3.4 Dividing Polynomials
p160 8–14, 17, 18, 21–37
If you remember doing long division, the process involved determining how many times one number went into another. That result would be placed on top, then multiplied by the divisor, subtracted from what you started with, and the whole process would repeat.
125
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4)500
4
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10
8
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20
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0
At the end, you would either get 0, meaning it divided evenly, or a remainder, an extra part that could not be broken down any further. In the example above we divided by 4, so possible remainders would be 3, 2, or 1.
We’re going to take the same process and apply it polynomials, which sounds worse than it actually is. I’ll include only one example below (thanks Wikipedia), but the whiteboard notes will have all the examples we go through in class
Example
Let’s divide $x^3-2x^2-4$ by $x-3$. Somethings to note before we start.
- Fill in empty terms with zeros. We are missing an $x$ term, but a place holder will benefit us.
- We’ll only look at the leading term your divisor, which is $x-3$ in this case.
Our first step has us looking to see how many times $x$ goes into $x^3$. You can determine that by dividing the two, and the result goes up top, typically above the term with the matching degree.
\[\begin{array}{l} {\color{White} x-3\ )\ x^3 - 2}x^2\\ x-3\ \overline{)\ x^3 - 2x^2 + 0x - 4}\\ {\color{White} x-3\ )\ } x^3 - 3x^2 \end{array}\]Now we multiply our result by the divisor, putting the result underneath. Make sure you line up terms with the same degree, because the next step is subtraction.
\[\begin{array}{l} {\color{White} x-3\ )\ x^3 - 2}x^2\\ x-3\ \overline{)\ x^3 - 2x^2 + 0x - 4}\\ {\color{White} x-3\ )\ } \underline{x^3 - 3x^2}\\ {\color{White} x-3\ )\ 0x^3} + {\color{White}}x^2 + 0x \end{array}\]It’s not displayed here, but I advise you to put parentheses around what you are subtracting and along with a minus sign in front, so $-(x^3-3x^2)$. A reminder that subtraction is happening is helpful.
Now the process will repeat itself. Divide $x^2$ and $x$, put the result up top, multiply, and subtract.
\[\begin{array}{r} x^2 + {\color{White}1}x {\color{White} {} + 3}\\ x-3\ \overline{)\ x^3 - 2x^2 + 0x - 4}\\ \underline{x^3 - 3x^2 {\color{White} {} + 0x - 4}}\\ +x^2 + 0x {\color{White} {} - 4}\\ \underline{+x^2 - 3x {\color{White} {} - 4}}\\ +3x - 4\\ \end{array}\]And then once more since we can still divide $3x$ by $x$.
\[\begin{array}{r} x^2 + {\color{White}1}x + 3\\ x-3\ \overline{)\ x^3 - 2x^2 + 0x - 4}\\ \underline{x^3 - 3x^2 {\color{White} {} + 0x - 4}}\\ +x^2 + 0x {\color{White} {} - 4}\\ \underline{+x^2 - 3x {\color{White} {} - 4}}\\ +3x - 4\\ \underline{+3x - 9}\\ +5 \end{array}\]We’ve hit a point where we can no longer divide. $5/x$ doesn’t reduce down anymore, so we have our remainder. Here’s how you can write your solution.
\[\frac{x^3-2x^2-4}{x-3} = x^2 + x + 3 + \frac{5}{x-3}\]The remainder might look strange since it’s likely you’ve never written that way before. Consider $5/2$. You’ll get a remainder of 1.
\[\frac{5}{2} = 2 + \frac{1}{2}\]The other way to write it is a bit cleaner since it eliminates the need for fractions. Multiply both sides by your divisor to get
\[x^3-2x^2-4 = (x-3)(x^2 + x + 3) + 5\]Maybe by looking at that version you might have figured out where we are going with this: we’re going to use this to help factor polynomials that otherwise look unfactorable.
The Remainder Theorem
Let’s take our result from above and plug 3 in for $x$. I’m choosing 3 on purpose since that will yield a zero in the binomial.
\[\begin{align*} (3)^3-2(3)^2-4 &= [(3)-3][(3)^2 + (3) + 3] + 5 \\ 27-18-4 &= (0)(15) + 5 \\ 5 &= 5 \end{align*}\]Written as function notation where $f$ is our original function, $q$ is thw quotient, $r$ the remainder, and we are dividing by $(x-k)$.
\[\begin{align*} \frac{f(x)}{x-k} &= q(x) + \frac{r}{x-k} \\ f(x) &= (x-k)q(x) + r \\ f({\color{red}k}) &= ({\color{red}k}-k)q(x) + r \\ f(k) &= r \\ \end{align*}\]The polynomial remainder theorem states that if a polynomial $f(x)$ is divided by $(x-k)$, then the remainder is $r=f(k)$.
OK, great. If we need to evaluate a polynomial at value $k$, we can instead just divide by $(x-k)$ and use the remainder instead. Except dividing isn’t exactly quick. So, what if we had a quicker way to divide?
Synthetic division
Long division is long, but repetitive, and if anything is repetitive, odds are we can find a quicker way to do it. That’s what synthetic division aims to do. Speed up the process of a dividing polynomials, but specifically only when dividing by something in the form $(x-k)$.
Below is the setup for $x^3-12x^2-42$ being divided by $x-3$. Inside the box are our coefficients, including a zero for the missing term. outside is $k$. Don’t forget the form is $(x-k)$, so watch your sign.
\[\begin{array}{cc} \begin{array}{r} \\ 3 \\ \end{array} & \begin{array}{|rrrr} \ 1 & -12 & 0 & -42 \\ & & & \\ \hline \end{array} \end{array}\]The first step is to simply drop the first coefficient down below the bar.
\[\begin{array}{cc} \begin{array}{r} \\ 3 \\ \\ \end{array} & \begin{array}{|rrrr} \color{blue}1 & -12 & 0 & -42 \\ & & & \\ \hline \color{blue}1 & & & \\ \end{array} \end{array}\]Then, multiply what you brought down by $k$ and write that under your next coefficient.
\[\begin{array}{cc} \begin{array}{r} \\ \color{grey}3 \\ \\ \end{array} & \begin{array}{|rrrr} 1 & -12 & 0 & -42 \\ & \color{brown}3 & & \\ \hline \color{blue}1 & & & \\ \end{array} \end{array}\]Add the pair and write it under the bar.
\[\begin{array}{cc} \begin{array}{c} \\ 3 \\ \\ \end{array} & \begin{array}{|rrrr} 1 & \color{green}-12 & 0 & -42 \\ & \color{green}3 & & \\ \hline 1 & \color{green}-9 & & \\ \end{array} \end{array}\]And then the process repeats. Multiply what’s now below the bar by $k$, write it under the next coefficient, add, multiply, add, multiply, …
Eventually, you’ll end up with
\[\begin{array}{cc} \begin{array}{c} \\ 3 \\ \\ \end{array} & \begin{array}{|rrrr} 1 & -12 & 0 & -42 \\ & 3 & -27 & -81 \\ \hline 1 & -9 & -27 & -123 \end{array} \end{array}\]What’s below the bar is our quotient and remainder. The last term is the remainder, and other terms are the coefficients in degree order. The solution to the example above is $q(x)=x^2-9x-27$ and $r(x)=-123$, or $x^2-9x-27-\frac{123}{x-3}$.
For another example, here’s an animated one curtesy of Wikipedia.
