How do you divide $\frac{6 x^{3} - 16 x^{2} + 17 x - 6}{3 x - 2}$ $(6x^3-16x^2+17x-6) / (3x-2)$ ?

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Answer 1 · 2018-05-07T06:05:24

$\frac{6 x^{3} - 16 x^{2} + 17 x - 6}{3 x - 2} = 2 x^{2} - 4 x + 3$ $(6x^3-16x^2+17x-6)/(3x-2)=2x^2-4x+3$

Use long division method.

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How do you divide (6x^3-16x^2+17x-6) / (3x-2)6x3−16x2+17x−63x−2(6x^3-16x^2+17x-6) / (3x-2)?