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\begin{document}
\chapter{Exponents and Polynomials}
\section{Definition of Exponents}
When we study Geometry course, we take a square to find an area,
and take a cube to find a volumn.
The square and cube are special cases of \emph{exponents}.
In this section, we introduce the formal definition
of integer (positive , zero and negative) exponents,
and discuss their properties.
\bigskip
\begin{tabular}{|rcl|}\hline
\multicolumn{3}{|l|}{} \\
\multicolumn{3}{|l|}{\large \bf Definition} \\
\multicolumn{3}{|l|}{} \\
\multicolumn{3}{|l|}
{If $x$ is nonzero real number and $n$ is a positive integer,
then} \\
\multicolumn{3}{|l|}{} \\
$x^n$ & = & $ \underbrace{x \cdot x \cdots x}_{n \hbox{ times}} $ \\
\multicolumn{3}{|l|}{} \\
$x^{-n}$ & = & $\frac{1}{x^n}$ \\
\multicolumn{3}{|l|}{} \\
$x^0 $ & = & $1 $ \\
\multicolumn{3}{|l|}{} \\
\hline
\end{tabular}
\bigskip
In the exponent form $x^n$, $x$ is called the {\it base}
and $n$ is called the {\it power index} or {\it exponent}.
In this section, we will be concerned with the simplification
of expressions with exponents. We start from some simple
examples and make some generalizations in order to obtain
important properties.
\begin{exa}
{\rm Write the product $x^3 \cdot x^5$ with a single exponent.}
\end{exa}
\noindent {\bf SOLUTION:}
\begin{eqnarray*}
x^3 \cdot x^5 & = & (x \cdot x \cdot x) \cdot
(x \cdot x \cdot x \cdot x \cdot x) \\
& = & x \cdot x \cdot x \cdot x \cdot x \cdot x \cdot x \cdot x
= x^8 . \hskip .2in \Box
\end{eqnarray*}
\begin{exa}
{\rm Write the product $x^3 \cdot x^{-5}$ with a single exponent.}
\end{exa}
\noindent {\bf SOLUTION:}
\begin{eqnarray*}
x^3 \cdot x^{-5} & = & x^3 \cdot \frac{1}{x^5} =
\frac{x \cdot x \cdot x} {x \cdot x \cdot x \cdot x \cdot x} \\
& = & \frac{1}{ x \cdot x } = \frac{1} {x^2} = x^{-2}.
\hskip .2in \Box
\end{eqnarray*}
Therefore, we will have the first property for exponent:
$x^m \cdot x^n = x ^{m+n} $ for any nonzero $x$ and any interges
$m$ and $n$.
\begin{exa}
{\rm Write the quotient $\frac{x^3}{x^5}$ with a single exponent.}
\end{exa}
\noindent {\bf SOLUTION:}
\begin{eqnarray*}
\frac{x^3}{x^5} = \frac{x \cdot x \cdot x}
{x \cdot x \cdot x \cdot x \cdot x} = \frac{1} {x^2}
= x^{-2}.
\hskip .2in \Box
\end{eqnarray*}
\begin{exa}
{\rm Write the quotient $\frac{x^3}{x^{-5}}$ with a single exponent.}
\end{exa}
\noindent {\bf SOLUTION:}
\begin{eqnarray*}
\frac{x^3}{x^{-5}} = \frac{x^3}{\frac{1} {x^5} } =
(x^3) \cdot (x^5) = x^8.
\hskip .2in \Box
\end{eqnarray*}
Therefore, we will have the second property for exponent:
$ \frac{x^m} {x^n} = x ^{m-n} $ for any nonzero $x$ and any interges
$m$ and $n$.
\begin{exa}
{\rm Write the product $(x^3)^4$ with a single exponent.}
\end{exa}
\noindent {\bf SOLUTION:}
\begin{eqnarray*}
(x^3)^4 & = & (x \cdot x \cdot x) \cdot (x \cdot x \cdot x) \cdot
(x \cdot x \cdot x) \cdot (x \cdot x \cdot x) \\
& = & \underbrace{ x \cdot x \cdots x}_{12 \hbox{ times}} = x^{12}.
\hskip .2in \Box
\end{eqnarray*}
Therefore, we will have the third property for exponent:
$ (x^m)^n = x ^{mn} $ for any nonzero $x$ and any interges
$m$ and $n$.
\begin{exa}
{\rm Expand $ (3x)^4$ and then multiply.}
\end{exa}
\noindent {\bf SOLUTION:}
\begin{eqnarray*}
(3x)^4 & = & (3x) \cdot (3x) \cdot (3x) \cdot (3x) \\
& = & ( 3 \cdot 3 \cdot 3 \cdot 3) \cdot
(x \cdot x \cdot x \cdot x) = 3^4 \cdot x^4 = 81 x^4.
\hskip .2in \Box
\end{eqnarray*}
Therefore, we will have the forth property for exponent:
$ (x \cdot y)^n = x^n \cdot y^n $ for any nonzero $x$ and $y$,
and any interge $n$.
\begin{exa}
{\rm Expand and rewrite $ \left ( \frac{x}{y} \right ) ^3 $}
\end{exa}
\noindent {\bf SOLUTION:}
\begin{eqnarray*}
\left ( \frac{x}{y} \right ) ^3
& = & \left ( \frac{x}{y} \right ) \cdot
\left ( \frac{x}{y} \right ) \cdot
\left ( \frac{x}{y} \right ) =
\frac{ x \cdot x \cdot x }{ y \cdot y \cdot y} = \frac{x^3}{y^3}.
\hskip .2in \Box
\end{eqnarray*}
Therefore, we will have the fifth property for exponent:
$ \left ( \frac{x}{y} \right ) ^{n} = \frac{x^n}{y^n} $
for any nonzero $x$ and $y$,
and any interge $n$.
We can summarize these five properties in the following table.
\bigskip
\begin{tabular}{|lrcl|}
\hline
\multicolumn{4}{|l|}{} \\
\multicolumn{4}{|c|}{\large \bf Properties of Exponent} \\
\multicolumn{4}{|l|}{} \\
\multicolumn{4}{|l|}{If $x$, $y$ are nonzero and $m$, $n$ are integers } \\
\multicolumn{4}{|l|}{} \\
I. & $ x^n \cdot x^m $ & = & $ x^{m+n}$ \\
\multicolumn{4}{|l|}{} \\
II. & $ \frac{x^m} {x^n} $ & = & $ x^{m-n}$ \\
\multicolumn{4}{|l|}{} \\
III. & $ (x^n)^m $ & = & $ x^{mn}$ \\
\multicolumn{4}{|l|}{} \\
IV. & $ (x \cdot y)^n $ & = & $ x^n \cdot y^n $ \\
\multicolumn{4}{|l|}{} \\
V. & $ \left ( \frac{x}{y} \right ) ^{n} $ & = &
$ \frac{x^n}{y^n} $ \\
\multicolumn{4}{|l|}{} \\
\hline
\end{tabular}
\bigskip
In Chapter 8, we will find that the above properties are also true for any
rational $m$ and $n$.
\section{Polynomials and their operations}
An expression $ a x^n $, where $a$ is a certain real number,
$ x $ is a variable,
$n$ is a nonnegative integer, is called a {\it term} in one variable.
$a$ is the {\it coefficiant} of the term and
$n$ is the {\it degree} of the term.
\begin{exa}
{\rm Determine which of the following expressions are terms.
If yes, find its coefficient and degree.}
\end{exa}
\begin{tabular}{p{1.2in} p{1.2in} p{1.2in} p{1.2in} }
(a) $ \frac{1}{2} x^2$, &
(b) $ -\sqrt 2 x^3 $, &
(c) $ \sqrt 3 y^4 $, &
(d) $-5$, \\
\multicolumn{4}{}{} \\
(e) $ \frac{1}{2x^2}$, &
(f) $ \sqrt x$, &
(g) $ x - 6 x^2 $. &
(h) $ -4 xy $. \\
\end{tabular}
\bigskip
\noindent {\bf SOLUTION:}
(a) Yes. Its coefficient $ \frac{1}{2} $, degree $2$.
(b) Yes. Its coefficient $ - \sqrt{2} $, degree $3$.
(c) Yes. Its coefficient $ \sqrt 3 $, degree $3$, Here, variable is $y$.
(d) Yes. Its coefficient $ -5 $, degree $0$.
(e) No.
(f) No.
(g) No.
(h) Yes. It is a term in two variables $x$ and $y$. Its coefficient $ -4 $.
(Although we did not define term in several variables, this example can help
you understand it).
\hskip .2in $\Box$
\bigskip
A sum of several terms is called a {\it polynomial} in one variable.
For example, the (g) in the above example is not a term, but it is a
sum of two terms $x$ and $ -6 x^2$, therefore, it is a polynomial.
Since a term can be considered as a sum of one term, it is also a polynomial.
Such polynomials of one term are called {\it monomials}.
Although, we concentrade polynomials in one variable in this chapter, we will
mention polynomials in several variables occationally. Fortunally,
the definitions are similar. Hence we only show several examples:
\begin{exa}
{\rm Determine which of the following expressions are polynomial.}
\end{exa}
\begin{tabular}{p{2.5in} p{2.5in} }
(a) $ \frac{1}{2} x^2 y^3$, &
(b) $ x^2 - 6x + 9 - 4y^2 $, \\
\multicolumn{2}{}{} \\
(c) $ x^3 + y^3 + z^3 - 3 xyz $, &
(d) $ \frac{2x^2} {y} $, \\
\multicolumn{2}{}{} \\
(e) $ 5 \sqrt x y$, &
(f) $ x^2 - 5x + 6 $ \\
\end{tabular}
\bigskip
\noindent {\bf SOLUTION:}
(a) It is a polynomial with one term in two variables $x$ and $y$.
(b) It is a polynomial with four terms in two variables $x$ and $y$.
(c) It is a polynomial with four terms in three variables $x$
$y$ and $z$.
(d) It is not a polynomial.
(e) It is not a polynomial.
(f) It is a polynomial with three term in one variable $x$.
Since the addition is communitative, we have several ways
to present one polynomial.
For instance, $ x - 6 x^2 = - 6 x^2 + x $.
For any polynomial, we can write down
the term with the highest degree first,
the term with the second highest degree second, and so on.
This presentation of a polynomial is called the {\it standard form} of
a polynomial. The highest degree is called the {\it degree} of the polynomial.
The coefficient of the term with highest degree is called the
{\it leading coefficient} of the polynomial.
\begin{exa}
{\rm Rewrite the following polynomial in the standard form, and find its degree
and leading coefficient:}
$ x^2 - 5x^4 + 3 - 4x $,
\end{exa}
\noindent {\bf SOLUTION:}
$ x^2 - 5x^4 + 3 - 4x = -5x^4 + x^2 - 4x + 3$.
The degree is $4$, the leading coefficient is $-5$.
The polynomials has three operations: addition, subtraction
and multiplication.
In the next chapter we will discuss the division of polynomials.
But we cannot say the division is the forth operation of polynomials,
since after division, the answer is no longer a polynomial, for most cases.
It is totally different from addition, subtraction and multiplication, since
after addition, subtraction and multiplication, the answer
is still a polynomial, for all the cases.
This fact can be expressed in a professional way:
The set of all polynomials is closed under
addition, subtraction and multiplication, but is not closed under division.
\section{Basic Algebraic Formulas}
\section{Factoring: By GCF and By Grouping}
In Arithmatic courses, the readers have studied factoring integers.
In the rest of this chapter, we will study how to factor polynomials
and its applications.
In the previous section, we found that in order to do multiplication of
polynomials, in other words, expand a product of polynomials, we use
distribution rule first, then multiply products of two terms, then combine the
like-terms, eventually, we end up one single polynomial.
Pay attention here, the above procedure is not reversable.
Factoring a polynomial is rewriting (keep equal) the polynomial as a product
of two or more polynomials.
For example, $(x+2)(x-3)$ is a product, since the last operation is
multiplication, meanwhile $x^2 - x - 6$ is not a product,
since the last operation is not multiplication.
Although, they are equal each other.
If we write $(x+2)(x-3) = x^2 - x - 6$, we say we do multiplication.
If we write $x^2 - x - 6 = (x+2)(x-3) $, we say we factor the polynomial.
Generally speaking, facoring is the reverse of multiplication.
However, one cannot follow the above procedure backwards to factor
a polynomial, since the above procedure is not reversable.
This is why we need three sections to study factoring methods.
There exists no challeging question for multiplication,
since everabody with patience, caution and sufficient amount of time
definitely can do it,
meanwhile, so many challeging questions exist for factoring.
Patience, caution and sufficient amount of time are not enough.
One needs to be smart.
How to become smart, there is no a short cut, practice, practice
and more practice.
The following two examples show the relations and differences between
multiplication and facoring.
\begin{exa}
{\rm Multiply polynomials:}
$ (x-2)(x-3) $
\end{exa}
\noindent {\bf SOLUTION:}
\begin{eqnarray*}
(x-2)(x-3) & = & x \cdot (x-3) + (-2) \cdot (x-3) \\
& = & x \cdot x + x \cdot (-3) + (-2) \cdot x + (-2) \cdot (-3) \\
& = & x^2 - 3x - 2x + 6 = x^2 - 5x + 6 . \hskip .2in \Box
\end{eqnarray*}
\begin{exa}
{\rm Facor the polynomial:}
$ x^2 - 5x + 6 $
\end{exa}
\noindent {\bf SOLUTION:}
If your memory is not too short, you definitely still remember the previous
example. Therefore, you can obtain the answer immediately:
$ x^2 - 5x + 6 = (x-2) \cdot (x-3) $.
\hskip .2in $\Box$
This method (if we can call it a method) is useless, since no multiplication
question will be provided before a factoring question.
Now, we will discuss the first method for factoring: factor by GCF
(Greatest Common Factor).
The {\bf greatest common factor} for a polynomial is the term with the
following three condition:
1. It is a factor of each term of the polynomial.
2. it has the largest integer exponents.
3. It has the maximum positive coefficient.
\begin{exa}
{\rm Find all common factors for the polynomial:}
$ -2x^2 + 6x $.
{\rm and determine its GCF.}
\end{exa}
\noindent {\bf SOLUTION:}
$-2x^2$ has the factors with positive coefficients:
$1$, $x$, $x^2$, $2$, $2x$, $2x^2$.
$ 6x$ has the factors with positive coefficients:
$1$, $x$, $2$, $2x$, $3$, $3x$, $6$, $6x$.
Therefore, $1$, $x$, $2$, $2x$ are common factors.
$2x$ is the GCF.
\hskip .2in $\Box$
\begin{exa}
{\rm Find all common factors for the polynomial:}
$ x^2 + 2 xy + y^2 $.
{\rm and determine its GCF.}
\end{exa}
\noindent {\bf SOLUTION:}
$x^2$ has the factors with positive coefficients:
$1$, $x$, $x^2$.
$ 2xy$ has the factors with positive coefficients:
$1$, $x$, $y$, $2$, $2x$, $2y$, $xy$, $2xy$.
$y^2$ has the factors with positive coefficients:
$1$, $y$, $y^2$.
Therefore, $1$ is the only common factor, it is also the GCF.
\hskip .2in $\Box$
\bigskip
If the GCF of a polynomial is not $1$, then it can be factored by GCF.
We show the procedure by the following examples.
\begin{exa}
{\rm Factor the polynomial:}
$ -2x^2 + 6x $.
\end{exa}
\noindent {\bf SOLUTION:}
$ -2x^2 + 6x = 2x \cdot (-x) + 2x \cdot 3 = 2x \cdot (-x + 3)$.
\hskip .2in $\Box$
\begin{exa}
{\rm Factor the polynomial:}
$ 24 a^5 b^4 - 16 a^3 b^5 - 8a^2 b^6 $.
\end{exa}
\noindent {\bf SOLUTION:}
\begin{eqnarray*}
24 a^5 b^4 - 16 a^3 b^5 - 8a^2 b^6 & = &
(8a^2 b^4) \cdot (3 a^3) + (8a^2 b^4) \cdot (- 2 a b)
+ (8a^2 b^4) \cdot (- b^2) \\
& = & (8a^2 b^4) \cdot (3 a^3 - 2 a b - b^2).
\hskip .2in \Box
\end{eqnarray*}
\begin{exa}
{\rm Factor the polynomial:}
$ x \cdot (a+b) + y \cdot (a+b) $.
\end{exa}
\noindent {\bf SOLUTION:}
$ x \cdot (a+b) + y \cdot (a+b)
= (a+b) \cdot x + (a+b) \cdot y
= (a+b) \cdot (x + y) $.
$\Box$
\bigskip
If the GCF of a polynomial is $1$, then it also can be factored by GCF.
From the following example, we can see this kind factoring is trivial.
Therefore, usually, we don't do that.
\begin{exa}
{\rm Factor the polynomial:}
$ x^2 + 2 xy + y^2 $.
\end{exa}
\noindent {\bf SOLUTION:}
$ x^2 + 2 xy + y^2 = 1 \cdot x^2 + 1 \cdot (2 xy) + 1 \cdot y^2 =
1 \cdot (x^2 + 2 xy + y^2) $.
\hskip .2in $\Box$
\bigskip
Many polynomials have $1$ as the GCF, therefore, they cannot be factored
by GCF.
However, the above example * shows that we can try to factor them by grouping.
\begin{exa}
{\rm Factor the polynomial:}
$ xa + xb + ya + yb $.
\end{exa}
\noindent {\bf SOLUTION:}
This is a polynomial with $4$ terms, we can group the first and second terms
together, the third and fourth terms together.
Then factor each group by GCF.
Then factor sum of two groups by GCF.
\begin{eqnarray*}
xa + xb + ya + yb & = & (xa + xb) + (ya + yb) \\
& = & x \cdot (a+b) + y \cdot (a+b) = (a+b) \cdot x + (a+b) \cdot y \\
& = & (a+b) \cdot (x + y).
\hskip .2in \Box
\end{eqnarray*}
\bigskip
For this polynomial, we also can group the first and third terms
together, the second and fourth terms together.
Eventually, we end up the same answer.
However, if we group the first and fourth terms
together, the second and third terms together, we cannot obtain anything.
From the example, we can understand that there exists no certain way to group.
The only way is to keep trying.
There also exists no guarenty that the method works for any polynomial.
However, more practices can build up some experience and insight for you.
\begin{exa}
{\rm Factor the polynomial:}
$ xa - xb - ya + yb $.
\end{exa}
\noindent {\bf SOLUTION:}
We can group the first and second terms
together, the third and fourth terms together.
\begin{eqnarray*}
xa - xb - ya + yb & = & (xa - xb) - (ya - yb) \\
& = & x \cdot (a-b) - y \cdot (a-b) = (a-b) \cdot x - (a-b) \cdot y \\
& = & (a-b) \cdot (x - y).
\hskip .2in \Box
\end{eqnarray*}
Notice in the above example,
$ - ya + yb = - (ya - yb) $ is correct, and
$ - ya + yb = - (ya + yb) $ is wrong.
Here is the place at which most students make mistakes.
\begin{exa}
{\rm Factor the polynomial:}
$ x^3 + 2x^2 - 9x - 18 $.
\end{exa}
\noindent {\bf SOLUTION:}
We can group the first and second terms
together, the third and fourth terms together.
\begin{eqnarray*}
x^3 + 2x^2 - 9x - 18 & = & (x^3 + 2x^2) - (9x + 18) \\
& = & x^2 \cdot (x+2) - 9 \cdot (x+2) = (x+2) (x^2-9).
\hskip .2in \Box
\end{eqnarray*}
Also Notice here,
$ - 9x - 18 = - (9x+18) $ is correct, and $ - 9x - 18 = - (9x-18) $ is wrong.
Our answer here $ (x+2) (x^2-9) $ is right.
However, it is not the best, since we can factor out $x^2-9$ further in the
nest sections.
\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*}
\end{document}
(
geocities.com/perryand)