How do you factor $\frac{3 x^{3} - 5 x^{2} + x - 6}{x - 2}$ $[3x^3-5x^2+x-6] / (x-2)$ ?

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How do you factor $\frac{3 x^{3} - 5 x^{2} + x - 6}{x - 2}$ $[3x^3-5x^2+x-6] / (x-2)$ ?

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Answer 1 · 2015-10-08T05:29:08

We have that

$\frac{3 x^{3} - 5 x^{2} + x - 6}{x - 2} = \frac{(x - 2) (3 x^{2} + x + 3)}{x - 2} = 3 x^{2} + x + 3$ $(3x^3-5x^2+x-6) / (x-2)=((x-2)(3x^2+x+3))/(x-2)=3x^2+x+3$

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How do you factor $\frac{3 x^{3} - 5 x^{2} + x - 6}{x - 2}$ $[3x^3-5x^2+x-6] / (x-2)$ ?

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How do you factor [3x^3-5x^2+x-6] / (x-2)3x3−5x2+x−6x−2[3x^3-5x^2+x-6] / (x-2)?

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How do you factor $\frac{3 x^{3} - 5 x^{2} + x - 6}{x - 2}$ $[3x^3-5x^2+x-6] / (x-2)$ ?