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

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

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Answer 1 · 2015-11-21T20:01:28

There are several ways to find:

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

Explanation:

We can factor by grouping:

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

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

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

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

So:

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

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

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

with no remainder.

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

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Explanation:

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How do you divide (12x^3-16x^2-27x+36)/ (3x-4)12x3−16x2−27x+363x−4(12x^3-16x^2-27x+36)/ (3x-4)?

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Explanation:

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