# How do you divide (-3x^4-9x^2+2x-3)/(x^2+4)?

Apr 28, 2017

$- 3 {x}^{2} + 3 + \frac{2 x - 15}{{x}^{2} + 4}$

#### Explanation:

$\textcolor{w h i t e}{\frac{000}{\textcolor{b l a c k}{{x}^{2} + 4}} \frac{000}{\textcolor{b l a c k}{\text{)} \overline{- 3 {x}^{4} - 9 {x}^{2} + 2 x - 3}}}}$

First, remember to fill in any powers we don't have in the divisor and dividend with a $0 x$ to that power

$\textcolor{w h i t e}{\frac{000}{\textcolor{b l a c k}{{x}^{2} + 0 x + 4}} \frac{000}{\textcolor{b l a c k}{\text{)} \overline{- 3 {x}^{4} + 0 {x}^{3} - 9 {x}^{2} + 2 x - 3}}}}$

Now divide the first term in the dividend with the first term in the divisor
(-3x^4-:x^2)
[Notice how the result from the division is written above its corresponding power $\left({x}^{2}\right)$]

$\textcolor{w h i t e}{\frac{000}{\textcolor{b l a c k}{{x}^{2} + 0 x + 4}} \frac{\textcolor{w h i t e}{000} \textcolor{b l a c k}{- 3 {x}^{2}}}{\textcolor{b l a c k}{\text{)} \overline{- 3 {x}^{4} + 0 {x}^{3} - 9 {x}^{2} + 2 x - 3}}}}$

Now multiply our divisor $\left({x}^{2} + 0 x + 4\right)$ with the result of our division $\left(- 3 {x}^{2}\right)$ and write it like so...

$\textcolor{w h i t e}{\frac{000}{\textcolor{b l a c k}{{x}^{2} + 0 x + 4}} \frac{\textcolor{w h i t e}{000} \textcolor{b l a c k}{- 3 {x}^{2}}}{\textcolor{b l a c k}{\text{)} \overline{- 3 {x}^{4} + 0 {x}^{3} - 9 {x}^{2} + 2 x - 3}}}}$
$\textcolor{w h i t e}{000000000000} - 3 {x}^{4} - 0 {x}^{3} - 12 {x}^{2}$

Remember just like regular long division, we subtract the result of our multiplication so the signs change

$\textcolor{w h i t e}{\frac{000}{\textcolor{b l a c k}{{x}^{2} + 0 x + 4}} \frac{\textcolor{w h i t e}{000} \textcolor{b l a c k}{- 3 {x}^{2}}}{\textcolor{b l a c k}{\text{)} \overline{- 3 {x}^{4} + 0 {x}^{3} - 9 {x}^{2} + 2 x - 3}}}}$
$\textcolor{w h i t e}{000000000000} + 3 {x}^{4} + 0 {x}^{3} + 12 {x}^{2}$

Now we get...

$\textcolor{w h i t e}{\frac{000}{\textcolor{b l a c k}{{x}^{2} + 0 x + 4}} \frac{\textcolor{w h i t e}{000} \textcolor{b l a c k}{- 3 {x}^{2}}}{\textcolor{b l a c k}{\text{)} \overline{- 3 {x}^{4} + 0 {x}^{3} - 9 {x}^{2} + 2 x - 3}}}}$
$\textcolor{w h i t e}{000000000000} + 3 {x}^{4} + 0 {x}^{3} + 12 {x}^{2}$
$\textcolor{w h i t e}{00000000000} \overline{\textcolor{w h i t e}{0000} 0 {x}^{4} + 0 {x}^{3} + 3 {x}^{2}}$

Bring down our remaining terms

$\textcolor{w h i t e}{\frac{000}{\textcolor{b l a c k}{{x}^{2} + 0 x + 4}} \frac{\textcolor{w h i t e}{000} \textcolor{b l a c k}{- 3 {x}^{2}}}{\textcolor{b l a c k}{\text{)} \overline{- 3 {x}^{4} + 0 {x}^{3} - 9 {x}^{2} + 2 x - 3}}}}$
$\textcolor{w h i t e}{000000000000} + 3 {x}^{4} + 0 {x}^{3} + 12 {x}^{2}$
$\textcolor{w h i t e}{00000000000} \overline{\textcolor{w h i t e}{000000000000000} 3 {x}^{2}} + 2 x - 3$

Now divide the first term from the subtraction with the first term from the divisor, kind of like from regular long division $\left(3 {x}^{2} \div {x}^{2}\right)$

$\textcolor{w h i t e}{\frac{000}{\textcolor{b l a c k}{{x}^{2} + 0 x + 4}} \frac{\textcolor{w h i t e}{0000000} \textcolor{b l a c k}{- 3 {x}^{2} + 3}}{\textcolor{b l a c k}{\text{)} \overline{- 3 {x}^{4} + 0 {x}^{3} - 9 {x}^{2} + 2 x - 3}}}}$
$\textcolor{w h i t e}{000000000000} + 3 {x}^{4} + 0 {x}^{3} + 12 {x}^{2}$
$\textcolor{w h i t e}{00000000000} \overline{\textcolor{w h i t e}{000000000000000} 3 {x}^{2}} + 2 x - 3$

and again, multiply the divisor $\left({x}^{2} + 0 x + 4\right)$ with the result from our division $\left(3\right)$

$\textcolor{w h i t e}{\frac{000}{\textcolor{b l a c k}{{x}^{2} + 0 x + 4}} \frac{\textcolor{w h i t e}{0000000} \textcolor{b l a c k}{- 3 {x}^{2} + 3}}{\textcolor{b l a c k}{\text{)} \overline{- 3 {x}^{4} + 0 {x}^{3} - 9 {x}^{2} + 2 x - 3}}}}$
$\textcolor{w h i t e}{000000000000} + 3 {x}^{4} + 0 {x}^{3} + 12 {x}^{2}$
$\textcolor{w h i t e}{00000000000} \overline{\textcolor{w h i t e}{000000000000000} 3 {x}^{2}} + 2 x - 3$
$\textcolor{w h i t e}{00000000000000000000000000} 3 {x}^{2} + 0 x + 12$

[Remember we are subtracting so don't forget to change the signs!]

$\textcolor{w h i t e}{\frac{000}{\textcolor{b l a c k}{{x}^{2} + 0 x + 4}} \frac{\textcolor{w h i t e}{0000000} \textcolor{b l a c k}{- 3 {x}^{2} + 3}}{\textcolor{b l a c k}{\text{)} \overline{- 3 {x}^{4} + 0 {x}^{3} - 9 {x}^{2} + 2 x - 3}}}}$
$\textcolor{w h i t e}{000000000000} + 3 {x}^{4} + 0 {x}^{3} + 12 {x}^{2}$
$\textcolor{w h i t e}{00000000000} \overline{\textcolor{w h i t e}{000000000000000} 3 {x}^{2}} + 2 x - 3$
$\textcolor{w h i t e}{000000000000000000000000} - 3 {x}^{2} - 0 x - 12$

$\textcolor{w h i t e}{\frac{000}{\textcolor{b l a c k}{{x}^{2} + 0 x + 4}} \frac{\textcolor{w h i t e}{0000000} \textcolor{b l a c k}{- 3 {x}^{2} + 3}}{\textcolor{b l a c k}{\text{)} \overline{- 3 {x}^{4} + 0 {x}^{3} - 9 {x}^{2} + 2 x - 3}}}}$
$\textcolor{w h i t e}{000000000000} + 3 {x}^{4} + 0 {x}^{3} + 12 {x}^{2}$
$\textcolor{w h i t e}{00000000000} \overline{\textcolor{w h i t e}{000000000000000} 3 {x}^{2}} + 2 x - 3$
$\textcolor{w h i t e}{000000000000000000000000} - 3 {x}^{2} - 0 x - 12$
$\textcolor{w h i t e}{000000000000000000000000} \overline{\textcolor{w h i t e}{000} 0 {x}^{2} + 2 x - 15}$

$\textcolor{w h i t e}{\frac{000}{\textcolor{b l a c k}{{x}^{2} + 0 x + 4}} \frac{\textcolor{w h i t e}{0000000} \textcolor{b l a c k}{- 3 {x}^{2} + 3}}{\textcolor{b l a c k}{\text{)} \overline{- 3 {x}^{4} + 0 {x}^{3} - 9 {x}^{2} + 2 x - 3}}}}$
$\textcolor{w h i t e}{000000000000} + 3 {x}^{4} + 0 {x}^{3} + 12 {x}^{2}$
$\textcolor{w h i t e}{00000000000} \overline{\textcolor{w h i t e}{000000000000000} 3 {x}^{2}} + 2 x - 3$
$\textcolor{w h i t e}{000000000000000000000000} - 3 {x}^{2} - 0 x - 12$
$\textcolor{w h i t e}{000000000000000000000000} \overline{\textcolor{w h i t e}{000000000} 2 x - 15}$

Now, we are left with $2 x - 15$. Before you jump in and try to divide again, notice how we have gotten to the point where the first term in our divisor $\left({x}^{2}\right)$ is larger than the first term from our subtraction $\left(2 x\right)$.
This means that $- 3 {x}^{2} + 3$ is our quotient and $2 x - 15$ is our remainder.

When dividing polynomials, our answer is defined as the quotient $\textcolor{red}{\left(- 3 {x}^{2} + 3\right)}$. Plus the remainder $\textcolor{g r e e n}{\left(2 x - 15\right)}$ divided by the divisor $\textcolor{b l u e}{\left({x}^{2} + 4\right)}$

So...

$\textcolor{red}{- 3 {x}^{2} + 3} + \frac{\textcolor{g r e e n}{2 x - 15}}{\textcolor{b l u e}{{x}^{2} + 4}}$