# How do you find the remainder when (3x^4 + 2x^3 – x^2 + 2x – 9) ÷ (x + 2)?

Nov 9, 2015

Start with dividing the term with the highest exponent by the term with the highest exponent, so here, divide $3 {x}^{4}$ by $x$.
$3 {x}^{4} \div x = 3 {x}^{3}$
Multiply $3 {x}^{3}$ with $x + 2$: $3 {x}^{3} \left(x + 2\right) = 3 {x}^{4} + 6 {x}^{3}$.
Your ${x}^{3}$ term is $2 {x}^{3}$, so to compute your remainder for the ${x}^{3}$ term, compute $2 {x}^{3} - 6 {x}^{3} = - 4 {x}^{3}$.

$3 {x}^{4} + 2 {x}^{3} - {x}^{2} + 2 x - 9 = \textcolor{g r e e n}{3 {x}^{3} \left(x + 2\right)} \textcolor{red}{- 4 {x}^{3} - {x}^{2} + 2 x - 9}$.

The green coloured terms are already done, now take care of the red coloured terms.

Again, divide the term with the highest exponent:
$\left(- 4 {x}^{3}\right) \div x = - 4 {x}^{2}$
Multiply with $x + 2$:
$- 4 {x}^{2} \left(x + 2\right) = - 4 {x}^{3} - 8 {x}^{2}$
The remainder for your ${x}^{2}$ term is:
$- {x}^{2} - \left(- 8 {x}^{2}\right) = 7 {x}^{2}$.

So,
$3 {x}^{4} + 2 {x}^{3} - {x}^{2} + 2 x - 9$
$= \textcolor{g r e e n}{3 {x}^{3} \left(x + 2\right)} - 4 {x}^{3} - {x}^{2} + 2 x - 9$
$= \textcolor{g r e e n}{3 {x}^{3} \left(x + 2\right) - 4 {x}^{2} \left(x + 2\right)} \textcolor{red}{+ 7 {x}^{2} + 2 x - 9} .$

Repeate the procedure with the remaining red term.
Divide: $7 {x}^{2} \div x = 7 x$
Multiply: $7 x \left(x + 2\right) = 7 {x}^{2} + 14 x$
Remainder: $2 x - 14 x = - 12 x$

In total:

$3 {x}^{4} + 2 {x}^{3} - {x}^{2} + 2 x - 9$
$= \textcolor{g r e e n}{3 {x}^{3} \left(x + 2\right)} - 4 {x}^{3} - {x}^{2} + 2 x - 9$
$= \textcolor{g r e e n}{3 {x}^{3} \left(x + 2\right) - 4 {x}^{2} \left(x + 2\right)} + 7 {x}^{2} + 2 x - 9$
$= \textcolor{g r e e n}{3 {x}^{3} \left(x + 2\right) - 4 {x}^{2} \left(x + 2\right) + 7 x \left(x + 2\right)} \textcolor{red}{- 12 x - 9} .$

Last step: try to divide $- 12 x - 9$ by $x + 2$:

Divide: $- 12 x \div x = - 12$
Multiply: $- 12 \left(x + 2\right) = - 12 x - 24$
Remainder: $- 9 - \left(- 24\right) = 15$

As the degree of the remainder ($0$, since no $x$ or rather ${x}^{0}$) is lower than the degree of the denominator (degree $1$, highest exponent ${x}^{1}$), you can stop here.

The result is:

$3 {x}^{4} + 2 {x}^{3} - {x}^{2} + 2 x - 9$
$= \textcolor{g r e e n}{3 {x}^{3} \left(x + 2\right)} - 4 {x}^{3} - {x}^{2} + 2 x - 9$
$= \textcolor{g r e e n}{3 {x}^{3} \left(x + 2\right) - 4 {x}^{2} \left(x + 2\right)} + 7 {x}^{2} + 2 x - 9$
$= \textcolor{g r e e n}{3 {x}^{3} \left(x + 2\right) - 4 {x}^{2} \left(x + 2\right) + 7 x \left(x + 2\right)} - 12 x - 9$
$= \textcolor{g r e e n}{3 {x}^{3} \left(x + 2\right) - 4 {x}^{2} \left(x + 2\right) + 7 x \left(x + 2\right) - 12 \left(x + 2\right)} + 15$

So, the result of
$\left(3 {x}^{4} + 2 {x}^{3} - {x}^{2} + 2 x - 9\right) \div \left(x + 2\right)$ is
$3 {x}^{3} - 4 {x}^{2} + 7 x - 12$ with the remainder $15$, or:

$\frac{3 {x}^{4} + 2 {x}^{3} - {x}^{2} + 2 x - 9}{x + 2} = 3 {x}^{3} - 4 {x}^{2} + 7 x - 12 + \frac{15}{x + 2}$

Hope that this helped!