How do you long divide #(9x^4-2-6x-x^2)/(3x-1)#?

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Answer 1 · 2015-06-15T21:46:50

#(3x^3 + x^2 - 2)-4/(3x-1)#

Let's put the polynomial in its usual order

#P(x) = 9x^4 - x^2 - 6x - 2#

First, divide #9x^4# by #3x => 3x^3#

Now subtract #3x^3(3x-1) = 9x^4 - 3x^3# to #P(x)#, you obtain #P_1(x)=3x^3 - x^2 - 6x - 2#

Now we procede the same way: divide #3x^3# by #3x => x^2#

Now subtract #x^2(3x-1) = 3x^3 - x^2# to #P_1(x)#, you obtain #P_2(x)= -6x - 2#

Now divide #-6x# by #3x => -2#

Now subtract #-2(3x-1)=-6x+2# to #6x - 2#, you obtain -4, which means

#P(x)=(3x-1)(3x^3 + x^2 - 2) - 4#

Or, if you want to write it in fractions, #P(x)=(3x^3 + x^2 - 2)-4/(3x-1)#

(NB: maybe there's an error: knowing this kind of exercises, I'm pretty sure they want the reminder to be 0, so maybe #P(x)=9x^4 - x^2 - 6x + 2#)