# How do you use synthetic division to divide (4x^2 - 2x + 6) (2x - 3) ^ -1?

Oct 24, 2015

$\left(2 x + 2\right)$ with remainder $= 12$

#### Explanation:

$\left(4 {x}^{2} - 2 x + 6\right) \cdot {\left(2 x - 3\right)}^{- 1}$
is equivalent to $\left(4 {x}^{2} - 2 x + 6\right) \div \left(2 x - 3\right)$

Using synthetic division (note that there may be slight variations in the structure depending upon how this is being taught):

Dividend: $\textcolor{red}{4} {x}^{2} \textcolor{red}{- 2} x \textcolor{red}{+ 6}$
Divisor: $\textcolor{b l u e}{2} x \textcolor{b l u e}{- 3}$

{: (,"|",color(red)(4),color(red)(-2),color(red)(+6)), (,"|",,6,6), (,,"-----","-----","-----"), (/color(blue)(2),"|",4,4,12), (color(white)("XX")color(cyan)(+3),"|",2,2,) :}
Note the change in the sign of the constant term of the divisor.

"Bring down" the $4$
Divide by the $/ 2$ to get $2$
Multiply by $+ 3$ to get $6$

{: (,"|",4,-2,+6), (,"|",,6,), (,,"-----","-----","-----"), (/2,"|",4,,), (color(white)("XX")+3,"|",2,,) :}

Add the $- 2$ and $6$ to get $4$
Divide by the $/ 2$ to get $2$
Miultiply by $+ 3$ to get $6$

{: (,"|",4,-2,+6), (,"|",,6,6), (,,"-----","-----","-----"), (/2,"|",4,4,), (color(white)("XX")+3,"|",2,2,) :}

Add the $+ 6$ and $6$ to get the final remainder $12$

{: (,,x^2,x^1,x^0), (,"|",4,-2,+6), (,"|",,6,6), (,,"-----","-----","-----"), (/2,"|",4,4,color(green)((12))), (color(white)("XX")+3,"|",2,2,), (,,x^1,x^0,) :}
Note: the additional rows showing corresponding powers of $x$ are not really part of the synthetic division (but hopefully add some clarity).