\hfill \hfill } %} \newtheorem{theorem}{Theorem} \newtheorem{acknowledgement}[theorem]{Acknowledgement} \newtheorem{algorithm}[theorem]{Algorithm} \newtheorem{axiom}[theorem]{Axiom} \newtheorem{case}[theorem]{Case} \newtheorem{claim}[theorem]{Claim} \newtheorem{conclusion}[theorem]{Conclusion} \newtheorem{condition}[theorem]{Condition} \newtheorem{conjecture}[theorem]{Conjecture} \newtheorem{corollary}[theorem]{Corollary} \newtheorem{criterion}[theorem]{Criterion} \newtheorem{definition}[theorem]{Definition} \newtheorem{example}[theorem]{Example} \newtheorem{exercise}[theorem]{Exercise} \newtheorem{lemma}[theorem]{Lemma} \newtheorem{notation}[theorem]{Notation} \newtheorem{problem}[theorem]{Problem} \newtheorem{proposition}[theorem]{Proposition} \newtheorem{remark}[theorem]{Remark} \newtheorem{solution}[theorem]{Solution} \newtheorem{summary}[theorem]{Summary} \newenvironment{proof}[1][Proof]{\noindent\textbf{#1.} }{\ \rule{0.5em}{0.5em}} \input{tcilatex} \geometry{left=0.5in,right=0.5in,top=0.5in,bottom=0.4in} \pagestyle{fancy} \lhead{\color{blue} \large Lecture Notes} \chead{\Large Factoring by Trial and Error} \rhead{\small page \thepage} \lfoot{\small \copyright \; \; Hidegkuti, 2018} \rfoot{\small Last revised: December 26, 2018} \cfoot{} \textwidth 7.6in \textheight 9.75in \setlength{\headheight}{28pt} \setlength{\parindent}{0pt} \setlength{\parskip}{0.11in} \newcommand{\defstart}{ \begin{boxedminipage}{7in} \setlength{\fboxrule}{15pt} \setlength{\fboxsep}{15pt} \setlength{\rightskip}{20pt} \setlength{\leftskip}{20pt} \color{white} . \color{black} } \newcommand{\defcont}{ \setlength{\hangindent}{57pt} } \newcommand{\definside}{ \setlength{\hangindent}{57pt} \setlength{\parindent}{57pt} } \newcommand{\defend}{ \end{boxedminipage} \setlength{\parindent}{0pt} \setlength{\hangindent}{0pt} \setlength{\parskip}{0.06in} \setstretch{1.08} } \newcommand{\exstart}{ \setlength{\parindent}{0pt} \setlength{\hangindent}{60pt} \setlength{\parskip}{0.04in} \setstretch{1.08} } \newcommand{\excont}{ \setlength{\parindent}{60pt} \setlength{\hangindent}{75pt} } \newcommand{\solstart}{ \setlength{\parindent}{0pt} \setlength{\hangindent}{63pt} } \newcommand{\solcont}{ \setlength{\parindent}{63pt} \setlength{\hangindent}{63pt} } \newcommand{\solrestart}{ \setlength{\parindent}{48pt} \setlength{\hangindent}{63pt} } \newcommand{\sol}{ \setlength{\parindent}{0pt} \setlength{\hangindent}{50pt} } \newcommand{\solinside}{ \setlength{\parindent}{50pt} \setlength{\hangindent}{50pt} \setstretch{1.08} } \newcommand{\vissza}{ \setlength{\parindent}{0pt} \setlength{\hangindent}{0pt} \setlength{\parskip}{0.1in} \setstretch{1.08} \normalsize \setcounter{enumi}{0}\RESUME } \newcommand{\matek}{ \setstretch{1.04} \normalsize \setlength{\parskip}{0.04in} \setlength{\parindent}{0pt} \setlength{\hangindent}{0pt} \setcounter{enumi}{0}\RESUME } \setstretch{1.06} \begin{document} %TCIMACRO{\TeXButton{\vissza}{\vissza}}% %BeginExpansion \vissza% %EndExpansion For all polynomials, factoring is unique. \ For example, the expression $% x^{2}-16$ can \textit{only} be factored as $\left( x+4\right) \left( x-4\right) $. \ As long as we insist to completely factor a polynomial, there is just one correct form. \ This is different from equations that can have more than one solution. \vspace{0.08in} Because of its uniqueness, we are not forced to develop mathods to systematically find all factored form; if we found one, we found \textit{it}% . \ Because of this, we are allowed to use trial and error to 'stumble into' the factored form. \ \textbf{Trial and error} (or by inspection) refers to a factoring method where we make educated guesses that allow us to quickly factor a quadratic expression.\vspace{0.08in} In what follows, we will only focus on expressions in which the coefficient of the quadraic term is $1$. \ Suppose we expand the expression \ $\left( x+a\right) \left( x+b\right) $. \ \begin{equation*} \left( x+a\right) \left( x+b\right) =x^{2}+bx+ax+ab=x^{2}+\left( b+a\right) x+ab \end{equation*}% The result is $x^{2}+\left( a+b\right) x+ab$. \ Notice that the linear coefficient is the sum of $a$ and $b$, and the number term is the product of $a$ and $b$. \ We can use these facts to quickly factor a quadratc expression such as $x^{2}+7x+12$. \ We will start with the easiest case: when all signs are +.\vspace{0.08in}\vspace{0.08in} \textbf{Case 1: \ All signs are + in the expression to be factored.\vspace{% 0.08in}} %TCIMACRO{\TeXButton{\exstart}{\exstart}}% %BeginExpansion \exstart% %EndExpansion \textbf{Example 1.} \ Completely factor the expression $x^{2}+7x+12$.\vspace{% 0.08in} %TCIMACRO{\TeXButton{\sol}{\sol}}% %BeginExpansion \sol% %EndExpansion \textbf{Solution: }\ If $x^{2}+7x+12$ is factored into $\left( x+a\right) \left( x+b\right) $, then factoring is just a matter of finding $a$ and $b$. \ Consider the equation $a+b=7$. \ This equation has infinitely many solutions. \ For any value of $a$ there is a value of $b$ that works. \ If $% a=1$, then $b=6$, if\ $a=10$, then $b=-3$, and so on. This is not the case with the equation $ab=12$. \ As long as we are looking for integer values, there is a finite list of how the product of two numbers is $12$. \ Because of this, we will \textbf{always start with the number term} that is the product of $a$ and $b$.\vspace{0.08in} %TCIMACRO{\TeXButton{\solinside}{\solinside}}% %BeginExpansion \solinside% %EndExpansion We can quickly list all the pairs of positive numbers with a product of $12$.% \vspace{0.08in}\ %TCIMACRO{\TeXButton{\solinside}{\solinside}}% %BeginExpansion \solinside% %EndExpansion %TCIMACRO{\TeXButton{\begin{minipage}{1.2in}}{\begin{minipage}{1.2in}}}% %BeginExpansion \begin{minipage}{1.2in}% %EndExpansion \begin{tabular}{lll} & $12$ & \\ $1$ & & $12$ \\ $2$ & & $6$ \\ $3$ & & $4$% \end{tabular}% %TCIMACRO{\TeXButton{\end{minipage}}{\end{minipage}}}% %BeginExpansion \end{minipage}% %EndExpansion %TCIMACRO{% %\TeXButton{\begin{minipage}{5.4in}}{\begin{minipage}{5.4in} %\setstretch{1.2} %\setlength{\leftskip}{10pt}}}% %BeginExpansion \begin{minipage}{5.4in} \setstretch{1.2} \setlength{\leftskip}{10pt}% %EndExpansion Now we consider the three pairs as candidates for $a$ and $b$. \ We are looking for the pair with sum $7$. \ Clearly that is $3$ and $4$.\vspace{% 0.04in} \ Once we found $a$ and $b$ with product $12$ and sum $7$, we have the factored form: \ $x^{2}+7x+12=$ \fbox{$\left( x+3\right) \left( x+4\right) $}% %TCIMACRO{\TeXButton{\end{minipage}}{\end{minipage}}}% %BeginExpansion \end{minipage}% %EndExpansion \vspace{0.08in}\vspace{0.08in} %TCIMACRO{\TeXButton{\vissza}{\vissza}}% %BeginExpansion \vissza% %EndExpansion Naturally, things are not always as simple. \ Consider now the second case, when the last term is positive but the second term is negative. \ \textbf{% \vspace{0.08in}} \textbf{Case 2: \ The third sign is + in the expression to be factored, and the second sign is negative.\vspace{0.08in}} %TCIMACRO{\TeXButton{\exstart}{\exstart}}% %BeginExpansion \exstart% %EndExpansion \textbf{Example 2.} \ Completely factor the expression $x^{2}-17x+30$.% \vspace{0.08in} %TCIMACRO{\TeXButton{\sol}{\sol}}% %BeginExpansion \sol% %EndExpansion \textbf{Solution: }\ We are looking for two numbers $a$ and $b$ with a product of $30$ and a sum of $-17$. \ A positive product indicates that $a$ and $b$ are either both positive or both negative. \ Because of the negative second sign, both positive is impossible. \ Therefore, we are looking for two negative numbers. %TCIMACRO{\TeXButton{\solinside}{\solinside}}% %BeginExpansion \solinside% %EndExpansion As always, we start with the equation $ab=30$. \ We list all the pairs of negative numbers with a product of $30$.\vspace{0.08in}\ %TCIMACRO{\TeXButton{\solinside}{\solinside}}% %BeginExpansion \solinside% %EndExpansion %TCIMACRO{\TeXButton{\begin{minipage}{1.2in}}{\begin{minipage}{1.2in}}}% %BeginExpansion \begin{minipage}{1.2in}% %EndExpansion \begin{tabular}{lll} & $30$ & \\ $-1$ & & $-30$ \\ $-2$ & & $-15$ \\ $-3$ & & $-10$ \\ $-5$ & & $-6$% \end{tabular}% %TCIMACRO{\TeXButton{\end{minipage}}{\end{minipage}}}% %BeginExpansion \end{minipage}% %EndExpansion %TCIMACRO{% %\TeXButton{\begin{minipage}{5.4in}}{\begin{minipage}{5.4in} %\setstretch{1.2} %\setlength{\leftskip}{10pt}}}% %BeginExpansion \begin{minipage}{5.4in} \setstretch{1.2} \setlength{\leftskip}{10pt}% %EndExpansion Now we consider these pairs as candidates for $a$ and $b$. \ We are looking for the pair with sum $-17$. \ Clearly that is $-2$ and $-15$.\vspace{0.04in} \ Once we found $a$ and $b$ with product $30$ and sum $-17$, we have the factored form: \ $x^{2}-17x+30=$ \fbox{$\left( x-2\right) \left( x-15\right) $}% %TCIMACRO{\TeXButton{\end{minipage}}{\end{minipage}}}% %BeginExpansion \end{minipage}% %EndExpansion \vspace{0.08in}\vspace{0.08in} %TCIMACRO{\TeXButton{\vissza}{\vissza}}% %BeginExpansion \vissza% %EndExpansion \textbf{Case 3: \ The third sign is - in the expression to be factored.% \vspace{0.08in}} %TCIMACRO{\TeXButton{\exstart}{\exstart}}% %BeginExpansion \exstart% %EndExpansion \textbf{Example 3.} \ Completely factor the expression $x^{2}-2x-48$.\vspace{% 0.08in} %TCIMACRO{\TeXButton{\sol}{\sol}}% %BeginExpansion \sol% %EndExpansion \textbf{Solution: }\ We are looking for two numbers $a$ and $b$ with a product of $-48$ and a sum of $-2$. \ A negative product indicates that one of $a$ and $b$ is positive and the other is negative. \ Now we inspect the second sign. \ If the sum of a positive and a negative number is negative, then between the two of them, the negative one has the greater absolute value. \ These observations will make our task much easier. %TCIMACRO{\TeXButton{\solinside}{\solinside}}% %BeginExpansion \solinside% %EndExpansion As always, we start with the equation $ab=-48$. \ We list all the pairs of positive numbers with a product of $48$, and put a $-$ sign in front of the greater one.\vspace{0.08in}\ %TCIMACRO{\TeXButton{\solinside}{\solinside}}% %BeginExpansion \solinside% %EndExpansion %TCIMACRO{\TeXButton{\begin{minipage}{1.2in}}{\begin{minipage}{1.2in}}}% %BeginExpansion \begin{minipage}{1.2in}% %EndExpansion \begin{tabular}{lll} & $-48$ & \\ $1$ & & $-48$ \\ $2$ & & $-24$ \\ $3$ & & $-16$ \\ $4$ & & $-12$ \\ $6$ & & $-8$% \end{tabular}% %TCIMACRO{\TeXButton{\end{minipage}}{\end{minipage}}}% %BeginExpansion \end{minipage}% %EndExpansion %TCIMACRO{% %\TeXButton{\begin{minipage}{5.4in}}{\begin{minipage}{5.4in} %\setstretch{1.2} %\setlength{\leftskip}{10pt}}}% %BeginExpansion \begin{minipage}{5.4in} \setstretch{1.2} \setlength{\leftskip}{10pt}% %EndExpansion Now we consider these pairs as candidates for $a$ and $b$. \ We are looking for the pair with sum $-2$. \ Clearly that is $-8$ and $6$.\vspace{0.04in} \ Once we found $a$ and $b$ with product $-48$ and sum $-2$, we have the factored form: \ $x^{2}-2x-48=$ \fbox{$\left( x-8\right) \left( x+6\right) $}% \vspace{0.08in}% %TCIMACRO{\TeXButton{\end{minipage}}{\end{minipage}}}% %BeginExpansion \end{minipage}% %EndExpansion \vspace{0.08in}\vspace{0.08in} %TCIMACRO{\TeXButton{\vissza}{\vissza}}% %BeginExpansion \vissza% %EndExpansion In the next example, the second term is positive.\vspace{0.08in} %TCIMACRO{\TeXButton{\exstart}{\exstart}}% %BeginExpansion \exstart% %EndExpansion \textbf{Example 4.} \ Completely factor the expression $x^{2}+11x-60$.% \vspace{0.08in} %TCIMACRO{\TeXButton{\sol}{\sol}}% %BeginExpansion \sol% %EndExpansion \textbf{Solution: }\ We are looking for two numbers $a$ and $b$ with a product of $-60$ and a sum $11$. \ A negative product indicates that one of $% a$ and $b$ is positive and the other is negative. \ Now we inspect the second sign. \ If the sum of a positive and a negative number is positive, then between the two of them, the negative one has the smaller absolute value. \ These observations will make our task much easier. %TCIMACRO{\TeXButton{\solinside}{\solinside}}% %BeginExpansion \solinside% %EndExpansion We start with the equation $ab=-60$. \ We list all the pairs of positive numbers with a product of $60$, and put a $-$ sign in front of the smaller one.\vspace{0.08in}\ %TCIMACRO{\TeXButton{\solinside}{\solinside}}% %BeginExpansion \solinside% %EndExpansion %TCIMACRO{\TeXButton{\begin{minipage}{1.2in}}{\begin{minipage}{1.2in}}}% %BeginExpansion \begin{minipage}{1.2in}% %EndExpansion \begin{tabular}{lll} & $-60$ & \\ $-1$ & & $60$ \\ $-2$ & & $30$ \\ $-3$ & & $20$ \\ $-4$ & & $15$ \\ $-5$ & & $12$ \\ $-6$ & & $10$% \end{tabular}% %TCIMACRO{\TeXButton{\end{minipage}}{\end{minipage}}}% %BeginExpansion \end{minipage}% %EndExpansion %TCIMACRO{% %\TeXButton{\begin{minipage}{5.4in}}{\begin{minipage}{5.4in} %\setstretch{1.2} %\setlength{\leftskip}{10pt}}}% %BeginExpansion \begin{minipage}{5.4in} \setstretch{1.2} \setlength{\leftskip}{10pt}% %EndExpansion Now we consider these pairs as candidates for $a$ and $b$. \ We are looking for the pair with sum $11$. \ Clearly that is $-4$ and $15$.\vspace{0.04in} \ Therefore, \ $x^{2}+11x-60=$ \fbox{$\left( x+15\right) \left( x-4\right) $}% \vspace{0.08in} We can check by mutiplication: \ $\left( x+15\right) \left( x-4\right) =x^{2}-4x+15x-60=x^{2}+11x-60$, so our solution is correct.% %TCIMACRO{\TeXButton{\end{minipage}}{\end{minipage}}}% %BeginExpansion \end{minipage}% %EndExpansion \vspace{0.08in}\vspace{0.08in} %TCIMACRO{\TeXButton{\vissza}{\vissza}}% %BeginExpansion \vissza% %EndExpansion This method is quick and easy. \ However, it only works for simple expressions that start with $x^{2}$. \ Consider for example the product $% \left( 2x+3\right) \left( x+5\right) =2x^{2}+10x+3x+15=2x^{2}+13x+15$. \ The middle term is clearly not the sum of $3$ and $5$. \ Because of uniqueness of factoring, trial and error is still a good approach, but the middle term is no longer just the sum.\vspace{0.08in} \pagebreak %TCIMACRO{\TeXButton{\exstart}{\exstart}}% %BeginExpansion \exstart% %EndExpansion \textbf{Example 5.} \ Solve the equation \ $\left( x-2\right) \left( x-4\right) =24$\vspace{0.08in} %TCIMACRO{\TeXButton{\sol}{\sol}}% %BeginExpansion \sol% %EndExpansion \textbf{Solution: }\ We might be tempted to use the factored form on the left-hand side, but it cannot be used because the other side is not zero. \ So we need to expand the product, reduce one side to zero and then factor.% \begin{eqnarray*} \left( x-2\right) \left( x-4\right) &=&24\text{ \ \ \ \ \ \ \ \ \ \ \ \ expand product} \\ x^{2}-6x+8 &=&24\text{ \ \ \ \ \ \ \ \ \ \ \ \ subtract }24 \\ x^{2}-6x-16 &=&0 \end{eqnarray*}% To factor $x^{2}-6x-16$, we need to find two integers $a$ and $b$ with product $-16$ and sum $-6$. The negative sign in $-16$ indicates that one is positive, the other one is negative. \ The negative sign in $-6$ indicates that the negative number has the greater absolute value. %TCIMACRO{\TeXButton{\solinside}{\solinside}}% %BeginExpansion \solinside% %EndExpansion %TCIMACRO{\TeXButton{\begin{minipage}{1.2in}}{\begin{minipage}{1.2in}}}% %BeginExpansion \begin{minipage}{1.2in}% %EndExpansion \begin{tabular}{lll} & $-16$ & \\ $1$ & & $-16$ \\ $2$ & & $-8$ \\ $4$ & & $-4$% \end{tabular}% %TCIMACRO{\TeXButton{\end{minipage}}{\end{minipage}}}% %BeginExpansion \end{minipage}% %EndExpansion %TCIMACRO{% %\TeXButton{\begin{minipage}{5.4in}}{\begin{minipage}{5.4in} %\setstretch{1.2} %\setlength{\leftskip}{10pt}}}% %BeginExpansion \begin{minipage}{5.4in} \setstretch{1.2} \setlength{\leftskip}{10pt}% %EndExpansion The only pair with sum $-6$ is $-8$ and $2$.\vspace{0.04in} \ Therefore, \ $% x^{2}-6x-16=$ $\left( x-8\right) \left( x+2\right) $. \ Applying the zero product rule, we obtain $x=8$ and $x=-2$.\vspace{0.08in}% %TCIMACRO{\TeXButton{\end{minipage}}{\end{minipage}}}% %BeginExpansion \end{minipage}% %EndExpansion \vspace{0.08in} %TCIMACRO{\TeXButton{\solinside}{\solinside}}% %BeginExpansion \solinside% %EndExpansion We check: \ if $x=8$, then \ LHS $=\left( 8-2\right) \left( 8-4\right) =6\cdot 4=24=$ RHS $\checkmark $\vspace{0.08in} %TCIMACRO{\TeXButton{\solinside}{\solinside}}% %BeginExpansion \solinside% %EndExpansion and if $x=-2$, then LHS $=\left( -2-2\right) \left( -2-4\right) =-4\left( -6\right) =24=$ RHS $\checkmark $ %TCIMACRO{\TeXButton{\solinside}{\solinside}}% %BeginExpansion \solinside% %EndExpansion So both \fbox{$8$ and $-2$} work.\vspace{0.08in}\vspace{0.08in} %TCIMACRO{\TeXButton{\vissza}{\vissza}}% %BeginExpansion \vissza% %EndExpansion If there is a leading coefficient, this method only works if it is also the greatest common factor and can be factored out. \ \vspace{0.08in} %TCIMACRO{\TeXButton{\exstart}{\exstart}}% %BeginExpansion \exstart% %EndExpansion \textbf{Example 6.} \ One side of a rectangle is $6$ feet shorter than twice another side. \ Find the sides of the rectangle if we also know that its area is $140\unit{ft}^{2}$.\vspace{0.08in} %TCIMACRO{\TeXButton{\sol}{\sol}}% %BeginExpansion \sol% %EndExpansion \textbf{Solution: }\ If we label one side by $x$, the other side is $2x-6$. \ The equation will express the area of the rectangle.% \begin{eqnarray*} x\left( 2x-6\right) &=&140\text{ \ \ \ \ \ distribute }x \\ 2x^{2}-6x &=&140\text{ \ \ \ \ \ subtract }140 \\ 2x^{2}-6x-140 &=&0\text{ \ \ \ \ \ \ \ \ \ factor out }2 \\ 2\left( x^{2}-3x-70\right) &=&0 \end{eqnarray*}% We will factor $x^{2}-3x-70$. \ We are looking for two integers with product $-70$ and sum $-3$. \ These are easily found: $-10$ and $7$. \ Therefore, $% x^{2}-3x-70=\left( x-10\right) \left( x+7\right) $. \ Back to the equation: \begin{equation*} 2\left( x-10\right) \left( x+7\right) =0\text{ \ \ \ }\Longrightarrow \text{ \ }x_{1}=10\text{ \ and \ }x_{2}=-7 \end{equation*}% The two solutions of this equation are $10$ and $-7$. \ Since we are looking for a distance and distances cannot be negative, $-7$ is easily ruled out. \ If the shorter side is $x$, then the other side is $2x-6=2\cdot 10-6=14$. \ So the two sides are \fbox{$10\unit{ft}$ and $14\unit{ft}$} long. \ \pagebreak %TCIMACRO{\TeXButton{\exstart}{\exstart}}% %BeginExpansion \exstart% %EndExpansion \textbf{Example 7.} \ Find all numbers that are exactly six less than their own square. \ \vspace{0.08in} %TCIMACRO{\TeXButton{\sol}{\sol}}% %BeginExpansion \sol% %EndExpansion \textbf{Solution: }\ If we label such a number by $x$, then the equation will be $x^{2}=x-6$.% \begin{eqnarray*} x^{2} &=&x-6\text{ \ \ \ \ \ \ \ subtract }x\text{ and add }6 \\ x^{2}-x-6 &=&0 \end{eqnarray*}% We quickly find $-3$ and $2$ as two numbers with sum $-1$ and product $-6$.% \begin{equation*} \left( x-3\right) \left( x+2\right) =0\text{ \ \ \ }\Longrightarrow \text{ \ }x_{1}=3\text{ \ and \ }x_{2}=-2 \end{equation*}% We check: \ $3$ is indeed $6$ less than $9$, and $-2$ is indeed less than $4$% . \ So our asnwer is \fbox{$-2$ and $3$} . \ Perhaps even more importantly, we also proved that there is no other number with this property. \ %TCIMACRO{\TeXButton{\vissza}{\vissza}}% %BeginExpansion \vissza% %EndExpansion \vspace{0.2in} \FRAME{itbpF}{0.5526in}{0.5734in}{0.1807in}{}{}{work.jpg}{\special{language "Scientific Word";type "GRAPHIC";maintain-aspect-ratio TRUE;display "USEDEF";valid_file "F";width 0.5526in;height 0.5734in;depth 0.1807in;original-width 4.6977in;original-height 4.875in;cropleft "0";croptop "1";cropright "1";cropbottom "0";filename 'work.jpg';file-properties "XNPEU";}}\ \ \ \ {\LARGE Practice Problems}% %TCIMACRO{\TeXButton{\vissza}{\vissza}}% %BeginExpansion \vissza% %EndExpansion %TCIMACRO{\TeXButton{ resume}{\setcounter{enumi}{0}\RESUME}}% %BeginExpansion \setcounter{enumi}{0}\RESUME% %EndExpansion \begin{enumerate} \item Completely factor each of the following using the trial and error method.% %TCIMACRO{\TeXButton{4 col begin}{\begin{multicols}{4}}}% %BeginExpansion \begin{multicols}{4}% %EndExpansion a) \ $x^{2}+2x-15$ b) \ $x^{2}-12x+32$ c) \ $x^{2}-2x-3$ d) \ $x^{2}-10x+25$ e) \ $x^{2}+9x+20$ f) \ $x^{2}-x-20$ g) \ $x^{2}-5x+6$ h) \ $x^{2}-5x-6$ i) \ $2x^{2}-8x-42$ j) $\ -3x^{2}-3x+6$ %TCIMACRO{\TeXButton{multcol end}{\end{multicols}}}% %BeginExpansion \end{multicols}% %EndExpansion \item Solve each of the following equations. Make sure to check your solutions.% %TCIMACRO{\TeXButton{3 col begin}{\begin{multicols}{3}}}% %BeginExpansion \begin{multicols}{3}% %EndExpansion a) \ $\left( w+5\right) \left( w-1\right) =0$ b) \ $\left( w+5\right) \left( w-1\right) =55$ c) \ $\left( x-2\right) \left( x+3\right) =50$ d) \ $5x^{3}=10x^{2}+75x$ e) \ $\left( 2x-1\right) ^{2}-x=3x\left( x-1\right) $ f) \ $\left( x+5\right) ^{2}+\left( x-1\right) ^{2}=\left( x+6\right) ^{2}+2$ %TCIMACRO{\TeXButton{multcol end}{\end{multicols}}}% %BeginExpansion \end{multicols}% %EndExpansion \item Find all numbers satisfying the given conditions. a) \ The square of the number is twenty greater than the number. b) \ The sum of the square of the number and three times the number is $70.$ \item a) \ One side of a rectangle is twelve feet shorter than three times another side. \ Find the sides of this rectangle if we also know that the area of this rectangle is $420\unit{ft}^{2}$. b) \ \ One side of a rectangle is twelve feet longer than three times another side. \ Find the sides of this rectangle if we also know that the area of this rectangle is $288\unit{ft}^{2}$. \end{enumerate} \pagebreak\ \ \ \ \FRAME{itbpF}{0.5518in}{0.5518in}{0.2006in}{}{}{answers.jpg}{\special% {language "Scientific Word";type "GRAPHIC";maintain-aspect-ratio TRUE;display "USEDEF";valid_file "F";width 0.5518in;height 0.5518in;depth 0.2006in;original-width 8.6455in;original-height 8.6455in;cropleft "0";croptop "1";cropright "1";cropbottom "0";filename 'answers.jpg';file-properties "XNPEU";}}{\Large \ \ \ \ Answers} {\large Discussion: \ } \vspace{0.04in} {\large Practice Problems}% %TCIMACRO{\TeXButton{\vissza}{\vissza}}% %BeginExpansion \vissza% %EndExpansion %TCIMACRO{\TeXButton{ resume}{\setcounter{enumi}{0}\RESUME}}% %BeginExpansion \setcounter{enumi}{0}\RESUME% %EndExpansion \begin{enumerate} \item a) \ $\left( x-3\right) \left( x+5\right) $ \ \ \ \ b) \ $\left( x-4\right) \left( x-8\right) $ \ \ \ \ c) \ $\left( x+1\right) \left( x-3\right) $ \ \ \ \ d) \ $\left( x-5\right) ^{2}$ \ \ \ \ e) \ $\left( x+4\right) \left( x+5\right) $ f) \ $\left( x+4\right) \left( x-5\right) $ \ \ \ \ \ g) \ $\left( x-2\right) \left( x-3\right) $ \ \ \ \ h) \ $\left( x-6\right) \left( x+1\right) $ \ \ \ \ i) \ $2\left( x-7\right) \left( x+3\right) $ \ \ \ j) \ $-3\left( x+2\right) \left( x-1\right) $ \item a) \ $-5,1$ \ \ \ \ \ b) \ $6,-10$ \ \ \ \ c) \ $7,-8$ \ \ \ \ \ \ d) \ $-3,0,5$ \ \ \ \ \ e) \ $1$ \ \ \ \ \ f) \ $6,-2$ \ \ \ \ \ 3. a) \ $5,-4$ \ \ \ \ \ b) \ $-10,7$ \item[4.] a) \ $14\unit{ft}$ by $30\unit{ft}$ \ \ \ \ b) $8\unit{ft}$ by $36% \unit{ft}$\ \ \ \vspace{0.4in}\vspace{5in}\vspace{0.4in} \end{enumerate} {\small %TCIMACRO{\TeXButton{\small}{\small}}% %BeginExpansion \small% %EndExpansion } \href{https://teaching.martahidegkuti.com/shared/lnotes/lecturenotes.html}{% For more documents like this, visit our page at\ https://teaching.martahidegkuti.com and click on Lecture Notes. \ E-mail questions or comments to mhidegkuti@ccc.edu.} \end{document}