# How can synthetic division be used to factor a polynomial?

Apr 11, 2015

Here is a reasonable Pre-Calculus example of synthetic division to illustrate the concept.

Let's say you had:

$2 {x}^{4} - 3 {x}^{3} - 5 {x}^{2} + 3 x + 8$

Like Joan said, there is a trial and error aspect to this.

Look at all the coefficients, and think about what a common factor might be.

• If you don't get a zero remainder, then the factor doesn't really work and you should try again.
• If the possible factors are all used up, maybe it isn't factorable.

Here, factors you might try include the ones that correspond to the fourth order coefficient ($2$) and the zeroth order coefficient ($8$).

• $8$ has factors of $1 , 2 , 4$, and $8$.
• $2$ has factors of $1$ and $2$.

So the possible factors might be said to be $\pm \frac{p}{q}$, where $p$ consists of the factors of the zeroth degree coefficient, and $q$ consists of the factors of the highest degree coefficient.

You can therefore have factors of:

$\pm \left[1 , 2 , 4 , 8 , \frac{1}{2}\right]$

So you can try all of these ($\frac{2}{2}$, $\frac{4}{2}$, and $\frac{8}{2}$ are duplicates). Recall that if $- a$ is used as what is written in the synthetic division process on the left corner, it corresponds to $x + a$.

We will use $- 1$ here. I tend to try $1$ and $- 1$ first, and go up in value, and try the fractions last.

$\underline{- 1 |} \text{ "2" "-3" "-5" "" "3" "" } 8$

Drop down the $2$, and multiply by the $- 1$ to get $- 2$.

$\underline{- 1 |} \text{ "2" "-3" "-5" "" "3" "" } 8$
$\underline{\text{ "" "" "" "-2" "" "" "" "" "" "" "" }}$
$\text{ "" } \textcolor{w h i t e}{.} 2$

Add $- 3$ and $- 2$, then multiply the resultant $- 5$ by $- 1$ again.

$\underline{- 1 |} \text{ "2" "-3" "-5" "" "3" "" } 8$
$\underline{\text{ "" "" "" "-2" "" "5" "" "" "" "" "" }}$
$\text{ "" "color(white)(.)2" } - 5$

Repeat until you're done.

Add $- 3$ and $- 2$, then multiply the resultant $- 1$ by $- 1$ again.

$\underline{- 1 |} \text{ "2" "-3" "-5" "" "3" "" } 8$
$\underline{\text{ "" "" "" "-2" "" "5" "color(white)(.)" "0color(white)(.)" "-3" }}$
$\text{ "" "color(white)(.)2" "-5" "" "0" "" "3" "" } 5$

Your answer here happens to be this, where 2 corresponds to $2 {x}^{3}$, since you divided a fourth-order polynomial by a first-order polynomial.

So, one way to express the result is:

$\frac{2 {x}^{4} - 3 {x}^{3} - 5 {x}^{2} + 3 x + 8}{x + 1}$

$= \textcolor{b l u e}{{\overbrace{2 {x}^{3} - 5 {x}^{2} + 0 x + 3}}^{\text{Quotient Term" + overbrace(5/(x+1))^"Remainder Term}}}$

where the $\frac{5}{x + 1}$ was written by saying that the last value below the horizontal bar (below $- 2 , 5 , 0 , - 3$), being $5$, is divided by the $x \pm a$ equation such that $x \pm a = 0$. So, $x + 1$ indicates that the factor we have just used is $- 1$.

(Naturally, if the remainder is $0$, you do not have the remainder fraction at the end.)

Apr 11, 2015

Synthetic division is usually used to find factors of polynomials of degree 3 that can't be factored by grouping. Synthetic division is also used to factor polynomials of degree 4 or higher. Generally you use synthetic division if no other methods will work since there is a trial and error aspect to using it. That said, here's the process... Find the possible rational factors of the polynomial by looking at the factors of the constant term over the factors of the lead coefficient.

The possible rational factors are what you should try, one at a time, using synthetic division with the polynomial to be factored. So, 2x^4+ ... -9 would have possible factors of + or - 1, 3, 9, 1/2, 3/2, 9/2. I ALWAYS start by checking 1n then -1, then further into the list of possibles. I only check the fractions if none of the whole numbers worked.

-Joan Brown