How do you divide #(12x^3-16x^2-27x+36)/ (3x-4)#?

Question

2152 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-11-21T20:01:28

There are several ways to find:

#(12x^3-16x^2-27x+36)/(3x-4) = 4x^2-9#

We can factor by grouping:

#12x^3-16x^2-27x+36#

#=(12x^3-16x^2)-(27x-36)#

#=4x^2(3x-4)-9(3x-4)#

#=(4x^2-9)(3x-4)#

So:

#(12x^3-16x^2-27x+36)/(3x-4) = 4x^2-9#

Alternatively, we can divide #12x^3-16x^2-27x+36# by #3x-4# by long dividing the coefficients:
enter image source here
to find:

#(12x^3-16x^2-27x+36)/(3x-4) = 4x^2-9#

with no remainder.