How do you divide #(x^4-2x^2+3)/(x-1)#?

1 Answer
Dec 3, 2015

#x^3 + x^2 -x - 1+ 2 / (x-1)#

Explanation:

I know that in some countries, a different notation of the polynomial long division is being used. I would like to use the notation that I'm familiar with, and I hope that it will be no problem to re-write the division in your notation if needed!

#color(white)(xii) (x^4 color(white)(xxxx) - 2x^2 color(white)(xxxxxx) + 3) -: (x- 1) = x^3 + x^2 -x - 1#
#-(x^4 - x^3)#
#color(white)(xii)(color(white)(xxxxxx))/()#
#color(white)(xxxxxx) x^3 - 2x^2#
#color(white)(xxxii) -(x^3 - color(white)(x) x^2)#
#color(white)(xxxxxii)(color(white)(xxxxxxx))/()#
#color(white)(xxxxxxxx) - x^2 + 0 * x#
#color(white)(xxxxxii) - (- x^2 + color(white)(xx) x color(white)(i))#
#color(white)(xxxxxxxii)(color(white)(xxxxxxxxxx))/()#
#color(white)(xxxxxxxxxxxxxx) -x color(white)(x) + 3#
#color(white)(xxxxxxxxxxxx) - (-x color(white)(x)+1)#
#color(white)(xxxxxxxxxxxxxxii)(color(white)(xxxxxxx))/()#
#color(white)(xxxxxxxxxxxxxxxxxxxxii) 2#

So, you have a remainder #2#, and in total, your solution is

#(x^4 - 2x^2 + 3)/(x-1) = x^3 + x^2 -x - 1+ 2 / (x-1)#