a)

As this question involves a number greater than $1$ before the $x^2$ value it has a factorised form $(ax+b)(cx+d)$.

To find $a$ and $c$, we need to consider the factors of $\var{a*c}$.

We are already given that one of them is $\var{a}$, so we know that the other one must be $\var{c}$.

This means our factorised equation must take the form

$$(\var{a}x+b)(\var{c}x+d)=0\text{.}$$

This expands to

$$\simplify{ {a*c}x^2 + ({a}*d+{c}*b)x + a*b}$$

So we must find two numbers which add together to make $\var{a*d+b*c}$, and multiply together to make $\var{b*d}$.

Therefore $b$ and $d$ must satisfy

$$\begin{align} b \times d &=\var{b*d}\\ \simplify{{a}d+{c}b} &= \var{a*d+b*c}\text{.} \end{align}$$

$b = \var{b}$ and $d = \var{d}$ satisfy these equations:

$$\begin{align} \var{b} \times \var{d} &=\var{b*d}\\ \simplify[]{ {a}*{d} + {b}*{c} } &= \var{a*d+b*c} \end{align}$$

So the factorised form of the equation is 

$$\simplify{({a}x+{b})({c}x+{d}) = 0} \text{.}$$

b)

$\simplify{({a}x+{b})({c}x+{d}) = 0}$ when either $\var{a}x+\var{b} = 0$ or $\var{c}x+ \var{d} = 0$.

So the roots of the equation are $\var[fractionnumbers]{-b/a}$ and $\var[fractionnumbers]{-d/c}$.