In general if the roots of a quadratic are given as $r$ and $s$, then the $(x-r)$ and $(x-s)$ are the factors of the quadratic equation, and not $(x+r)$ and $(x+s)$. This is because if we replace $x$ with $r$, then the expression needs to equal zero, which will not be the case if the quadratic is written as $(x+r)(x+s)$. I hope this makes sense.

Josh says

I believe your concepts are correct. Just one thing to note is that the original quadratic equation in the question is x^2 + Bx + C = 0.

And then your second step is (x – R)( x – S). However, I think it should have been (x + R)(x +S) in order for it to mimic the first question exactly.

Let me know if my logic is flawed? It might be that R and S could be negative but I guess we wouldn’t know that.

GMAT Quantum says

Hi Josh,

In general if the roots of a quadratic are given as $r$ and $s$, then the $(x-r)$ and $(x-s)$ are the factors of the quadratic equation, and not $(x+r)$ and $(x+s)$. This is because if we replace $x$ with $r$, then the expression needs to equal zero, which will not be the case if the quadratic is written as $(x+r)(x+s)$. I hope this makes sense.

Dabral