Answer:
The answer is no because
.
Step-by-step explanation:
If
is a factor of
, then
.
Let's check.
![(1)^{100}+4(1)^{99}+3](https://tex.z-dn.net/?f=%281%29%5E%7B100%7D%2B4%281%29%5E%7B99%7D%2B3)
![1+4(1)+3](https://tex.z-dn.net/?f=1%2B4%281%29%2B3)
![1+4+3](https://tex.z-dn.net/?f=1%2B4%2B3)
![5+3](https://tex.z-dn.net/?f=5%2B3)
![8](https://tex.z-dn.net/?f=8)
Since
, then
is not a factor of
.
======================================================
So anyways more on how this works:
If
where
is the remainder and
is the quotient of the division, then multiplying both sides by
gives
.
We see that
is a factor if the remainder is 0. (Just like we know 2 is a factor of 6 because when we divide 6 by 2 we get a remainder of 0.)
Anyways if we evaluate
for
we get:
![P(c)=Q(x)(c-c)+R](https://tex.z-dn.net/?f=P%28c%29%3DQ%28x%29%28c-c%29%2BR)
![P(c)=Q(x)(0)+R](https://tex.z-dn.net/?f=P%28c%29%3DQ%28x%29%280%29%2BR)
![P(c)=R](https://tex.z-dn.net/?f=P%28c%29%3DR)
So evaluating
for
will tell us the remainder of the quotient
without actually performing division.
The remainder will tell us if
is a factor of ![P](https://tex.z-dn.net/?f=P)
If the remainder is 0, then the answer is yes.
If the remainder is not 0, then the answer is no.
(The question to be answer "Is this a factor of this?")