How do you divide #5x^2  6x^3 + 1 + 7x# by #3x  4#?
1 Answer
#(6x^3 + 5x^2 + 7x + 1) : (3x  4) = 2x^2  x +1#
with remainder
Explanation:
First of all, order the terms by the power of
# 6 x^3 + 5x^2 + 7x + 1#
for the first term.
Now let me walk you through the polynomial long division.
You are basically doing the following operations:

Divide the dividend's term with the highest power by the divisor's term with the highest power. In your case, that's
#( 6 x^3) : (3x) = 2 x^2# 
Multiply the result, in your case
#2x^2# , with the divisor:#(2x^2) * (3x  4) = 6x^3 + 8x^2# 
Subtract the result from the last step from your divident:
#(6x^3 + 5x^2 + 7x + 1)  ( 6 x^3 + 8x^2) =  3x^2 + 7x + 1# 
Now, you can repeat all those steps with the term
#3x^2 + 7x + 1# as a new divident.... etc.
In total, your division should look like this:
#color(white)(xx) (6x^3 + 5x^2 + 7x + 1) : (3x  4) = 2x^2  x +1#
# ( 6 x^3 + 8 x^2)#
# color(white)(xx) color(white)(xxxxxxxxx) / #
# color(white)(xxxxxxx)  3 x^2 + 7x #
# color(white)(xxxxx) ( 3 x^2 + 4x) #
# color(white)(xxxxxxx) color(white)(xxxxxxxxx) / #
# color(white)(xxxxxxxxxxxxxx) 3x +1#
# color(white)(xxxxxxxxxxxii) (3x4)#
# color(white)(xxxxxxxxxxxxx) color(white)(xxxxxxxxx) / #
# color(white)(xxxxxxxxxxxxxxxxxx) 5#
This means that
#(6x^3 + 5x^2 + 7x + 1) : (3x  4) = 2x^2  x +1#
with remainder
Or, if you prefer a different notation,
#(6x^3 + 5x^2 + 7x + 1) : (3x  4) = 2x^2  x +1 + 5 / (3x  4)#