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)?

FlyEaglesFly Apr 24, 2019

#1**+2 **

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)

\(g(x)=7(x+5)(x-2)(x+3)(x-4)(x-8)\\ f(x)=(x^8+x^5-4x)(x+3)(x-4)(x-8)\\\)

Melody Apr 25, 2019

#1**+2 **

Best Answer

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)

\(g(x)=7(x+5)(x-2)(x+3)(x-4)(x-8)\\ f(x)=(x^8+x^5-4x)(x+3)(x-4)(x-8)\\\)

Melody Apr 25, 2019