Since function A has 2 zeros, A has the form:
[tex]A(x)=f(x)(x-2)(x-4)[/tex]
Then, if we divide it by (x-2), we get:
[tex]\frac{A(x)}{(x-2)}=f(x)(x-4)[/tex]
And the remainder is zero.
Additionally, we know that A has 2 unique zeros: then, it cannot have a repeated factor of (x-4).
If we plot function A, we get:
The graph seems to be a polynomial of grade 3, and the leading term is negative.
Notice that
[tex]\lim _{x\to\infty}A(x)=-\infty[/tex]
According to the graph we got. On the other hand:
[tex]\lim _{x\to-\infty}A(x)=\infty[/tex]
Then. Function A matches:
1. When I am divided by (x-2) the remainder is 0.
2. f(4)=0
3. One of my end behaviors is: As x->-infinite, f(x)->infinite
As for function B.
The graph indicates that there are 3 unique zeros: x=1,2,4. That it represents a polynomial of grade 4 and the leading term is positive since the function grows when x->infinite and x->-infinite.
Thus, function B matches:
1. When I am divided by (x-2) the remainder is 0.
2. I have a positive leading coefficient.
3. f(4)=0
4. I have 3 roots
5. One of my end behaviors is: As x->infinite, f(x)->infinite
6. (x-1) is one of my factors
7. I am a Quartic
8. One of my end behaviors is: As x->-infinite, f(x)->infinite
Finally, regarding function C.
[tex]C(x)=x^3+4x^2-3x-18=(x-2)(x+3)^2[/tex]
Then, C(x) is a polynomial of grade 3, and its limits are:
[tex]\lim _{x\to\infty}C(x)=\infty,\lim _{x\to-\infty}C(x)=-\infty[/tex]
Then, the answers for function C are:
1. When I am divided by (x-2) the remainder is 0.
2. I have a positive leading coefficient. (leading coefficient is 1)
3. I have 3 roots
4. One of my end behaviors is: As x->infinite, f(x)->infinite
