a)

When we expand a factorised quadratic expression we obtain

$$(x+a)(x+b)=x^2+(a+b)x+ab\text{.}$$

This means when factorising the quadratic expression

$$\simplify{x^2+{r1+r2}x+{r1*r2}=0}$$

we need to find two values that add together to make $\var{r1+r2}$ and multiply together to make $\var{r1*r2}$.

$$\begin{align} \var{r1} \times \var{r2}&=\var{r1*r2}\\ \var{r1}+\var{r2}&=\var{r1+r2}\\ \end{align}$$

Therefore using $\var{r1}$ and $\var{r2}$ we can write out the factorised equation

$$\simplify{(x+{r1})(x+{r2})}=0\text{.}$$

b)

In order to find the values of $x$ we need to factorise the questions like in part a). To do this we need to find factors of $\var{r1*r2}$ that add together to give $\var{r1+r2}$, which could be $\var{r1}$ and $\var{r2}$. 

This would give us a factorised equation of

$$\simplify {(x+{r1})(x+{r2})}=0\text{.}$$

In order to solve for $x$ we need $x$ values that would mean that

$$\begin{align} (\simplify {(x+{r1})})&=0 \\ \text{or}&\\ (\simplify{(x+{r2})})&=0\text{.} \end{align}$$

If at least one of the factorised brackets equals $0$ then our equation is satisfied because $0\times(x+a)=0$.

Therefore our possible $x$ values are 

$$\begin{align} x_1&=\var{-r2}\\ x_2&=\var{-r1}\text{.} \end{align}$$