# How do you divide (x^4+3x^3-12x-6 ) / ( x^3+2x^2-4 )?

Mar 30, 2016

$x + 1 - \frac{2 {x}^{2} + 8 x + 2 x}{{x}^{3} + 2 x - 4}$

#### Explanation:

Only divide using the most significant figures ie ?-:color(green)(x^3)

$\text{ "color(green)(x^3+2x^2-4)" |} \overline{{x}^{4} + 3 {x}^{3} - 12 x - 6}$

'~~~~~~~~~~~~~~~~~~~~~~~~~

$\textcolor{b l u e}{\text{Step 1 }} {x}^{4} / {x}^{3} = \textcolor{m a \ge n t a}{x}$

$\text{ } \textcolor{m a \ge n t a}{x}$
Write as: $\text{ "color(green)(x^3+2x^2-4)" |} \overline{{x}^{4} + 3 {x}^{3} - 12 x - 6}$

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b l u e}{\text{Step 2 }}$

$\textcolor{g r e e n}{{x}^{3} + 2 {x}^{2} - 4}$
underline(" "color(magenta)(x))" "->"Multiply"
""color(blue)(x^4+2x^3-4x)

Write as:
$\text{ } x$
$\text{ "color(green)(x^3+2x^2-4)" |} \overline{{x}^{4} + 3 {x}^{3} - 12 x - 6}$
$\text{ "underline(color(blue)(x^4+2x^3-4x))" "->} S u b t r a c t$
$\text{ "0+color(red)(x^3)-8x-6" bring down the 6}$
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

$\textcolor{b l u e}{\text{Step 3 }}$

$\frac{\textcolor{red}{{x}^{3}}}{\textcolor{g r e e n}{{x}^{3}}} = \textcolor{m a \ge n t a}{+ 1}$

Write as:
$\text{ } x \textcolor{m a \ge n t a}{+ 1}$
$\text{ "color(green)(x^3+2x^2-4)" |} \overline{{x}^{4} + 3 {x}^{3} - 12 x - 6}$
$\text{ "underline(x^4+2x^3-4x)" "->} S u b t r a c t$
$\text{ } 0 + \textcolor{red}{{x}^{3}} - 8 x - 6$

$\textcolor{g r e e n}{{x}^{3} + 2 {x}^{2} - 4}$
underline(" "color(magenta)(+1))" "->"Multiply"
""color(blue)(x^3+2x^2-4)
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

$\textcolor{b l u e}{\text{Step 4 }}$

Write as:
$\text{ } x + 1$
$\text{ "color(green)(x^3+2x^2-4)" |} \overline{{x}^{4} + 3 {x}^{3} - 12 x - 6}$
$\text{ "underline(x^4+2x^3-4x)" "->} S u b t r a c t$
$\text{ "0+x^3} - 8 x - 6$
$\text{ "underline(color(blue)(x^3+2x^2-4))" } \to S u b t r a c t$
These do not line up any more as we have an extra term. For convenience change the order. This can be sorted out later.
Write as:
$\text{ } x + 1$
$\text{ "color(green)(x^3+2x^2-4)" |} \overline{{x}^{4} + 3 {x}^{3} - 12 x - 6}$
$\text{ "underline(x^4+2x^3-4x)" "->} S u b t r a c t$
$\text{ "0+x^3} - 8 x - 6$
$\text{ "underline(color(blue)(x^3+0color(white)(.)-4+2x^2))" } \to S u b t r a c t$
$\text{ } 0 \textcolor{w h i t e}{.} - 8 x - 2 - 2 {x}^{2}$>>>>

The remainder is:$\text{ } - 2 {x}^{2} - 8 x - 2$ We can not divide any further as its most significant value is less than the ${x}^{3}$ in ${x}^{3} + 2 {x}^{2} - 4$

So we express it as $\frac{- 2 {x}^{2} - 8 x - 2}{{x}^{3} + 2 {x}^{2} - 4} = - \frac{2 {x}^{2} + 8 x + 2}{{x}^{3} + 2 x - 4}$
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

$\implies \frac{{x}^{4} + 3 {x}^{3} - 12 x - 6}{{x}^{3} + 2 x - 4} \text{ " =" } x + 1 - \frac{2 {x}^{2} + 8 x + 2}{{x}^{3} + 2 x - 4}$