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)?
Next time make sure you delete all the meaningless $ signs. So that your question presents properly.
\(\text{Let F(x) and G(x) be polynomials such that }\\ f(x)=F(x)(x+3)(x-4)(x-8)\\ g(x)=G(x)(x+3)(x-4)(x-8)(x+5)(x-2)\\ \frac{g(x)}{f(x)}=\frac{G(x)(x+3)(x-4)(x-8)(x+5)(x-2)}{F(x)(x+3)(x-4)(x-8)}\\ \frac{g(x)}{f(x)}=\frac{G(x)(x+5)(x-2)}{F(x)}\\ \)
g(x) is only divisable by f(x) if F(x) is a factor of G(x)(x+5)(x-2)
for instance in the following example g(x) is not divisable by f(x)
\(g(x)=7(x+5)(x-2)(x+3)(x-4)(x-8)\\ f(x)=(x^8+x^5-4x)(x+3)(x-4)(x-8)\\\)
Next time make sure you delete all the meaningless $ signs. So that your question presents properly.
\(\text{Let F(x) and G(x) be polynomials such that }\\ f(x)=F(x)(x+3)(x-4)(x-8)\\ g(x)=G(x)(x+3)(x-4)(x-8)(x+5)(x-2)\\ \frac{g(x)}{f(x)}=\frac{G(x)(x+3)(x-4)(x-8)(x+5)(x-2)}{F(x)(x+3)(x-4)(x-8)}\\ \frac{g(x)}{f(x)}=\frac{G(x)(x+5)(x-2)}{F(x)}\\ \)
g(x) is only divisable by f(x) if F(x) is a factor of G(x)(x+5)(x-2)
for instance in the following example g(x) is not divisable by f(x)
\(g(x)=7(x+5)(x-2)(x+3)(x-4)(x-8)\\ f(x)=(x^8+x^5-4x)(x+3)(x-4)(x-8)\\\)