how to long divide algebraic fraction with same power of x in both numerator and denominator

Question

7x+3/4x+9

this is an algebraic fraction. divide expressing your answer in the form Q(x)+((R(x)/D(x)).

Answer 1

You can write $7x+3=(4x+9)+(3x-6)=\cfrac 74(4x+9)-12\cfrac {3}4$ depending whether you want integer coefficients or rational ones. Note that in the second case the remainder is of lower degree than the divisor. The second is canonical in the sense that for a given $d(x)$, where the coefficients come from a field, we can write $p(x)=q(x)d(x)+r(x)$ in a unique way with the degree of $r(x)$ less than the degree of $d(x)$.

This form gives $$\frac {p(x)}{d(x)}=q(x)+\frac {r(x)}{d(x)}$$

The integers form an integral domain rather than a field, so if we are dealing with integer coefficients we can't always divide. In this case $\frac 74$ is not an integer. But it is a rational number, and the rationals form a field. This is why you get two different answers, depending on context.

Answer 2

Assuming you mean $\frac {7x+3}{4x+9}$, because the degrees are the same your $Q(x)$ will be a constant. The ratio of the highest powers is $\frac 74$, so that will be the constant. This gives $\frac{7x+3}{4x+9}=\frac 74 + \frac {3-\frac {63}4}{4x+9}=\frac 74 -\frac {51}{16x+36}$. Note that the degree of the numerator in the fraction is lower than the degree of the denominator.