Synthetic division is simply a shortcut to polynomial long division. But it only works for monic linear factors, i.e. those of the form $(x-\alpha)$.
When you divide a polynomial $P(x)$ by a non-monic linear factor like $(ax-b), a \neq 1$, this is what you get:
$$P(x) = Q(x)(ax-b) + R$$
where $Q(x)$ is the quotient polynomial and $R$ is the remainder (it will be a constant term).
That can also be expressed as :
$$P(x) = aQ(x)(x-\frac{b}{a}) + R$$
Note that when you're doing synthetic division by $\displaystyle (x-\frac{b}{a})$, you're actually getting $aQ(x)$ as the quotient. Hence the quotient $Q(x)$when you divide by $(ax-b)$ has to be determined by dividing $aQ(x)$ by $a$. That's why you're doing that division step. Note also that the remainder $R$ is unaltered, which is why you don't have to do anything to it.
In your specific example, $a=2, b = 3$. You're doing the synthetic division by $\displaystyle (x-\frac{3}{2})$ and I hope it should now be clear why you need to divide the quotient you get by $2$ to get the quotient when you divide by $(2x-3)$.
Plug in $1$ and $-1$ to get two values of $r(x)$ which is linear. From there you can get what $a,b$ are in $ax+b.$
Since
$$f(x)=g(x)(x+1)(x-1)+r(x)$$
we have
$$ f(1)=g(1)(1+1)(1-1)+r(1)=r(1)=-10$$
$$ f(-1)=g(1)(-1+1)(-1-1)+r(-1)=r(-1)=16$$
We know the remainder is of degree $1$, so
$r(x)=ax+b$
and now we know,
$$r(1)=ax+b=a+b=-10$$
$$r(-1)=ax+b=-a+b=16$$
so, solve
$$a+b=-10$$
$$-a+b=16$$
which yields, $a=-13$ $b=3$, so
$$r(x)=-13x+3$$
Best Answer
119 represents the remainder, and $x^3+x^2+7x+30$ represents the quotient.