If you don't have time or don't feel like readings lot (or what ever reason you may have) there is an abridged version at the bottom that is much easier to read.
In this note, I will prove that for all polynomials , with and x=0 is not a root, that after being divided by GCD of the coefficients () or multiplied by a factor of such that the new coefficients are both integral and co-prime, the function with the smallest possible integral roots has the form where p is a constant and if and only if has all group wise co-prime integer coefficients.
(In the short, I'm proving that the coefficients are reversed if the roots are inverted (AKA: becomes )
By Vieta's, we have for
Now, assume that we didn't know that had the reverse coefficients of
Thus
However, we know that the roots of are where .
For , we have
If we multiply both the top and bottom of every fraction on the R.H.S. by every root other than the one in its denominator, we will have (This is too ugly to show in "proper" math format here. So I'll show a simpler and proper version below)
It can be easily observed that where p is a constant and .
This process can be repeated to show that where (NOTE: this is the one time where rather than because there is no nor
We can see that the ratio's of the is equal to the ratio's of the in reverse order. Going back to our other equation . Assuming that has all co-prime coefficients, could be reducible by a factor of h. However, this leads to two contradictions: 1. every other ratio would have to be reduced by h or would be changed. 2. If every other coefficient was reduced then some would become non-integers if they were all group wise co-prime (if they weren't then this would contradict a condition earlier).
Also since we're looking for an with the smallest possible co-prime integer coefficients, should be irreducible. These statements imply that and .
Once again, we can use this logic to prove that where .
Thus we have proved that for integral coefficients of ,
If doesn't have integer coefficients, then we can multiply by a factor of such that the coefficients become integral, thus our final result is .
Polynomial has roots p,q,r.
Polynomial has roots
-\dfrac{g}{h}=\dfrac{1}{p}+\dfrac{1}{q}+\dfrac{1}{r}=\dfrac{(pq)}{(pq)r}+\dfrac{(rp)}{(rp)q}+\dfrac{(pq)}{(pq)r}=\dfrac{pq+qr+rp}{pqr}=\Rightarrow -\dfrac{\left(\dfrac{c}{a}\right)}{\left(\dfrac{d}{a}\right)}=\boxed{\color \pink{-\dfrac{c}{d}}}
\dfrac{j}{h}=\dfrac{1}{pq}+\dfrac{1}{qr}+\dfrac{1}{rp}=\dfrac{(r)}{(r)pq}+\dfrac{p}{(p)qr}+\dfrac{(q)}{(q)rp}=\dfrac{p+q+r}{pqr}=\dfrac{\left(\dfrac{b}{a}\right)}{\left(\dfrac{d}{a}\right)}=\boxed{\color \green{\dfrac{b}{d}}}
-\dfrac{k}{h}=\dfrac{1}{pqr}=\dfrac{1}{\left(-\dfrac{d}{a}\right)}=\boxed{\color \red{-\dfrac{a}{d}}}
Assuming a monic polynomial (we can divide by h because dividing by a constant won't change the roots)
has equivalent roots to x^3+\dfrac{g}{h}x^2+\dfrac{j}{h}x+\dfrac{k}{h}=x^3+\color \pink{\dfrac{c}{d}}x^2+\color \green{\dfrac{b}{d}}x+\color \red{\dfrac{a}{d}}
Multiplying through to rationalize the denominators by d.
Easy Math Editor
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Top NewestThanks bruh
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Chee Bra.
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Note: You should state that x=0 is not a solution, or that equivalently a0=0. This would also justify allowing you to divide by a0.
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Ahh, thanks, thats a good point.
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Why not tranform the equation by the substitution x=y1 Its a 1 line proof after that.
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That transformation is certainly a much quicker approach. Can you add more details about what you did?
For a problem along the same lines, check out Squaring the roots of a polynomial.
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I'm not sure exactly what you mean, but by your substitution, since we're solving for roots, isn't x=∞
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Why x=∞?
Aren't we just inverting the roots?
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\x=y1. And since we're finding roots y=0
Because you substituteLog in to reply
I just used y to take a different variable. I did not mean the y in y=f(x)
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