Fo the equation #5x^5 + 13x^4 - 2x^2 -6, x-1# , How to divide the first expression by the second?

One method is to redistribute the terms in the dividend into multiples of the divisor, which you can do like this:

#5x^5+13x^4-2x^2-6#

#=5x^5-5x^4+18x^4-2x^2-6#

#=5x^5-5x^4+18x^4-18x^3+18x^3-2x^2-6#

#=5x^5-5x^4+18x^4-18x^3+18x^3-18x^2+16x^2-6#

#=5x^5-5x^4+18x^4-18x^3+18x^3-18x^2+16x^2-16x+16x-6#

#=5x^5-5x^4+18x^4-18x^3+18x^3-18x^2+16x^2-16x+16x-16+10#

#=(x-1)(5x^4+18x^3+18x^2+16x+16)+10#

So:

#(5x^5+13x^4-2x^2-6)/(x-1) = 5x^4+18x^3+18x^2+16x+16+10/(x-1)#

Alternative method

The way I would do the division would be to start writing the factorisation and add terms one at a time...

The first term of the quotient must be #5x^4# in order that when multiplied by #x# it gives the leading term #5x^5#, so start to write down:

#5x^5+13x^4-2x^2-6 = (x-1)(5x^4...#

Then note that #(-1)(5x^4)# will give us #-5x^4#, but we want #13x^4#. So we can deduce that the next term in the quotient is #18x^3#:

#5x^5+13x^4-2x^2-6 = (x-1)(5x^4+18x^3...#

This will result in a term #-18x^3# which we don't want, so the next term in our quotient is #18x^2#...

#5x^5+13x^4-2x^2-6 = (x-1)(5x^4+18x^3+18x^2...#

This will result in a term #-18x^2#, but we want #-2x^2#. So the next term we need for our quotient is #16x#...

#5x^5+13x^4-2x^2-6 = (x-1)(5x^4+18x^3+18x^2+16x...#

This will give result in #-16x#, but there is no #x# term in the dividend, so we need a constant term #16# to cancel it out...

#5x^5+13x^4-2x^2-6 = (x-1)(5x^4+18x^3+18x^2+16x+16)...#

Finally, note that #(-1)(16) = -16#, but we want #-6#, so we need to add a remainder term #10#...

#5x^5+13x^4-2x^2-6 = (x-1)(5x^4+18x^3+18x^2+16x+16)+10#

This takes a lot longer to describe than to do. With a bit of practice you may find this the most convenient method.

Shortcut method is synthetic division. To do this, write the coefficients of all the terms, including the missing ones, starting with the highest degree term in a single row. The missing terms here are #x^3# and #x#. Their coefficients would be o s. To the left of this row write the value of x ,obtained by equating the divisor to 0. In this case it would be x=1

            1|   5         13           0         -2       0           -6

 Now divide the first number 5 by 1 and the write the quotient 5 below the next number 13.

Add them to get 18.

Now divide this number by 1 and write the quotient 18 below the next number 0. continue this process till the last digit. The final display would be as follows

            1|   5         13           0          -2      0            -6
                            _5______18____18___16_____16___________
                             18           18        16     16          10

The Last digit obtained is 10. This is the Remainder. The quotient would be # 5x^4 +18x^3 +18x^2 +16x +16#