# How do you use synthetic division to divide 3x^3 - x^2 + 8x-10 by x-1?

Jun 21, 2015

This will be a bit hard to write out here, but let's see. I'm assuming that this is a typo. You may have meant $3 {x}^{3}$. If it was $3 {x}^{2}$, you wouldn't need synthetic division and you could just factor this to get $x = 1 , x = - 5$.

Assuming $3 {x}^{3}$... First, you let the coefficients of each degree be used in the division ($3 , - 1 , 8 , - 10$).

Then, dividing by $x - 1$ implies that you use $1$ in your upper left (if it was $x + 1$, put $- 1$). So, draw the bottom and right sides of a square, put $1$ inside it, and then write $\text{3" " -1" " 8" " -10}$ to the right.

$\text{1} | |$ $\text{3 " "-1 " "8 " "-10 }$
$+$
$\text{ " }$$- - - - -$

First, bring the $3$ down to the bottom, and multiply it by the $1$. Put that $3$ below $- 1$.

$\text{1} | |$ $\text{3 " "-1 " "8 " "-10 }$
$+$ $\text{ " " "" "3}$
$\text{ " }$$- - - - -$
$\text{ " " "3}$

$\text{1} | |$ $\text{3 " "-1 " "8 " "-10 }$
$+$ $\text{ " " "" "3}$
$\text{ " }$$- - - - -$
$\text{ " " "3" "" " 2}$

Repeat a few times once you've figured out the simple pattern ($\text{divisor"*"new sum}$, put result under next-lowest degree term, add to get another $\text{new sum}$, repeat).

$\text{1} | |$ $\text{3 " "-1 " "8 " "-10 }$
$+$ $\text{ " " "" "3" "" " 2}$
$\text{ " }$$- - - - -$
$\text{ " " "3" "" " 2" "" " 10}$

(All that really happened here was $1 \cdot 2 = 2$ and $8 + 2 = 10$.)

$\text{1} | |$ $\text{3 " "-1 " "8 " "-10 }$
$+$ $\text{ " " "" "3" "" " 2" " 10}$
$\text{ " }$$- - - - -$
$\text{ " " "3" "" " 2" "" " 10" " 0}$

(Then to finish it up, $1 \cdot 10 = 10$, and $- 10 + 10 = 0$.)

You know you can stop when you reach the far right and you have no spot left below the original dividend ($\text{3" " -1" " 8" " -10}$) to insert a product.

Since presumably you started with a cubic, the answer is a quadratic. Thus, you have:

$3 {x}^{2} + 2 x + 10 {x}^{0} + \left(r = 0\right)$

Indeed, if you multiply them together, you get the original back:

$\left(3 {x}^{2} + 2 x + 10\right) \left(x - 1\right) = 3 {x}^{3} - 3 {x}^{2} + 2 {x}^{2} - 2 x + 10 x - 10 = 3 {x}^{3} - {x}^{2} + 8 x - 10$

So the answer would be $3 {x}^{2} + 2 x + 10$. In a general case, you could write it like this:

Let $f \left(x\right) = a {x}^{3} + b {x}^{2} + c x + d$ and $g \left(x\right) = x - h$. If the solution is $h \left(x\right)$, and a remainder is $r \left(x\right)$, then:

$h \left(x\right) = f \frac{x}{g} \left(x\right) + \frac{r \left(x\right)}{g \left(x\right)}$

So you would have:
$h \left(x\right) = \text{answer} = \frac{3 {x}^{3} - {x}^{2} + 8 x - 10}{x - 1} \textcolor{b l u e}{= 3 {x}^{2} + 2 x + 10 + \frac{r \left(x\right)}{x - 1}}$

where $r \left(x\right) = 0$.

Had you meant $3 {x}^{2} - {x}^{2} + 8 x - 10$, you could have just done the quadratic equation. Or, factor.

$3 {x}^{2} - {x}^{2} + 8 x - 10 = 2 {x}^{2} + 8 x - 10 \implies \left(2 x - 2\right) \left(x + 5\right) = 0$
$\implies x = 1 , x = - 5$