So far I've covered rational roots, multiple roots, polys over complex field.
However, for the FToA you don't need a proof - it's beyond the scope of the course (do a search for Gauss and fundamental theorem of algebra on Google if you're curious).
Rational roots proof:
Let P(x) be some polynomial in x with integral coefficients,
i.e. P(x)=a<sub>n</sub>x<sup>n</sup>+a<sub>n-1</sub>x<sup>n-1</sup>+...+a<sub>1</sub>x+a<sub>0</sub> (just a generic polynomial, but with integer coefficients)
Now, suppose x=r/s is a RATIONAL zero, where r and s are mutually prime, then r must be a factor of a<sub>0</sub> and s must be a factor of a<sub>n</sub>.
i.e. r is a factor of the coefficient of the last term, and s is a factor of the coefficient of the first term. Mutually prime means that r and s have no factors in common.
Proof: We are supposing x=r/s is a mutually zero of P(x). .'. P(r/s)=0 (remainder theorem)
i.e.:
P(r/s)=a<sub>n</sub>(r/s)<sup>n</sup>+a<sub>n-1</sub>(r/s)<sup>n-1</sup>+...+a<sub>1</sub>(r/s)+a<sub>0</sub>=0
Multiply both sides by s<sup>n</sup>:
P(r/s)=a<sub>n</sub>r<sup>n</sup>+a<sub>n-1</sub>r<sup>n-1</sup>.s+...+a<sub>1</sub>r.s<sup>n-1</sup>+a<sub>0</sub>.s<sup>n</sup>=0
take a<sub>0</sub>.s<sup>n</sup> from both sides:
a<sub>n</sub>r<sup>n</sup>+a<sub>n-1</sub>r<sup>n-1</sup>.s+...+a<sub>1</sub>r.s<sup>n-1</sup>=-a<sub>0</sub>.s<sup>n</sup>
Now notice that on the LHS at least r is a factor of EVERY term. This means that the RHS r is ALSO a factor of the LHS.
i.e. r is a factor of -a<sub>0</sub>.s<sup>n</sup>
So r is a factor of either s^n or a<sub>0</sub>. However, at the start we said that r and s are mutually prime, so r cannot be a factor of s^n, leaving only a<sub>0</sub>
.'. r is a factor of a<sub>0</sub>
The same method is used to prove s is a factor of a<sub>n</sub>.
The proof for multiple roots (or in this case factors) is thankfully a bit easier:
Theorem: if (x-b) is a factor of P(x) and a factor of multiplicity r times, then (x-b) is also a factor of P'(x) and a factor of multiplicity r-1 times.
Proof:
Let P(x)=(x-b)<sup>r</sup>.Q(x)
P'(x)={(x-b)<sup>r</sup>}'.Q(x) + Q'(x).(x-b)<sup>r</sup> [quotient rule]
P'(x)=r(x-b)<sup>r-1</sup>.Q(x) + Q'(x).(x-b)<sup>r</sup>
P'(x)=(x-b)<sup>r-1</sup>{r.Q(x)+Q'(x).(x-b)}
Which obviously means (x-b) is a factor of P'(x) of one degree less than it is in P(x)
i.e. (x-b) is a factor of multiplicity (r-1) of P'(x). Q.E.D.
Factor and root are VERY loosely interchangable. Factor is in the form (x-b), root is in the form x=b.
Obviously you can also keep going this way for 2nd derivatives, all the way to nth derivatives... though you'll hit the trivial case of multiplicity (n-n) sooner or later.
