# How do you use the remainder theorem to determine the remainder when the polynomial (x^3+2x^2-3x+9)div(x+3)?

Oct 21, 2017

The remainder is $\textcolor{red}{9}$

#### Explanation:

The remainder theorem states :

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

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

and

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

We apply this theorem

f(x)=(x^3+2x^2-3x+9

Therefore,

f(-3)=((-3)^3+2(-3)^2-3(-3)+9

$= - 27 + 18 + 9 + 9 = 9$

The remainder is $= 9$

Let's perform the synthetic division to confirm the results

$\textcolor{w h i t e}{a a}$$- 3$$\textcolor{w h i t e}{a a a a a}$$|$$\textcolor{w h i t e}{a a a}$$1$$\textcolor{w h i t e}{a a a a a}$$2$$\textcolor{w h i t e}{a a a a a a}$$- 3$$\textcolor{w h i t e}{a a a a a}$$9$
$\textcolor{w h i t e}{a a a a a a a a a a a a}$$- - - - - - - - - - - -$

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

$\textcolor{w h i t e}{a a a a}$$\textcolor{w h i t e}{a a a a a a}$$|$$\textcolor{w h i t e}{a a a}$$1$$\textcolor{w h i t e}{a a a a}$$- 1$$\textcolor{w h i t e}{a a a a a a a}$$0$$\textcolor{w h i t e}{a a a a a}$$\textcolor{red}{9}$

The remainder is $\textcolor{red}{9}$ and the quotient is $= {x}^{2} - x$

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

Oct 21, 2017

The remainder is $9$. See explanation.

#### Explanation:

According to the Remainder Threorem the remainder of division $P \left(x\right) \div \left(x - a\right)$ is equal to $P \left(a\right)$.

Here $a = - 3$, so the remainder is:

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

$= - 27 + 18 + 9 + 9 = 9$