\documentclass[12pt]{book} \usepackage{amsmath} \usepackage{latexsym} \newcounter{example} \newtheorem{thm}{Theorem}[section] \newtheorem{rem}[thm]{Remark} \newtheorem{exa}[example]{Example} \newtheorem{defn}[thm]{Definition} \begin{document} \chapter{Exponents and Polynomials} \section{Factoring: Factor Quadratic Trinomials} Polynomials with one term are called {\it monomials}. Similarly, polynomials with two terms are called {\it binomials} and polynomials with three terms are called {\it trinomials}. In this section, we discuss two methods of factoring trinomials $ ax^2 + bx + c$, where $a$, $b$ and $c$ are integers. \bigskip {\bf Factoring trinomials by trial and error} \bigskip Suppose we can factor it as $ ax^2 + bx + c = (kx+l)(mx+n) $, where $k$, $l$, $m$ and $n$ are integers. Then \begin{eqnarray*} ax^2 + bx + c & = & (kx+l)(mx+n) \\ & = & (km)x^2 + (kn+lm) x + ln. \end{eqnarray*} Therefore, we have three equations $ km = a$, $ ln = c$ and $ kn+lm = b $ for four integer unknowns $k$, $l$, $m$ and $n$. It is not easy to solve them algebraically. Howerver, from these equations, we know $k$ and $m$ are factors of $a$ and $ km = a$, $l$ and $n$ are factors of $c$ and $ ln = c$. From this fact, we can list all possible pairs $kx+l$ and $mx+n$, then determine which one matches the middle term equation $ kn+lm = b $. When doing this, do not forget some of these integer may be negative. In order to save your time, we can assume $k$ is always positive. \begin{exa} {\rm Factor the polynomial:} $ x^2 - 2x - 15 $. \end{exa} \noindent {\bf SOLUTION:} Since $a=1$, $k$ and $m$ are factors of $1$ and $km=1$. Since $c=-15$, $l$ and $n$ are factors of $-15$ and $ln=-15$. The following table lists all possible factors: \bigskip \begin{tabular}{| c | c | c | c |} \hline \hbox{} & \hbox{} & \hbox{} & \hbox{} \\ {\bf Possible factors} & {\bf First term} & {\bf Middle term} & {\bf Last term} \\ \hbox{} & \hbox{} & \hbox{} & \hbox{} \\ $(x+1)(x-15) $ & $ 1x^2 $ & $ -14x $ & $ -15 $ \\ \hbox{} & \hbox{} & \hbox{} & \hbox{} \\ $(x-1)(x+15) $ & $ 1x^2 $ & $ +14x $ & $ -15 $ \\ \hbox{} & \hbox{} & \hbox{} & \hbox{} \\ $(x+3)(x-5) $ & $ 1x^2 $ & $ -2x $ & $ -15 $ \\ \hbox{} & \hbox{} & \hbox{} & \hbox{} \\ $(x-3)(x+5) $ & $ 1x^2 $ & $ +2x $ & $ -15 $ \\ \hbox{} & \hbox{} & \hbox{} & \hbox{} \\ \hline \end{tabular} \bigskip We find that the third factoring matches. Therefore, the answer is $x^2 -2x -15 = (x+3)(x-5). \hskip .2in \Box $ \begin{exa} {\rm Factor the polynomial:} $ x^2 + x - 12 $. \end{exa} \noindent {\bf SOLUTION:} Since $a=1$, $k$ and $m$ are factors of $1$ and $km=1$. Since $c=-12$, $l$ and $n$ are factors of $-12$ and $ln=-12$. The following table lists all possible factors: \bigskip \begin{tabular}{| c | c | c | c |} \hline \hbox{} & \hbox{} & \hbox{} & \hbox{} \\ {\bf Possible factors} & {\bf First term} & {\bf Middle term} & {\bf Last term} \\ \hbox{} & \hbox{} & \hbox{} & \hbox{} \\ $(x+1)(x-12) $ & $ 1x^2 $ & $ -11x $ & $ -12 $ \\ \hbox{} & \hbox{} & \hbox{} & \hbox{} \\ $(x-1)(x+12) $ & $ 1x^2 $ & $ +11x $ & $ -12 $ \\ \hbox{} & \hbox{} & \hbox{} & \hbox{} \\ $(x+2)(x-6) $ & $ 1x^2 $ & $ -4x $ & $ -12 $ \\ \hbox{} & \hbox{} & \hbox{} & \hbox{} \\ $(x-2)(x+6) $ & $ 1x^2 $ & $ +4x $ & $ -12 $ \\ \hbox{} & \hbox{} & \hbox{} & \hbox{} \\ $(x+3)(x-4) $ & $ 1x^2 $ & $ -1x $ & $ -12 $ \\ \hbox{} & \hbox{} & \hbox{} & \hbox{} \\ $(x-3)(x+4) $ & $ 1x^2 $ & $ +1x $ & $ -12 $ \\ \hbox{} & \hbox{} & \hbox{} & \hbox{} \\ \hline \end{tabular} \bigskip We find that the last factoring matches. Therefore, the answer is $x^2 + x -12 = (x-3)(x+4). \hskip .2in \Box $ \begin{exa} {\rm Factor the polynomial:} $ 3x^2 + 5x - 2 $. \end{exa} \noindent {\bf SOLUTION:} Since $a=3$, $k$ and $m$ are factors of $3$ and $km=3$. Therefore, there are two answers for $k$ and $m$: $ \{ k=1, m=3 \} $ and $ \{ k=3, m=1 \} $. But the second one cannot provide anything new, because we can commute two binomials to obtain the first one. Thus, we can assume $k=1$ and $m=3$ without loss of generality. Since $c=-2$, $l$ and $n$ are factors of $-2$ and $ln=-2$. The following table lists all possible factors: \bigskip \begin{tabular}{| c | c | c | c |} \hline \hbox{} & \hbox{} & \hbox{} & \hbox{} \\ {\bf Possible factors} & {\bf First term} & {\bf Middle term} & {\bf Last term} \\ \hbox{} & \hbox{} & \hbox{} & \hbox{} \\ $(x+1)(3x-2) $ & $ 3x^2 $ & $ +1x $ & $ -2 $ \\ \hbox{} & \hbox{} & \hbox{} & \hbox{} \\ $(x-1)(3x+2) $ & $ 3x^2 $ & $ -1x $ & $ -2 $ \\ \hbox{} & \hbox{} & \hbox{} & \hbox{} \\ $(x+2)(3x-1) $ & $ 3x^2 $ & $ +5x $ & $ -2 $ \\ \hbox{} & \hbox{} & \hbox{} & \hbox{} \\ $(x-2)(3x+1) $ & $ 3x^2 $ & $ -5x $ & $ -2 $ \\ \hbox{} & \hbox{} & \hbox{} & \hbox{} \\ \hline \end{tabular} \bigskip We find that the third factoring matches. Therefore, the answer is $3x^2 + 5x -2 = (x+2)(3x-1). \hskip .2in \Box $ \begin{exa} {\rm Factor the polynomial:} $ x^2 - 8x + 5 $. \end{exa} \noindent {\bf SOLUTION:} Since $a=1$, $k$ and $m$ are factors of $1$ and $km=1$. Since $c=5$, $l$ and $n$ are factors of $5$ and $ln=5$. The following table lists all possible factors: \bigskip \begin{tabular}{| c | c | c | c |} \hline \hbox{} & \hbox{} & \hbox{} & \hbox{} \\ {\bf Possible factors} & {\bf First term} & {\bf Middle term} & {\bf Last term} \\ \hbox{} & \hbox{} & \hbox{} & \hbox{} \\ $(x+1)(x+5) $ & $ x^2 $ & $ +6x $ & $ 5 $ \\ \hbox{} & \hbox{} & \hbox{} & \hbox{} \\ $(x-1)(x-5) $ & $ x^2 $ & $ -6x $ & $ 5 $ \\ \hbox{} & \hbox{} & \hbox{} & \hbox{} \\ \hline \end{tabular} \bigskip We find that no one matches. Therefore, the trinomial $x^2 -8x + 5$ is not factorable relative to integers. Although, it can be factored in reals. But this is not what we want to discuss in this section. \hskip .2in $ \Box $ Some readers may be scared by the long tables in these examples. You are not along. Actually, the authors of this book had the same feeling when they were new learners of the topic. But after several (about ten) examples, most experirnced learners can find the correct answer without listing all factorings. Fortunately, there exists another method of factoring trinomials. The reader can decide which one is better for him (her). \bigskip {\bf Factoring trinomials by grouping} \bigskip We start this method by an easy example: \begin{exa} {\rm Factor the polynomial:} $ 3x^2 + 2x -12x - 8$. \end{exa} \noindent {\bf SOLUTION:} \begin{eqnarray*} 3x^2 + 6x - 4x - 8 & = & (3x^2 + 6x) - (4x + 8) \\ & = & 3x(x + 2) - 4(3x + 2) \\ & = & (x + 2)(3x - 4) \hskip .2in \Box \end{eqnarray*} But, the question usually is not given this way. It is given like this: factor trinomial $ 3x^2 + 2x - 8$. Splitting the middle term $ 2x = 6x - 4x $ is essencial for factoring. However, there exist thousands splittings: $ 2x = 6x - 4x $, $ 2x = 10x - 8x $, $ 2x = x + x $, $ \cdots $. Which one can help us factor? Here are the steps we have to follow in order to factor the trinomial $ ax^2 + bx + c$. {\bf Step 1:} Form the product $ac$ (with signs). {\bf Step 2:} Find a pair of integers, whose product is $ac$ and sum is $b$ (with signs). {\bf Step 3:} Rewrite the trinomial by splitting the middle term accordingly. {\bf Step 4:} Factor by grouping. Let us redo the above example by this new method. \begin{exa} {\rm Factor the polynomial:} $ 3x^2 + 2x - 8$. \end{exa} \noindent {\bf SOLUTION:} $ac = -24$. \bigskip \begin{tabular}{c c c c c c } \multicolumn{4}{c}{\bf Pair} && \multicolumn{1}{c}{\bf Sum} \\ 1 && -24 &&& 1 +(-24) = -23 \\ -1 && 24 &&& -1 +(24) = 23 \\ 2 && -12 &&& 2 +(-12) = -10 \\ -2 && 12 &&& -2 + 12 = 10 \\ 3 && -8 &&& 3 +(-8) = -5 \\ -3 && 8 &&& -3 + 8 = 5 \\ 4 && -6 &&& 4 +(-6) = -2 \\ -4 && 6 &&& -4 + 6 = 2 \\ \end{tabular} \bigskip The last pair matches. Then we can rewrite accordingly. \begin{eqnarray*} 3x^2 + 2x - 8 & = & 3x^2 - 4x + 6x - 8 = (3x^2 - 4x) + (6x - 8) \\ & = & x(3x - 4) + 2(3x-4) = (3x-4)(x+2) \hskip .2in \Box \end{eqnarray*} In the above example, the table is still too long. We can make it shorter. Notice that the first pair $1$ and $-24$ whose sum $-23$ does not match the the coeficient $2$ of the middle term. Then the second pair $-1$ and $24$ whose sum $23$, which is the opposite of $-23$, does not match the coeficient $2$ of the middle term either. Therefore, we don't have to try the second pair. Similarly, after trying the third pair, we can ignore the forth pair. and so on. Let us redo the question again. \begin{exa} {\rm Factor the polynomial:} $ 3x^2 + 2x - 8$. \end{exa} \noindent {\bf SOLUTION:} $ac = -24$. \bigskip \begin{tabular}{c c c c c c } \multicolumn{4}{c}{\bf Pair} && \multicolumn{1}{c}{\bf Sum} \\ 1 && -24 &&& 1 +(-24) = -23 \\ 2 && -12 &&& 2 +(-12) = -10 \\ 3 && -8 &&& 3 +(-8) = -5 \\ 4 && -6 &&& 4 +(-6) = -2 \\ -4 && 6 &&& -4 + 6 = 2 \\ \end{tabular} \bigskip The last pair matches. Then we can rewrite accordingly. \begin{eqnarray*} 3x^2 + 2x - 8 & = & 3x^2 - 4x + 6x - 8 = (3x^2 - 4x) + (6x - 8) \\ & = & x(3x - 4) + 2(3x-4) = (3x-4)(x+2) \hskip .2in \Box \end{eqnarray*} Notice that the absolute values of these sums is decreasing. After finding the first sum absolute value $23$ is far away from the coeficient $2$ of the middle term. we can skip the second pair, even the third pair. This fact also enable us to save time. Let us try more examples. \begin{exa} {\rm Factor the polynomial:} $ 15x^2 + 11x - 12$. \end{exa} \noindent {\bf SOLUTION:} $ac = -180$. \bigskip \begin{tabular}{c c c c c c } \multicolumn{4}{c}{\bf Pair} && \multicolumn{1}{c}{\bf Sum} \\ 1 && -180 &&& 1 +(-180) = -179 \\ 2 && -90 &&& 2 +(-90) = -88 \\ 3 && -60 &&& 3 +(-60) = -57 \\ 4 && -45 &&& 4 +(-45) = -41 \\ 5 && -36 &&& 5 +(-36) = -31 \\ 6 && -30 &&& 6 +(-30) = -24 \\ 9 && -20 &&& 9 +(-20) = -11 \\ -9 && 20 &&& -9 + (20) = 11 \\\ \end{tabular} \bigskip The last pair matches. Then we can rewrite accordingly. \begin{eqnarray*} 15x^2 + 11x - 12 & = & 15x^2 + 20x - 9x - 12 = (15x^2 + 20x) - (9x + 12) \\ & = & 5x(3x + 4) - 3(3x + 4) = (3x + 4)(5x - 3) \hskip .2in \Box \end{eqnarray*} We redo the example by skipping several pairs. \begin{exa} {\rm Factor the polynomial:} $ 15x^2 + 11x - 12$. \end{exa} \noindent {\bf SOLUTION:} $ac = -180$. \bigskip \begin{tabular}{c c c c c c } \multicolumn{4}{c}{\bf Pair} && \multicolumn{1}{c}{\bf Sum} \\ 1 && -180 &&& 1 +(-180) = -179 \\ \multicolumn{5}{c}{\bf Far away. Jump} \\ 6 && -30 &&& 6 +(-30) = -24 \\ 9 && -20 &&& 9 +(-20) = -11 \\ -9 && 20 &&& -9 + (20) = 11 \\\ \end{tabular} \bigskip The last pair matches. Then we can rewrite accordingly. \begin{eqnarray*} 15x^2 + 11x - 12 & = & 15x^2 + 20x - 9x - 12 = (15x^2 + 20x) - (9x + 12) \\ & = & 5x(3x + 4) - 3(3x + 4) = (3x + 4)(5x - 3) \hskip .2in \Box \end{eqnarray*} These two methods are also useful for polynomials with two variables $ax^2 + bxy + cy^2$. We only provide one example. \begin{exa} {\rm Factor the polynomial:} $ 20x^2 -23xy + 6y^2 $. \end{exa} \noindent {\bf SOLUTION:} $ac = 120$. \bigskip \begin{tabular}{c c c c c c } \multicolumn{4}{c}{\bf Pair} && \multicolumn{1}{c}{\bf Sum} \\ 1 && 120 &&& 1 + 120 = 121 \\ \multicolumn{5}{c}{\bf Far away. Jump} \\ 6 && 20 &&& 6 + 20 = 26 \\ 8 && 15 &&& 8 + 15 = 23 \\ -8 && -15 &&& -8 +(-15)= -23 \\ \end{tabular} \bigskip The last pair matches. Then we can rewrite accordingly. \begin{eqnarray*} 20x^2 -23xy + 6y^2 & = & 20x^2 -8xy - 15xy + 6y^2 = (20x^2 -8xy) - (15xy - 6y^2) \\ & = & 4x(5x-2y) - 3y(5x-2y) = (5x-2y)(4x-3y) \hskip .2in \Box \end{eqnarray*} \section{Factoring: By Formulas} We mention five special multiplication formulas in the previous section. Now we discuss their application for factoring. Let us list these formulas again: \begin{tabular}{p{2.5in} p{2.5in} } $ A^2 - B^2 = (A-B)(A+B) $ & Difference of squares \\ $ A^2 + 2AB + B^2 = (A+B)^2 $ & Square of sum \\ $ A^2 - 2AB + B^2 = (A-B)^2 $ & Square of difference \\ $ A^3 - B^3 = (A-B)(A^2 + AB + B^2) $ & Difference of cubes \\ $ A^3 + B^3 = (A+B)(A^2 - AB + B^2) $ & Sum of cubes \\ \end{tabular} \bigskip When use them to factor polynomials, pay attention for the follwing important points: (i) Don't forget factor GCF first, if GCF is not $1$. (ii) Determine which formula you want to use and rewrite some terms accordingly. (iii) Identify what is your $A$ and what is your $B$. Let us start from the easy examples. \begin{exa} {\rm Factor the polynomial:} $ x^2 - y^2 $. \end{exa} \noindent {\bf SOLUTION:} GCF is $1$, use the difference of squares formula, $x$ is $A$, $y$ is $B$. We can factor it immediately. $ x^2 - y^2 = (x-y)(x+y) \hskip .2in \Box $ \begin{exa} {\rm Factor the polynomial:} $ x^2 - 4 $. \end{exa} \noindent {\bf SOLUTION:} GCF is still $1$, still use the difference of squares formula, but we have to rewrite $4=2^2$. Therefore, $x$ is $A$, $2$ is $B$. We can factor it immediately. $ x^2 - 4 = x^2 - 2^2 = (x-2)(x+2) \hskip .2in \Box $ \begin{exa} {\rm Factor the polynomial:} $ 2x^2 - 18y^2 $. \end{exa} \noindent {\bf SOLUTION:} GCF is $2$, still use the difference of squares formula, but after factoring GCF $2$, the coefficient $9$ bothers us, we have to rewrite $9y^2 = 3^2 y^2 = (3y)^2 $. Therefore, $x$ is $A$, $(3y)$ is $B$. We can factor it immediately. $ 2x^2 - 18y^2 = 2[x^2 - (3y)^2] = 2(x-3y)(x+3y) \hskip .2in \Box $ After you are familiar with those three points, we can offer more difficult examples: \begin{exa} {\rm Factor the polynomial:} $ 8x^2 - 24xy + 18y^2 $. \end{exa} \noindent {\bf SOLUTION:} \begin{eqnarray*} 8x^2 - 24xy + 18y^2 & = & 2 (4x^2 -12xy + 9y^2) = 2 [(2x)^2 - 2 (2x)(3y) + (3y)^2] \\ & = & 2 (2x - 3y)^2 \hskip .2in \Box \end{eqnarray*} \begin{exa} {\rm Factor the polynomial:} $ 2x^2 + 2 xy + \frac{1}{2} y^2 $. \end{exa} \noindent {\bf SOLUTION:} \begin{eqnarray*} 2x^2 + 2xy + \frac{1}{2} y^2 & = & \frac{1}{2} (4x^2 + 4xy + y^2) = \frac{1}{2} [(2x)^2 + 2(2x)y + y^2] \\ & = & \frac{1}{2} (2x + y)^2 \hskip .2in \Box \end{eqnarray*} \begin{exa} {\rm Factor the polynomial:} $ 27x^3 + 8y^3 $. \end{exa} \noindent {\bf SOLUTION:} \begin{eqnarray*} 27x^3 + 8y^3 & = & 3^3 x^3 + 2^3 y^3 = (3x)^3 + (2y)^3 \\ & = & (3x + 2y) (9x^2 + 6xy + 4y^2) \hskip .2in \Box \end{eqnarray*} \begin{exa} {\rm Factor the polynomial:} $ x^6 - 64 $. \end{exa} \noindent {\bf SOLUTION:} We have to use three formulas for this question. \begin{eqnarray*} x^6 - 64 & = & (x^3)^2 - 8^2 = (x^3 - 8)(x^3 + 8) \\ & = & (x^3 - 2^3)(x^3 + 2^3) = (x-2)(x^2 + 2x + 4) (x+2)(x^2 - 2x + 4) \hskip .2in \Box \end{eqnarray*} \begin{exa} {\rm Factor the polynomial:} $ x^2 - 10x + 25 - y^2 $. \end{exa} \noindent {\bf SOLUTION:} No formula has four terms. But the first three term is the square of difference, and the last one is $-y^2$. We have to use the square of difference formula first, then use difference of squares formula second for this question. \begin{eqnarray*} x^2 - 10x + 25 - y^2 & = & (x^2 - 10x + 25) - y^2 = (x-5)^2 - y^2 \\ & = & (x-5-y) (x-5+y) \hskip .2in \Box \end{eqnarray*} \begin{exa} {\rm Factor the polynomial:} $ 20x^2 - 45y^2 + 210y - 245$. \end{exa} \noindent {\bf SOLUTION:} \begin{eqnarray*} 20x^2 - 45y^2 + 210y - 245 & = & 5 (4x^2 - 9y^2 + 42y -49) = 5 [4x^2 - (9y^2 - 42y + 49)] \\ & = & 5 [(2x)^2 - ((3y)^2 - 2(3y)7 + 7^2)] = 5 [(2x)^2 - (3y -7)^2] \\ & = & 5 (2x - 3y + 7)(2x + 3y - 7) \hskip .2in \Box \end{eqnarray*} When factoring polynomials, rewriting skill is essential. Rewriting an expression is to change its form but keep them equal. From the above examples, the reader can realize how important it is. \end{document}