<s>[INST] Using the remainder theorem to prove a quadratic is a factor of a polynomial For example, if I have $P(x) = 3x^4 + 5x^3 -17x^2 -13x + 6$ then to show that $x^2 + x - 6$ is a factor I individually show that $x+3$ and $x-2$ are factors using the factor theorem (i.e. $P(-3) = 0$ and $P(2) = 0$).
However, if $x+3$ and $x-2$ are individually factors, why does that conclude their product would be a factor?
Because I have learnt if two individual numbers are factors, they also need to be "coprime" for the product to be a factor.
(For example: $12$ is divisible by both $6$ and $12$ but not their product $72$)
How would you know that $x+3$ and $x-2$ are coprime? E.g. for $x=7$ they do not seem to be?
Thanks in advance!
 [/INST] $x+3$ and $x-2$ (or any two distinct monic polynomials of degree $1$) are coprime as polynomials.  That is, there is no polynomial of positive degree that divides both of them.  Indeed, $(x-2)-(x+3)=-5$.  If $P(x)$ is a polynomial that is divisible by both $x-2$ and $x+3$, i.e. there are polynomials $q(x)$ and $r(x)$ such that $P(x) = (x-2) q(x) = (x+3) r(x)$, we have
$$ (x+3) r(x) = P(x) = (x-2)(q(x)) = (x+3) q(x) -5 q(x)$$
so $$ q(x) = (x+3)(q(x)-r(x))/5$$
and $$P(x) = (x-2) q(x) = (x-2)(x+3)(q(x)-r(x))/5$$
is divisible by $(x-2)(x+3)$.
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