I do not even know how to start on this problem, I really need help. I do not get it at all. I do not understand what this problem is really asking for. Please explain your answer step by step and in a detailed manner. Thank you very much.

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 by 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 arbitrary 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.)

Guest Aug 3, 2019

#1**+2 **

I really need help. I do not get it at all. I do not understand what this problem is really asking for.

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 by 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 arbitrary 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.

**Hi Guest!**

**Part1:**

\(f(x)=(x+3)\cdot (x-4)\cdot (x-8)\)

This function f(x) has the function value 0 at x = -3, 4 and 8.

\(g(y)=(x+5)\cdot (x+3)\cdot (x-2)\cdot (x-4)\cdot (x-8)\)

This function g(x) has the function value 0 at x = -5, -3, 2, 4 and 8.

\(\frac{g(x)}{f(x)}=\frac{(x+5)\cdot (x+3)\cdot (x-2)\cdot (x-4)\cdot (x-8)}{(x+3)\cdot (x-4)\cdot (x-8)}=(x+5)\cdot (x-2)\)

g(x) is divisible by f(x) without remainder.

**Part2:**

**\(If\\ \color{blue}\{x|f(x)=0\}\subset \{x|g(x)=0\}\\ then\ (g)x\ is\ divisible\ by\ (f)x\ without\ remainder.\)**

!

asinus Aug 3, 2019

#3**+2 **

**Hello!** There is something on that.

I correct:

\(If\\ \color{blue}\{Number\ of\ all\ x|f(x)=0\}\subset \{ Number\ of\ all\ x|g(x)=0\}\\ then\ (g)x\ is\ divisible\ by\ (f)x\ without\ remainder. \\ A\ specific\ Zero\ can\ occur\ multiple\ times,\\ can\ be\ one,\ two\ or\ n\ times.\ n\in \mathbb N.\)

\(This\ claim\ is\ valid\ for\ parabolic\ functions\ only.\)

**Greetings**

!

asinus Aug 3, 2019