# How do you divide (2x^3+9x^2-9x+1) / (2x-3)  using polynomial long division?

$\frac{2 {x}^{3} + 9 {x}^{2} - 9 x + 1}{2 x - 3} = {x}^{2} + 6 x + \frac{9}{2} + \frac{\frac{29}{2}}{2 x - 3}$

#### Explanation:

We divide by long division method
$\text{ " " " " " } \underline{{x}^{2} + 6 x + \frac{9}{2}}$
$2 x - 3 \lceiling 2 {x}^{3} + 9 {x}^{2} - 9 x + 1$
" " " " " "underline(2x^3-3x^2" " " " " " " ")
$\text{ " " " " " " " " } 12 {x}^{2} - 9 x + 1$
" " " " " " " " " "underline(12x^2-18x" " " )
$\text{ " " " " " " " " " " " " " } + 9 x + 1$
" " " " " " " " " " " " " "underline(+9x-27/2" " " )
$\text{ " " " " " " " " " " " " " " " " " } + \frac{29}{2}$

The result is

$\frac{2 {x}^{3} + 9 {x}^{2} - 9 x + 1}{2 x - 3} = {x}^{2} + 6 x + \frac{9}{2} + \frac{\frac{29}{2}}{2 x - 3}$

Checking:

$\text{Divisor x Quotient"+"Remainder"="Dividend}$

$\left(2 x - 3\right) \left({x}^{2} + 6 x + \frac{9}{2}\right) + \frac{29}{2} = 2 {x}^{3} + 12 {x}^{2} + 9 x - 3 {x}^{2} - 18 x - \frac{27}{2} + \frac{29}{2}$

$\left(2 x - 3\right) \left({x}^{2} + 6 x + \frac{9}{2}\right) + \frac{29}{2} = 2 {x}^{3} + 9 {x}^{2} - 9 x - \frac{27}{2} + \frac{29}{2}$

$\left(2 x - 3\right) \left({x}^{2} + 6 x + \frac{9}{2}\right) + \frac{29}{2} = 2 {x}^{3} + 9 {x}^{2} - 9 x + \frac{2}{2}$

$\left(2 x - 3\right) \left({x}^{2} + 6 x + \frac{9}{2}\right) + \frac{29}{2} = 2 {x}^{3} + 9 {x}^{2} - 9 x + 1$

God bless....I hope the explanation is useful.