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 $P(x)=9x^4 - x^2 - 6x + 2$ )

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