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

Nov 21, 2015

There are several ways to find:

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

#### Explanation:

We can factor by grouping:

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

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

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

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

So:

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

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

to find:

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

with no remainder.