Solving polynomial equations by decomposition












0












$begingroup$


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 $a+b=0implies a=0 quadtext{and}, b=0$ why?






polynomials systems-of-equations







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    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






  • 1




    $begingroup$
    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
















0












$begingroup$


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 $a+b=0implies a=0 quadtext{and}, b=0$ why?






polynomials systems-of-equations







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    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






  • 1




    $begingroup$
    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














0












0








0





$begingroup$


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 $a+b=0implies a=0 quadtext{and}, b=0$ why?






polynomials systems-of-equations







share|cite|improve this question











$endgroup$




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 $a+b=0implies a=0 quadtext{and}, b=0$ why?






polynomials systems-of-equations




polynomials systems-of-equations






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 21 at 23:19







IamNotaMathematician

















asked Jan 21 at 22:38









IamNotaMathematicianIamNotaMathematician

33




33








  • 2




    $begingroup$
    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






  • 1




    $begingroup$
    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














  • 2




    $begingroup$
    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






  • 1




    $begingroup$
    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








2




2




$begingroup$
It does not follow from $a+b=0$ that $a=0$ and $b=0$.
$endgroup$
– J. W. Tanner
Jan 21 at 22:40






$begingroup$
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




$begingroup$
@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




1




1




$begingroup$
It could be, for example, that $a=1$ and $b=-1$.
$endgroup$
– J. W. Tanner
Jan 21 at 22:52




$begingroup$
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






$begingroup$
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




$begingroup$
@PicaudVincent but we accept only the values that satisfy both equations
$endgroup$
– IamNotaMathematician
Jan 21 at 22:57










2 Answers
2






active

oldest

votes


















0












$begingroup$

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 $x=frac{7pmsqrt{37-4k}}{2}$, and then insert than into the quintic for solutions, but this results in:



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



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.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    I think that is what I was looking for thank you
    $endgroup$
    – IamNotaMathematician
    Jan 21 at 23:34



















0












$begingroup$

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, $P_n(a) = P_k(a) + P_m(a) = 0 +0.$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









0












$begingroup$

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 $x=frac{7pmsqrt{37-4k}}{2}$, and then insert than into the quintic for solutions, but this results in:



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



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.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    I think that is what I was looking for thank you
    $endgroup$
    – IamNotaMathematician
    Jan 21 at 23:34
















0












$begingroup$

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 $x=frac{7pmsqrt{37-4k}}{2}$, and then insert than into the quintic for solutions, but this results in:



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



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.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    I think that is what I was looking for thank you
    $endgroup$
    – IamNotaMathematician
    Jan 21 at 23:34














0












0








0





$begingroup$

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 $x=frac{7pmsqrt{37-4k}}{2}$, and then insert than into the quintic for solutions, but this results in:



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



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.






share|cite|improve this answer









$endgroup$



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 $x=frac{7pmsqrt{37-4k}}{2}$, and then insert than into the quintic for solutions, but this results in:



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



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.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 21 at 23:25









Rhys HughesRhys Hughes

6,9011530




6,9011530












  • $begingroup$
    I think that is what I was looking for thank you
    $endgroup$
    – IamNotaMathematician
    Jan 21 at 23:34


















  • $begingroup$
    I think that is what I was looking for thank you
    $endgroup$
    – IamNotaMathematician
    Jan 21 at 23:34
















$begingroup$
I think that is what I was looking for thank you
$endgroup$
– IamNotaMathematician
Jan 21 at 23:34




$begingroup$
I think that is what I was looking for thank you
$endgroup$
– IamNotaMathematician
Jan 21 at 23:34











0












$begingroup$

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, $P_n(a) = P_k(a) + P_m(a) = 0 +0.$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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


















0












$begingroup$

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, $P_n(a) = P_k(a) + P_m(a) = 0 +0.$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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
















0












0








0





$begingroup$

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, $P_n(a) = P_k(a) + P_m(a) = 0 +0.$






share|cite|improve this answer











$endgroup$



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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edited Jan 21 at 22:58

























answered Jan 21 at 22:48









Jack PfaffingerJack Pfaffinger

162112




162112












  • $begingroup$
    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




















  • $begingroup$
    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


















$begingroup$
Yes I know but why it doesn't work how to prove that
$endgroup$
– IamNotaMathematician
Jan 21 at 22:49




$begingroup$
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




$begingroup$
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






$begingroup$
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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