This will be a bit hard to write out here, but let's see. I'm assuming that this is a typo. You may have meant #3x^3#. If it was #3x^2#, you wouldn't need synthetic division and you could just factor this to get #x = 1, x = -5#.

Assuming #3x^3#... First, you let the coefficients of each degree be used in the division (#3, -1, 8, -10#).

Then, dividing by #x - 1# implies that you use #1# in your upper left (if it was #x+1#, put #-1#). So, draw the bottom and right sides of a square, put #1# inside it, and then write #"3" " -1" " 8" " -10"# to the right.

#"1" ||# #"3 " "-1 " "8 " "-10 "#

#+#

#"" " "##-----#

First, bring the #3# down to the bottom, and multiply it by the #1#. Put that #3# below #-1#.

#"1" ||# #"3 " "-1 " "8 " "-10 "#

#+# #"" " " "" "3"#

#"" " "##-----#

#"" " " "3"#

Then add it up:

#"1" ||# #"3 " "-1 " "8 " "-10 "#

#+# #"" " " "" "3"#

#"" " "##-----#

#"" " " "3" "" " 2"#

Repeat a few times once you've figured out the simple pattern (#"divisor"*"new sum"#, put result under next-lowest degree term, add to get another #"new sum"#, repeat).

#"1" ||# #"3 " "-1 " "8 " "-10 "#

#+# #"" " " "" "3" "" " 2"#

#"" " "##-----#

#"" " " "3" "" " 2" "" " 10"#

(All that really happened here was #1*2 = 2# and #8 + 2 = 10#.)

#"1" ||# #"3 " "-1 " "8 " "-10 "#

#+# #"" " " "" "3" "" " 2" " 10"#

#"" " "##-----#

#"" " " "3" "" " 2" "" " 10" " 0"#

(Then to finish it up, #1*10 = 10#, and #-10 + 10 = 0#.)

You know you can stop when you reach the far right and you have no spot left below the original dividend (#"3" " -1" " 8" " -10"#) to insert a product.

Since presumably you started with a cubic, the answer is a quadratic. Thus, you have:

#3x^2 + 2x + 10x^0 + (r=0)#

Indeed, if you multiply them together, you get the original back:

#(3x^2 + 2x + 10)(x-1) = 3x^3 - 3x^2 + 2x^2 - 2x + 10x - 10 = 3x^3 - x^2 + 8x - 10#

So the answer would be #3x^2 + 2x + 10#. In a general case, you could write it like this:

Let #f(x) = ax^3 + bx^2 + cx + d# and #g(x) = x - h#. If the solution is #h(x)#, and a remainder is #r(x)#, then:

#h(x) = f(x)/g(x) + (r(x))/(g(x))#

So you would have:

#h(x) = "answer" = (3x^3 - x^2 + 8x - 10)/(x-1) color(blue)(= 3x^2 + 2x + 10 + (r(x))/(x-1))#

where #r(x) = 0#.

Had you meant #3x^2 - x^2 + 8x - 10#, you could have just done the quadratic equation. Or, factor.

#3x^2 - x^2 + 8x - 10 = 2x^2 + 8x - 10 => (2x - 2)(x + 5) = 0#

#=> x = 1, x = -5#