# How do I find the quotient and remainder using synthetic division?

Jul 7, 2016

Let's take this as an example:

$f \left(x\right) = {x}^{3} - {x}^{2} + x - 6$

To find factors, notice that $6$ has factors of $\pm 1 , \pm 2 , \pm 3 , \pm 6$. Pick one and try synthetic division on it, and if you pick the right one (meaning that it divides), it'll give a remainder of $0$.

I pick $2$, so I am assuming that $x - 2$ divides ${x}^{3} - {x}^{2} + x - 6$. This means the result should be of the form:

color(blue)((x^3 - x^2 + x - 6)/(x-2) = q(x) + r(x)

where $q \left(x\right)$ is the quotient and $r \left(x\right)$ is the remainder.

$\textcolor{w h i t e}{\left[\left(\textcolor{b l a c k}{2} , \textcolor{b l a c k}{|} , \textcolor{b l a c k}{1} , \textcolor{b l a c k}{- 1} , \textcolor{b l a c k}{1} , \textcolor{b l a c k}{- 6}\right) , \left(\textcolor{b l a c k}{-} , \textcolor{b l a c k}{\text{),color(black)("_"),color(black)("_"),color(black)("_"),color(black)("_")),(color(black)(""),color(black)(""),color(black)(""),color(black)(""),color(black)(""),color(black)(}}\right)\right]}$

If you don't have a term, use $0$. So if you had ${x}^{3} - 6$, use $\text{1 0 0 "-"6}$.

The general process is:

• Bring down the first coefficient.
• Multiply it by the test factor (in this case, $2$), and store the result under the next coefficient.
• Subtract the result from this coefficient.
• Repeat steps 2 and 3 until you've subtracted the result from the last coefficient.

The result is one degree lower.

$\implies \textcolor{w h i t e}{\left[\left(\textcolor{b l a c k}{2} , \textcolor{b l a c k}{|} , \textcolor{b l a c k}{1} , \textcolor{b l a c k}{- 1} , \textcolor{b l a c k}{1} , \textcolor{b l a c k}{- 6}\right) , \left(\textcolor{b l a c k}{-} , \textcolor{b l a c k}{\text{),color(black)("_"),color(black)(ul(2)),color(black)("_"),color(black)("_")),(color(black)(""),color(black)(""),color(black)(1),color(black)(-3),color(black)(""),color(black)(}}\right)\right]}$

$\text{ }$

$\implies \textcolor{w h i t e}{\left[\left(\textcolor{b l a c k}{2} , \textcolor{b l a c k}{|} , \textcolor{b l a c k}{1} , \textcolor{b l a c k}{- 1} , \textcolor{b l a c k}{1} , \textcolor{b l a c k}{- 6}\right) , \left(\textcolor{b l a c k}{-} , \textcolor{b l a c k}{\text{),color(black)("_"),color(black)(ul(2)),color(black)(ul(-6)),color(black)("_")),(color(black)(""),color(black)(""),color(black)(1),color(black)(-3),color(black)(7),color(black)(}}\right)\right]}$

$\text{ }$

$\implies \textcolor{w h i t e}{\left[\left(\textcolor{b l a c k}{2} , \textcolor{b l a c k}{|} , \textcolor{b l a c k}{1} , \textcolor{b l a c k}{- 1} , \textcolor{b l a c k}{1} , \textcolor{b l a c k}{- 6}\right) , \left(\textcolor{b l a c k}{-} , \textcolor{b l a c k}{\text{),color(black)("_"),color(black)(ul(2)),color(black)(ul(-6)),color(black)(ul(14))),(color(black)(""),color(black)(}} , \textcolor{b l a c k}{\setminus m a t h b f \left(1\right)} , \textcolor{b l a c k}{\setminus m a t h b f \left(- 3\right)} , \textcolor{b l a c k}{\setminus m a t h b f \left(7\right)} , \textcolor{b l a c k}{\setminus m a t h b f \left(- 20\right)}\right)\right]}$

So, our result is $\textcolor{b l u e}{{x}^{2} - 3 x + 7 - \frac{20}{x - 2}}$ for $x \ne 2$. What we have is:

$\setminus m a t h b f \left(q \left(x\right) = {x}^{2} - 3 x + 7\right)$
$\setminus m a t h b f \left(r \left(x\right) = - \frac{20}{x - 2}\right)$

If $r \left(x\right) = 0$, then $q \left(x\right)$ is a quotient that divides $f \left(x\right) = {x}^{3} - {x}^{2} + x - 6$.