\documentclass[12pt]{article} \setlength{\topmargin}{-1in} \setlength{\oddsidemargin}{-.3in} \setlength{\textheight}{9in} \setlength{\textwidth}{7.2in} \usepackage{amsmath} \usepackage{amsfonts} \usepackage{amsthm} \usepackage{amstext} \usepackage{amssymb} % define some shortcuts for N,Z,Q,R,C,H \newcommand{\N}{\mathbb{N}} \newcommand{\Z}{\mathbb{Z}} \newcommand{\Q}{\mathbb{Q}} \newcommand{\R}{\mathbb{R}} \newcommand{\C}{\mathbb{C}} \newcommand{\HH}{\mathbb{H}} % define some short cuts for other things % arrows \newcommand{\ra}{\rightarrow} \newcommand{\Ra}{\Rightarrow} \newcommand{\lra}{\longrightarrow} \newcommand{\Lra}{\Longrightarrow} \newcommand{\la}{\leftarrow} \newcommand{\La}{\Leftarrow} \newcommand{\lla}{\longleftarrow} \newcommand{\Lla}{\Longleftarrow} \newcommand{\da}{\downarrow} \newcommand{\Da}{\Downarrow} \newcommand{\ua}{\uparrow} \newcommand{\Ua}{\Uparrow} \newcommand{\actson}{\curvearrowright} % mbox stuff \newcommand{\tr}{\mbox{tr}} \newcommand{\isom}{\mbox{Isom}} \newcommand{\id}{\mbox{id}} \newcommand{\const}{\mbox{const}} \newcommand{\diag}{\mbox{diag}} \newcommand{\dist}{\mbox{dist}} % others \newcommand{\sm}{\setminus} \newcommand{\lie}{\mathfrak} % theorems and such \newtheorem{problem}{Problem} \newtheorem{definition}{Definition}[section] \newtheorem{lemma}[definition]{Lemma} \newtheorem{property}[definition]{Property} \newtheorem{theorem}[definition]{Theorem} \newtheorem{corollary}[definition]{Corollary} \newtheorem{remark}[definition]{Remark} \newtheorem{proposition}[definition]{Proposition} \newtheorem{question}[definition]{Question} \begin{document} \title{Factoring and Division} \date{} \maketitle Using the quadratic formula, we can solve any quadratic equation. How do we go about solving an polynomial equation that is not a quadratic? To do this we need to learn (or re-learn, or review) division of polynomials. Lets divide using long division, we'll do this in class. \[ \frac{3x^3-19x^2+22x+24}{x-4} \] The thing to notice is that we get remainder 0. This means that $x-4$ is a factor of $3x^3-19x^2+22x+24$. Using the division we see that \[ 3x^3-19x^2+22x+24=(x-4)(3x^2-7x-6) \] Now if we want to factor this more, we only have to worry about $3x^2-7x-6$. This is the idea: to slowly factor the polynomial one factor at at time. Each step will will have to worry about an easier polynomial. There are only 2 hard things: the long division (we'll make it easier) and what do we divide by, (where did I come up with $x-4$). To make the division easier, we will use synthetic division. Heres how its done. In our above example, we want to divide by $x-4$. Notice that if $x=4$ then this is zero. So we'll think of dividing by $4$. Also we don't care about the $x$'s. The $x$'s are really just place holders, kind of like when we took away the $x,y,z$ when we solved systems of equations. So here is our setup: \[ \begin{array}{rrrrrr} 4 & \vline & 3 & -19 & 22 & 24 \\ \\ \hline \\ \end{array} \] Do you see where everything came from? Now the process is: \begin{enumerate} \item Bring the 3 down below the long line. \item \label{goto} Multiply the 4 by the 3 and put the 12 just below the -19. \item Add the -19 and the 12 and put the -7 below the long line. \item Goto step~\ref{goto} and repeat. \end{enumerate} This gives: % \[ \begin{array}{rrrrrr} 4 & \vline & 3 & -19 & 22 & 24 \\ & & & 12 & -28 & -24 \\ \hline & & 3 & -7 & -6 & \vline \;\;\;\; 0 \end{array} \] See how this matches up with the long division. Notice how I put a line before the zero -- its the remainder. See how this helps factor the polynomial? Notice that if the remainder is not zero, all that we've learned is that what we divided by is not a factor. Here are some to try. Do synthetic division. If the remainder is zero then factor the first polynomial. If you do factor it, you should check that it really works by multiplying back out. % \begin{enumerate} \item Divide $3x^3-11x^2-24x+20$ by $x-2$. \item Divide $3x^3-11x^2-24x+20$ by $x+2$. \item Divide $3x^3-11x^2-24x+20$ by $x-7$ \item Divide $x^3-7x-6$ by $x-3$. The $x^2$ term is $0$, so make sure you put a $0$ as a place holder. \item Divide $x^3-7x-6$ by $x+1$. \item Divide $x^3-7x-6$ by $x+2$. \item Divide $15x^4+2x^3-101x^2+80x-12$ by $x+2$. See how easy it is to divide big polynomials? \item \label{one} Fractions are a bit harder but not too bad: Divide $15x^4+2x^3-101x^2+80x-12$ by $x-\frac{1}{5}$. \item \label{two} Divide $15x^3+5x^2-100x+60$ by $x-2$. \item \label{three} Divide $15x^2+35x-30$ by $x+3$ \end{enumerate} The ones that you factored, you may be able to completely factor. The last three~(\ref{one},\ref{two},\ref{three}) are related. Do the division and see if you can see how they are related. Use this to solve the equation, by first factoring the left side: \[ 15x^4+2x^3-101x^2+80x-12 =0 \] Do you see how easy synthetic division is? Do you see how division and factoring are related? Unfortunately, there are only a couple of problems with synthetic division. The biggest one is that we can only divide by things like $x-a$ where $a$ is some number. What if we wanted to divide \[ \frac{x^3-3x^2+6x-4}{3x^2+2} \] We would have to use long division to do this since we are dividing by a polynomial with an $x^2$ in it. Heres another one that won't work: \[ \frac{x^3-3x^2+6x-4}{3x+2} \] In this case the coefficient of $x$ is 3 and we need it to be a 1. This won't be a real problem though. Another problem is: how do we decide what to divide by. Lets do an example to see if we can guess what to divide by. Lets solve the equation $10x^2+x-21=0$. After a bit of work, the left hand side factors to: $(2x+3)(5x-7)=0$. This means that our solutions are $x=-\frac{3}{2}$ or $x=\frac{7}{5}$. Notice that the numerators are both factors of 21 (the last coefficient) and the denominators are factors of 10 (the first coefficient). This is how it works. We look for solutions that are factors of the last coefficient divided by factors of the first coefficient. Lets try and solve the equation below. We need to first factor the left hand side. \[ 5x^3+7x^2-28x-12=0 \] The solutions that we look for are factors of 12 divided by factors of 5. The possibilities are: $\{ \pm 12, \pm 6, \pm 4, \pm 3, \pm 2, \pm 1, \pm \frac{12}{5}, \pm \frac{6}{5}, \pm \frac{4}{5}, \pm \frac{3}{5}, \pm \frac{2}{5}, \pm \frac{1}{5} \}$. Notice that the factors of 5 are 5 and 1. This looks like a long list, but synthetic division is easy so we can just try them all until we get one that has a zero remainder. We try the ones without the fractions first because they are easier. Since I know the answer, we'll pretend that we tried a bunch unsuccessfully (meaning we get a remainder that is not zero) and then we finally try 2. (This would be dividing by $x-2$.) % \[ \begin{array}{rrrrrr} 2 & \vline & 5 & 7 & -28 & -12 \\ & & & 10 & 34 & 12 \\ \hline & & 5 & 17 & 6 & \vline \;\;\;\; 0 \end{array} \] We get a remainder of 0, so we can factor our left hand side to get \[ (x-2)(5x^2+17x+6)=0 \] To finish solving, we need to factor the $5x^2+17x+6$. We can either do this by our old techniques or use the new ones. Our possibilities are $\{\pm 6, \pm 3, \pm 2, \pm 1, \pm \frac{6}{5}, \pm \frac{3}{5}, \pm \frac{2}{5}, \pm \frac{1}{5} \}$. When we set up the synthetic division, we can even work below the work that we did before. Lets divide by $-\frac{2}{5}$, so we are really testing $x+\frac{2}{5}$. % \[ \begin{array}{rrrrrr} 2 & \vline & 5 & 7 & -28 & -12 \\ & & & 10 & 34 & 12 \\ \hline -\frac{2}{5} & & 5 & 17 & 6 & \vline \;\;\;\; 0 \\ & & & -2 & -6 \\ \hline & & 5 & 15 & \vline \;\;\; 0 \\ \end{array} \] This gives: \[ (x-2)(x+\frac{2}{5})(5x+15)=0 \] So now we can solve this and get that $x=2$, $x=-\frac{2}{5}$ and $x=-3$. Here are some equations for you to solve. Remember that before you can factor you need to get 0 on one side. \begin{enumerate} \item $5x^3-19x^2=-16x-4$ \item $7x^2-x=7-x^3$ \item $12x^3+8x-x-1=0$ \item $2x^4+15x^3+24x^2-11x=30$ \item $x^3+2x^2-10x=15$. This is a bit trickier. Remember that the quadratic formula is in your bag of tricks \item $42x^3+11x^2-38x-10=5$. Maybe try $-\frac{5}{6}$ first. \end{enumerate} \section*{More ways to solve harder equations} The synthetic division methods work great until we run into a nasty equation that doesn't factor, or has some nasty decimals. In these cases perhaps the best thing to do is graph the equation. I'll run through what to do, but these are best done on a graphing calculator or on a computer. If you don't have access to either of these, these next examples will be more difficult. Lets solve $x^5+2x^4-x^3+5x+x-3=0$. If we graph the function $y=x^5+2x^4-x^3+5x+x-3$, then we want to find the spots where $y=0$. These are the $x$-intercepts. So we can look for these on the graph. I'm not going to go into detail on this but for this equation, you should find $x$-intercepts around the $x$-values of $-2.87$,$-.76$, and $.65$. One final method to solving a polynomial equation, and perhaps the best (although the others have quite a bit of merit) is by using a computer polynomial equation solving program. This sounds fancy, and it is. But its not so fancy that you don't have access to this. Many calculators have this built in. My Hewlett-Packard does. So does the TI-85. I'm not sure about others. If your calculator does have this, you should be able to figure it out using your calculator manual and some of the examples in this handout. To show you how nice this is, this is what my hp does. If I want to solve $-4x^5+5x^4+x^3=-2x^2+1$, I first get a 0 on one side (as usual). This gives: $-4x^5+5x^4+x^3+2x^2-1=0$. I then enter $[-4\;\; 5\;\; 1\;\; 2\;\; 0\;\; -1]$, you should be able to see where this comes from. I then hit the polynomial equation solver button and I get the answer: \begin{eqnarray*} \{(-.534286237328,0), (.543959886927,0), (-.165209905756,-.721348138311), \\ (-.165209905756,.721348138311), (1.57074616197,0) \} \end{eqnarray*} This is a list of the 5 solutions to the equation. This needs a little bit of deciphering, but if you remember complex numbers it should be ok. The first, second and third numbers represent real numbers and the middle two represent complex numbers. If your calculator has this capability, you should see what you get for this same equation. As a side note/story. When I was an undergraduate, taking a numerical methods class (solving equations with a computer), my professor talked about a fantastic program available on the computer that would solve any degree polynomial equation. He said that this program was so terrific that before we finished the semester, he wanted to make sure that everyone had a copy of this program. He even said that if we had to we should get a printed copy so that if we found ourselves in a situation where we needed it, we could enter the program by hand into a computer. (This was ridiculous, the program was huge, probably 50 pages long.) The point is that this is a powerful tool in solving equations. %\bibliographystyle{alpha} %\bibliography{mybib} \end{document}