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I have a very little background on mathematics. I have a very basic question about solving polynomial equations. if we have $$P_n(x) = 0$$ where $P_n(x)$ is a polynomial of degree $n$.For example: $$x^5-3x^3+x^2-7x+3 = 0$$

Why we cannot just decompose the polynomial equation to a system of equations solving them separately and the solutions to the original polynomial will be the intersection of the roots of each equation: $$\begin{cases} x^5-3x^3 = 0 \\ \text{and} \\x^2-7x+3=0 \end{cases}$$ There is a lot of possibilities (infinity?) to choose this sub-equations. I know this seems very stupid but could you tell me which rule of logic I broke.

As far as I know, if $a=0$ and $b=0$ then we can add them to form $a+b=0$ we cannot go the other way, that is $a+b=0\implies a=0 \quad\text{and}\, b=0$ why?

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    $\begingroup$ It does not follow from $a+b=0$ that $a=0$ and $b=0$. $\endgroup$ – J. W. Tanner Jan 21 at 22:40
  • $\begingroup$ @J.W.Tanner $a=0$ and $b=0$ then we can write $a+b=0$ why we cannot go the other way? $\endgroup$ – IamNotaMathematician Jan 21 at 22:45
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    $\begingroup$ It could be, for example, that $a=1$ and $b=-1$. $\endgroup$ – J. W. Tanner Jan 21 at 22:52
  • $\begingroup$ You have 2 equations for 1 unknown, apart very special cases you will get an incompatible system (no solution). Moreover not all solutions of a+b=0 are of the form a=0 and b=0 $\endgroup$ – Picaud Vincent Jan 21 at 22:53
  • $\begingroup$ @PicaudVincent but we accept only the values that satisfy both equations $\endgroup$ – IamNotaMathematician Jan 21 at 22:57
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We have that: $$a +b =0 \to a=-b$$ Which means it holds for $a=b=0$, but also that you can't go the other way, as you said.

Your system of equations should be: $$\begin{cases} x^5-3x^3 = k \\ \text{and} \\x^2-7x+3=-k \end{cases}$$

We can solve $x^2-7x+(3+k)=0$ for $x=\frac{7\pm\sqrt{37-4k}}{2}$, and then insert than into the quintic for solutions, but this results in:

$$\frac{(7\pm\sqrt{37-4k})^5}{32}-\frac{3(7\pm\sqrt{37-4k})^3}{8}-k=0$$

One question springs to mind, is this really any simpler? To my mind, not at all.

We can substitute $t=7+\sqrt{37-4k}$, but this requires the solution of:

$$t^5-12t^3+8t^2-112t+96=0$$ which is more complicated than the polynomial we are trying to solve.

All in all: is this a valid method? Absolutely. Is it practical? Absolutely not in my opinion.

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  • $\begingroup$ I think that is what I was looking for thank you $\endgroup$ – IamNotaMathematician Jan 21 at 23:34
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Note that this is not exactly the same question, for example what if $x^5 - 3x^3 = 1$ and $x^2 - 7x + 3 = -1$? Then the original equation would be satisfied, but neither of the original equations would be satisfied. However, they also take on these values at different values of $x$. Additionally, if you continued to break down the equation into systems of equations, you could eventually get to the case of a $0th$ order polynomial which could never be equal to $0$. In your particular example this would be looking for solutions of $3 = 0$ which clearly doesn't make sense.

One counter-example is enough to show that there are solutions not contained in the intersection of the zeros. As for the proving that the intersection of the zeros is also a solution to the original polynomial do as follows:

Let $P_n(x) = P_k(x) + P_m(k).$ Suppose $P_k(a) = 0,$ and $P_m(a) = 0$. Then, $P_n(a) = P_k(a) + P_m(a) = 0 +0.$

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  • $\begingroup$ Yes I know but why it doesn't work how to prove that $\endgroup$ – IamNotaMathematician Jan 21 at 22:49
  • $\begingroup$ You are correct however, that if both equations are $0$ for the same value of $x$, then that would be a solution. Consider for example $x^5 - 3x^3+x^2-7x = 0$. You can break it into: $x^5 - 3x^3 = 0,$$ and x^2-7x = 0.$ Both of which have solutions at $x = 0$, and so the original equation also has a solution at $x = 0$. $\endgroup$ – Jack Pfaffinger Jan 21 at 22:52
  • $\begingroup$ but how could you prove that there is no such decomposition not necessarily 2 terms at a time. in other words if I say for a given polynomial equation there is an equivalent system of simple equations how can you disprove that $\endgroup$ – IamNotaMathematician Jan 21 at 23:03

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