# How do you divide (2x^3 - 11x^2 + 12x + 9)/(x-3)?

Jul 27, 2018

The remainder is $0$ and the quotient is $= 2 {x}^{2} - 5 x - 3$

#### Explanation:

Let's perform the synthetic division

$\textcolor{w h i t e}{a a a a}$$3$$|$$\textcolor{w h i t e}{a a a a}$$2$$\textcolor{w h i t e}{a a a a}$$- 11$$\textcolor{w h i t e}{a a a a a a}$$12$$\textcolor{w h i t e}{a a a a a}$$9$

$\textcolor{w h i t e}{a a a a a}$$|$$\textcolor{w h i t e}{a a a a}$$\textcolor{w h i t e}{a a a a a a a}$$6$$\textcolor{w h i t e}{a a a a a}$$- 15$$\textcolor{w h i t e}{a a a}$$- 9$

$\textcolor{w h i t e}{a a a a a a a a a}$_________

$\textcolor{w h i t e}{a a a a a}$$|$$\textcolor{w h i t e}{a a a a}$$2$$\textcolor{w h i t e}{a a a a a}$$- 5$$\textcolor{w h i t e}{a a a a a}$$- 3$$\textcolor{w h i t e}{a a a a a}$$\textcolor{red}{0}$

The remainder is $0$ and the quotient is $= 2 {x}^{2} - 5 x - 3$

$\frac{2 {x}^{3} - 11 {x}^{2} + 12 x + 9}{x - 3} = 2 {x}^{2} - 5 x - 3$

Apply the remainder theorem

When a polynomial $f \left(x\right)$ is divided by $\left(x - c\right)$, we get

$f \left(x\right) = \left(x - c\right) q \left(x\right) + r$

Let $x = c$

Then,

$f \left(c\right) = 0 + r$

Here,

$f \left(x\right) = 2 {x}^{3} - 11 {x}^{2} + 12 x + 9$

Therefore,

$f \left(3\right) = 2 \cdot {3}^{3} - 11 \cdot {3}^{2} + 12 \cdot 3 + 9$

$= 54 - 99 + 36 + 9$

$= 0$

The remainder is $= 0$

Jul 27, 2018

$\left(2 {x}^{3} - 11 {x}^{2} + 12 x + 9\right) = \left(x - 3\right) \left(2 {x}^{2} - 5 x - 3\right) + \left(0\right)$

#### Explanation:

$\left(2 {x}^{3} - 11 {x}^{2} + 12 x + 9\right) \div \left(x - 3\right)$

We can divide this polynomial by using synthetic division

We have , $p \left(x\right) = 2 {x}^{3} - 11 {x}^{2} + 12 x + 9 \mathmr{and} \text{divisor :} x = 3$

We take ,coefficients of $p \left(x\right) \to 2 , - 11 , 12 , 9$

. $3 |$ $2 \textcolor{w h i t e}{\ldots .} - 11 \textcolor{w h i t e}{\ldots \ldots .} 12 \textcolor{w h i t e}{\ldots \ldots .} 9$
$\underline{\textcolor{w h i t e}{}} |$ ul(0color(white)( ..........)6color(white)(....)-15color(white)(...)-9
color(white)(......)2color(white)(......)-5color(white)(...)color(white)(..)-3color(white)(.....)color(violet)(ul|0|
We can see that , quotient polynomial :

$q \left(x\right) = 2 {x}^{2} - 5 x - 3 \mathmr{and} \text{the Remainder} = 0$

Hence ,

$\left(2 {x}^{3} - 11 {x}^{2} + 12 x + 9\right) = \left(x - 3\right) \left(2 {x}^{2} - 5 x - 3\right) + \left(0\right)$