How do you divide #2x^2-5x-3# by #2x-6# ?

1 Answer
George C.
Mar 4, 2017

#(2x^2-5x-3)/(2x-6) = x+1/2#

Explanation:

We can split the numerator up into parts which are monomial multiples of #(2x-6)#, then combine them like this:

#(2x^2-5x-3)/(2x-6) = (2x^2-6x+x-3)/(2x-6)#

#color(white)((2x^2-5x-3)/(2x-6)) = (x(2x-6)+1/2(2x-6))/(2x-6)#

#color(white)((2x^2-5x-3)/(2x-6)) = ((x+1/2)color(red)(cancel(color(black)((2x-6)))))/color(red)(cancel(color(black)((2x-6))))#

#color(white)((2x^2-5x-3)/(2x-6)) = x+1/2#

Related questions
See all questions in Division of Polynomials
Impact of this question
2110 views around the world
You can reuse this answer
Creative Commons License