How do you divide ( x^5 - x^3 + x - 2 x - 5)/(x - 2 )x5x3+x2x5x2?

1 Answer
May 4, 2016

Have a look at https://socratic.org/s/aubpbqz9. Not the same values but the process is the same. Unless I have gone wrong, your answer should be:
" Corrected solution"-> x^4+2x^3+3x^2+6x+11 +17/(x-2) Corrected solutionx4+2x3+3x2+6x+11+17x2

Explanation:

Using place holders such as 0x^20x2 which is the same as 00

" "x^4+2x^3+3x^2+6x+11 x4+2x3+3x2+6x+11
" "x-2bar(|" "color(magenta)(x^5+0x^4-x^3+0x^2-x-5)) x2¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ x5+0x4x3+0x2x5
color(brown)(x^4(x-2)->)" "underline(x^5-2x^4)" " larr" Subtract
" "0+2x^4-x^3
color(brown)(2x^3(x-2))->" "underline(2x^4-4x^3) larr Subtract
" "0+3x^3+0x^2
color(brown)(3x^2(x-2))->" "underline(3x^3-6x^2) larr Subtract
" "0+6x^2-x
color(brown)(6x(x-2))->" "underline(6x^2-12x)
" "0+11x-5
color(brown)(11(x-2))->" "underline(11x-22)
color(brown)(" Remainder") ->" "+17

x^4+2x^3+3x^2+6x+11 +17/(x-2)

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
color(blue)("Check")

" "x^4+2x^3+3x^2+6x+11 + 17/(x-2)

underline(color(white)(///////////////////////////////////////.)x-2)" "larr "Multiply"
x^5+2x^4+3x^2+6x^2+11x
underline(0x^5-2x^4-4x^3-6x^2-12x-22+A)" "larr" Add"
x^5+" "0-x^3+color(white)(...)0-color(white)(...)x-color(white)(..)22+ A

Where A= 17/(x-2)xx(x-2) = 17

x^5-x^3-x-22+17" "=" "color(magenta)(x^5-x^3-x-5)-> as required