentheses, so their sum is 7/4, not 7. – J. W. Tanner Jan 30 '19 at 04:21
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@JohnHughes: The sum of the roots is the opposite of the coefficient of the next highest power in the monic polynomial, no? – J. W. Tanner Jan 30 '19 at 04:32
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I figured it out. It is 7/4, and you don't have to do the monic polynomial thing. – SpencerLS Jan 30 '19 at 14:45
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Edited to reflect @J.W.Tanner's comments, and to fix up a couple of small glitches. Apologies for earlier sloppiness. – John Hughes Jan 30 '19 at 15:01