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This question is part of a calculus problem, involving the limit of a rational function. I modified it a little bit to make it more interesting. The question is:

$x^4+x^3-x^2-5x+4=0$, find all the solutions (real and complex) to this polynomial

This is my attempt, please do let me know if this is correct, if it can be improved and please share your approach to!

Since this is a quartic polynomial, factoring it is going to be challenging (if not impossible in some cases). However, I'm going to attempt to factor it into quadratics:

$$x^4+x^3-x^2-5x+4=0$$

$$x^2(x^2-1)+x^3-1-5x+5=0$$

$$x^2(x-1)(x+1)+(x-1)(x^2+x+1)-5(x-1)=0$$

$$(x-1)[(x^3+x^2+x^2+x-4)]=0$$

$$(x-1)(x^3+2x^2+x-4)=0$$

$$(x-1)(x^3-1+2x^2+x-3)=0$$

$$(x-1)[(x-1)(x^2+x+1)+x(2x+3)-1(2x+3)]=0$$

$$(x-1)^2(x^2+3x+4)=0$$

Therefore, $x=1$ and $x=\frac{-3\pm\iota\sqrt{7}}{2}$ are the solutions to this polynomial.

冥王 Hades
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    The problem composer may have intended that your first step would be to attempt to use the rational root theorem. If you had done this, you would have reduced the quartic to $(x-1)$ times a cubic. Then, repeating, you would have reduced the cubic to $(x-1)$ times a quadratic. So, you would have had $(x-1)^2$ times a quadratic, and you could then have immediately attacked the quadratic. It is probably not a coincidence that the quartic had two rational roots. – user2661923 Jan 23 '23 at 18:55
  • Did you try expanding your final expression to see if it matches the given one? – Karl Jan 23 '23 at 18:55
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    I’m voting to close this question because there is no question here. – coudy Jan 23 '23 at 18:55
  • @Karl There were some problems with the original post, I've edited the mistakes now – 冥王 Hades Jan 23 '23 at 19:00
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    @coudy isn't he asking to find the roots of the polynomial? – Sine of the Time Jan 23 '23 at 19:06
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    Yes, more specifically I'm asking if there are any different ways to solve this and if my own attempt was correct @SineoftheTime – 冥王 Hades Jan 23 '23 at 19:07
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    @coudy Read the OP, and my comment, more carefully please. Lastly, this isn't the place to announce whether you wish to close a question or not. If you wish to close it, cast your vote and leave. – 冥王 Hades Jan 23 '23 at 19:08
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    I don't think the question should be closed – Sine of the Time Jan 23 '23 at 19:08
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    Close votes $\neq$ down votes. That aside, my question meets neither criteria for close votes. If you still wish to close it, that is fine cast your vote and leave. Please do not clog the comments section with any that isn't directly related to the problem. Thank you @coudy – 冥王 Hades Jan 23 '23 at 19:15

2 Answers2

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If you look carefully, you'll notice that the sum of the coefficients of the polynomial is equal to $0$. Then, it is obvious that $x = 1$ is a root of the polynomial. Now, let $p(x)$ be our polynomial. By doing $\frac{p(x)}{x-1}$ you'll get $x^3+2x^2+x-4$. The sum of the coefficients is $0$ again. Now divide it by $x-1$, obtaining $x^2 + 3x + 4$, which is clearly irreducible over $\mathbb R$. You can compute the solutions of this last polynomial using the well-known formula for quadratic equations. Hope this helps you somehow.

juanan
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    Excellent approach! I didn't think to check for the sums of the polynomial. +1 – 冥王 Hades Jan 23 '23 at 19:11
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    Sometimes it is better to look a little bit closer to the polynomial because you might be able to reduce the difficulty of the problem ;) – juanan Jan 23 '23 at 19:12
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You have asked for an alternative solutions. Either you can use the rational root theorem, or you can do an observation about the sum of coefficients, or just guess that $1$ is a root, and then do a polynomial division. The sum of coefficients appears to be "ad hoc", or a little bit like a trick. The rational root theorem is a general theorem, but of course it requires to check several possibilities.

Another way is to consider the quartic modulo $2$, then it is $$ x^4+x^3-x^2-5x+4=x^4+x^3+x^2+x= x(x-1)^3, $$ which gives $x=1$ as a candidate also over $\Bbb Q$. Of course the other candidate $x=0$ does not work.

Dietrich Burde
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