How do you divide #(x-2)^2/(x^2-5x+9)#?

1 Answer
Jun 14, 2016

#(x^2-4x+4)/(x^2-5x+9)" "=" " 1 +(x-5)/(x^2-5x+9)#

Explanation:

Look at https://socratic.org/s/avjGBRiF as a further example of the method.

Write as #(color(blue)(x^2-4x+4))/(color(green)(x^2-5x+9))#

#"Numerator "->" "color(blue)( x^2-4x+4)#
#color(magenta)(1xx)color(green)((x^2-5x+9)) ->" "ul(x^2-5x+9)" "larr" Subtract"#
#"Remainder" ->" " 0+color(white)(.)x-5#

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
So #(x^2-4x+4)/(x^2-5x+9)" "=" "color(magenta)( 1) +(x-5)/(x^2-5x+9)#