How do you divide #(x^3-8x^2+17x-10) -:(x-5)#?

N.B the quotient is the answer.

If you're unfamiliar with this technique, all you are really doing is choosing a number (can contain a variable, e.g #x# which you multiply by divisor, (#x-5#), so that you get an output that is equal to the first term of your polynomial, in this case #x^3#.

For this method you want to work left to right, so the first thing I want to remove is the #x^3# term, so I need to multiply entire #(x-5)# (divisor) term by something that will get me an #x^3# term. In this case it will be #x^2#. Subtract this value from the polynomial. You then move on to the next term of the polynomial (moving right) until you reach constants (integers). At this point, multiplying the divisor by anything will just add more #x# variables, which will get you further from the solution. Anything left over can just be left as a "remainder". In this case, it is 0.

#"one way is to use the divisor as a factor in the numerator"#

#"consider the numerator"#

#color(red)(x^2)(x-5)color(magenta)(+5x^2)-8x^2+17x-10#

#=color(red)(x^2)(x-5)color(red)(-3x)(x-5)color(magenta)(-15x)+17x-10#

#=color(red)(x^2)(x-5)color(red)(-3x)(x-5)color(red)(+2)(x-5)color(magenta)(+10)-10#

#=color(red)(x^2)(x-5)color(red)(-3x)(x-5)color(red)(+2)(x-5)+0#

#rArr(x^3-8x^2+17x-10)/(x-5)#

#=(cancel((x-5))(color(red)(x^2-3x+2)))/cancel((x-5))#

#=x^2-3x+2#