How do you divide #(4x^3+3x^2+8x+12)/(3x-4) #?

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Answer 1 · 2017-12-12T22:19:15

#4x^3 +x^2-5#

So to begin, lets break down this problem to its lowest form

#(4*x^3 +3*x^2+8*x+12)/(3*x-4)=#

We can eliminate any like terms, see any? How about

#(4*x^3 +cancel(3)*x^2+8*cancel(x)+12)/(cancel(3)*cancel(x)-4)=#

Now we can get rid of #-4# by dividing it with the top

#(4*x^3 +x^2+cancel(20)^color(red)(-5))/cancel(-4)=#

So, we are left with #4x^3 +x^2-5# and that is its simplest form! There you have it! Hope this helped!
~Chandler Dowd