How do you divide #(x^3-10x^2+30x+3)/(x-5)#?

Question

Eli Assouline · Accepted Answer

Use long division or synthetic division.

First step: Realize that #x# needs to be multiplied by #x^2# in order to be the #x^3# in #x^3-10x^2+30x+3#. Then, multiply the #x^2# throughout #x-5# to get #x^3-5x^2# and SUBTRACT it (remember, the signs will change) from #x^3-10x^2+30x+3#.

You are left with #-5x^2+30x+3#, but only the first term is important for now. Like you did with the #x^3# in the first step, realize that you will need to add a #-5x# on top so that the #x(-5x)=-5x^2#. Continue multiplying and subtracting until you get to 28, which is the remainder.

All together, the answer would be written as:
#color(blue)(x^2-5x+5+28/(x-5)#

Eli Assouline · Answer

undefined To divide (x^3-10x^2+30x+3) by (x-5), you can use long division or synthetic division. Here is the solution using long division:

x^2 - 5x + 25
    _______________________
x - 5 | x^3 - 10x^2 + 30x + 3
         - (x^3 - 5x^2)
         _________________
                  -5x^2 + 30x
                  - (-5x^2 + 25x)
                  _________________
                             5x + 3
                             - (5x + 25)
                             _________________
                                    -22

Therefore, the quotient is x^2 - 5x + 25 and the remainder is -22.