How do you divide $(12x^3-16x^2-27x+36)/ (3x-4)$ ?

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Answer 1 · 2015-11-21T20:01:28

There are several ways to find:

$(12x^3-16x^2-27x+36)/(3x-4) = 4x^2-9$

We can factor by grouping:

$12x^3-16x^2-27x+36$

$=(12x^3-16x^2)-(27x-36)$

$=4x^2(3x-4)-9(3x-4)$

$=(4x^2-9)(3x-4)$

So:

$(12x^3-16x^2-27x+36)/(3x-4) = 4x^2-9$

Alternatively, we can divide $12x^3-16x^2-27x+36$ by $3x-4$ by long dividing the coefficients:
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to find:

$(12x^3-16x^2-27x+36)/(3x-4) = 4x^2-9$

with no remainder.

How do you divide (12x^3-16x^2-27x+36)/ (3x-4)(12x^3-16x^2-27x+36)/ (3x-4)?