In my textbook is the following text:
Vieta's Formulas can be used to relate the sum and product of the roots of a polynomial to its coefficients. The simplest application of this is with quadratics.
If we have a quadratic $x^2+ax+b=0$ with solutions $p$ and $q$, then we know that we can factor it as $x^2+ax+b=(x-p)(x-q)$ (Note that the first term is $x^2$, not $ax^2$.)
Using the distributive property to expand the right side we now have $x^2+ax+b=x^2-(p+q)x+pq$
My question is:
Why $x^2+ax+b=(x-p)(x-q)$, and not $x^2+ax+b=(x+p)(x+q)$?
Because $p$ and $q$ are roots and not $-p$ and $-q$.
If $x^2+ax+b=(x+p)(x+q)$ then since $p$ is a root of $x^2+px+q$, we need $p^2+ap+b=2p(p+q)$ and $p^2+ap+b=0$, which gives $2p(p+q)=0$,
which not always true.
By the way, $x^2+ax+b=(x-p)(x-q)$ for $x=p$ and $x=q$ is true, which says that it's true for all reals $x$.