Part 1
Let f(x) and g(x) be polynomials.
Suppose f(x)=0 for exactly three values of x: namely, x=-3,4, and 8
Suppose g(x)=0 for exactly five values of x: namely, x=-5,-3,2,4, and 8
Is it necessarily true that g(x) is divisible bt f(x)? If so, carefully explain why. If not, give an example where g(x) is not divisible by f(x).
Part 2
Generalize: for arbritary polynomials f(x) and g(x), what do we need to know about the zeroes (including complex zeroes) of f(x) and g(x) to infer that g(x) is divisible by f(x)?
(If your answer to Part 1 was "yes", then stating the generalization should be straightforward. If your answer to Part 1 was "no", then try to salvage the idea by imposing extra conditions as needed. Either way, prove your generalization.)
+23 For the first part, if one or more of the roots of f(x) has a multiplicity of 2 or greater and g(x) has no roots greater than multiplicity 1. an example polynomial you could use for f(x) is:
f(x) = (x+3)(x+3)(x-4)(x-8)
and for g(x):
g(x) = (x+5)(x+3)(x-2)(x-4)(x-8)
I'll let you expand those if you need to.
For the second part, f(x) evenly divides g(x) if all the overlapping roots have the same multiplicity. I'm not too good at writing out proofs though, so you can try to prove my statement yourself or wait for someone else to.
Hope I was of some help!
i dont think i understand what the second part is asking. thank you for your answer for the first part though, it really helped!
+23 The second part is asking "what more specific information do we need about the roots in order to know for certain if f(x) divides g(x) evenly?" The reason why you need to know the multiplicity (the number of times the roots shows up in the factorization) is because a polinomial could be something like (x+10)3 which has -10 as a zero and another polynomial could be (x+10)(x+17). it also has a zero at 10, but (x+10)3 obviously does not go into the other polynomial. This is not a proof because it is not general, but i hope it can further put you on the right track. Sorry for not getting back sooner, hope this is still useful!
