$F[x]/(p(x))$ contains the roots of $p(x)$

Question

The following theorem and exercise are from "Abstract Algebra, An Introduction, 3rd Edition, Thomas W. Hungerford"

Corollary 4.19

Let $F$ be a field and let $f(x) \in F[x]$ be a polynomial of degree 2 or 3. Then $f(x)$ is irreducible if and only if f(x) has no roots in $F$.

Exercise 5.3.8

Let $F$ be a field. If $p(x)$ is an irreducible quadratic polynomial in $F[x]$, show that $F[x]/(p(x))$ contains all the roots of $p(x)$

This is highly confusing to me. If $p(x)$ is irreducible and quadratic, then it has no roots. So there is nothing to prove. On the other hand, if we see $p$ as $p: F[x]/(p(x)) \to F[x]/(p(x))$, then obviously all the roots of $p(x)$ are in $F[x]/(p(x))$. What am I missing here?

Answer 1

Although $p(x)$ does not have roots in $F$, it may have roots in a larger field $F'$ containing $F$.

What they're asking you to show is that $F'=\dfrac{F[t]}{\langle p(t)\rangle}$ is a field with elements $k_1,k_2,k_3$ such that $p(x)=k_3(x-k_1)(x-k_2)$.

Answer 2

In $A =F[x]/I$ with $I=\langle p(x)\rangle$ take the residue class $\overline x=x+I$ of $x$.

Then $p(\overline x)=p(x+I) =p(x)+I = I =0+I =\overline 0$ as claimed.