\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 the AC-method} \rhead{\small page \thepage} \lfoot{\small \copyright \; \; Hidegkuti, 2018} \rfoot{\small Last revised: December 27, 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 Factoring by the AC-method is extremely useful because it adresses the most difficult situation, factoring a general quadratic expression with a leading coefficient, such as $6x^{2}-5x-4$. \ Trial and error still works, but it is more difficult because there are more possibilities to be considered. \ The AC-method is a neat and powerful method to quickly factor a trinomial. \ The main steps are: we cleverly take apart the linear term into two parts and then factor by grouping.\vspace{0.08in} In case of a general quadratic equation, we often denote the coefficients by $a$, $b$, and $c$, where the trinomial is $ax^{2}+bx+c$. \ Notice the addition. \ This means that coefficients also carry the negative signs if there are any. \ In the case of $6x^{2}-5x-4$, $a=6$, $b=-5$, and $c=-4$.% \vspace{0.08in} \ %TCIMACRO{\TeXButton{\exstart}{\exstart}}% %BeginExpansion \exstart% %EndExpansion \textbf{Example 1.} \ Completely factor the expression $6x^{2}-5x-4$.\vspace{% 0.08in} %TCIMACRO{\TeXButton{\sol}{\sol}}% %BeginExpansion \sol% %EndExpansion \textbf{Solution: }\ We will use the AC-method. \ The first step is to re-write the middle term, $-5x$ into a sum of two terms in such a way that grouping would work after that. \ To do that, we need to find two numbers $p$ and $q$ with a sum of $-5$ and a product that is the same as the product $ac $ (hence the name, AC-method.) \ In this particular case, $ac=6\left( -4\right) =-24$.\vspace{0.08in} \ So we need to find two numbers, $p$ and $q$% , such that \begin{equation*} p+q=-5\text{ \ and \ \ }pq=-24 \end{equation*}% This sounds very similar to the trial and error method. Indeed, we will apply the same method to find $p$ and $q$. \ A negative product $pq$ indicates that one of $p$ and $q$ is positive and the other is negative. \ The negative sum $-5$ indicates that between the two numbers $p$ and $q$, the negative one has a greater absolute value. \ These observations will make our task much easier. \ \ \ There are infinitely many integer pair solutions for $p+q$. \ There are just a few ways to solve the second equation \ for integers, and so we start there. %TCIMACRO{\TeXButton{\solinside}{\solinside}}% %BeginExpansion \solinside% %EndExpansion We list all the pairs of positive numbers with a product of $24$, and take the opposite of the greater one in each pair. \ We are looking for the pair with sum $-5$.\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} & $-24$ & \\ $1$ & & $-24$ \\ $2$ & & $-12$ \\ $3$ & & $-8$ \\ $4$ & & $-6$% \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 $p$ and $q$. \ We are looking for the pair with sum $-5$. \ Clearly that is $-8$ and $3$.\vspace{0.04in} \ Once we found $p$ and $q$ with product $-24$ and sum $-5$, we can rewrite the middle term, $-5x$ as $-8x+3x$ or $3x-8x$ and factor by grouping.% %TCIMACRO{\TeXButton{\end{minipage}}{\end{minipage}}}% %BeginExpansion \end{minipage}% %EndExpansion \begin{eqnarray*} 6x^{2}-5x-4 &=&6x^{2}-8x+3x-4 \\ &=&2x\left( 3x-4\right) +1\left( 3x-4\right) \\ &=&\left( 2x+1\right) \left( 3x-4\right) \end{eqnarray*}% and so the factored form is \fbox{$\left( 2x+1\right) \left( 3x-4\right) $}. \ We can check our solution by multiplying back: %TCIMACRO{\TeXButton{\solinside}{\solinside}}% %BeginExpansion \solinside% %EndExpansion $\left( 2x+1\right) \left( 3x-4\right) =6x^{2}-8x+3x-4=6x^{2}-5x-4$ and so our solution is correct.\vspace{0.08in}\vspace{0.08in} \pagebreak %TCIMACRO{\TeXButton{\exstart}{\exstart}}% %BeginExpansion \exstart% %EndExpansion \textbf{Example 2.} \ Completely factor the expression $10x^{2}-19x+6$.% \vspace{0.08in} %TCIMACRO{\TeXButton{\sol}{\sol}}% %BeginExpansion \sol% %EndExpansion \textbf{Solution: }\ In this case, $a=10$, $b=-19$, and $c=6$. \ The product $ac$ is $60$. \ We are looking for two numbers $p$ and $q$ with a product of $60$ and a sum of $-19$. \ \ A positive product indicates that both $p$ and $% q$ are positive, or both of them negative. \ The negative sum indicates that both $p$ and $q$ are negative. \ %TCIMACRO{\TeXButton{\solinside}{\solinside}}% %BeginExpansion \solinside% %EndExpansion As always, we start with $pq=60$. \ We list all the pairs of negative numbers with a product of $60$.\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 The only pair with a sum $-19$ is $-15$ and $-4$. \ Now we know how to re-write the linear term, $-19x$ and then factor by grouping.\vspace{0.8in}% %TCIMACRO{\TeXButton{\end{minipage}}{\end{minipage}}}% %BeginExpansion \end{minipage}% %EndExpansion \begin{eqnarray*} 10x^{2}-19x+6 &=&10x^{2}-15x-4x+6\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ factor by grouping} \\ &=&5x\left( 2x-3\right) -2\left( 2x-3\right) \\ &=&\left( 5x-2\right) \left( 2x-3\right) \end{eqnarray*}% Therefore, $10x^{2}-19x+6=$ \fbox{$\left( 5x-2\right) \left( 2x-3\right) $}$% . $ \ We check by multiplying back:\vspace{0.08in} %TCIMACRO{\TeXButton{\solinside}{\solinside}}% %BeginExpansion \solinside% %EndExpansion $\left( 5x-2\right) \left( 2x-3\right) =10x^{2}-15x-4x+6=\allowbreak 10x^{2}-19x+6$ and so our solution is correct.\vspace{0.08in}\vspace{0.08in} %TCIMACRO{\TeXButton{\vissza}{\vissza}}% %BeginExpansion \vissza% %EndExpansion What happens if we cannot find a suitable pair of numbers $p$ and $q$? \ If that is the case, the trinomial cannot be factored using integer coefficients. \ At this point, we will just say that the expression is \textbf{prime} or \textbf{irreducible}. \ We are still responsible however, to factor out the GCF if there is any.\vspace{0.08in} %TCIMACRO{\TeXButton{\exstart}{\exstart}}% %BeginExpansion \exstart% %EndExpansion \textbf{Example 3.} \ Completely factor the expression $3x^{2}-4x+2$.\vspace{% 0.08in} %TCIMACRO{\TeXButton{\sol}{\sol}}% %BeginExpansion \sol% %EndExpansion \textbf{Solution: }\ In this case, $ac=6$. \ Therefore, we are looking for two numbers $p$ and $q$ with a product of $6$ and a sum $-4$. \ A positive product indicates that both $p$ and $q$ are negative or both are positive. \ The negative sum $-4$ indicates that both $p$ and $q$ are negative. \ %TCIMACRO{\TeXButton{\solinside}{\solinside}}% %BeginExpansion \solinside% %EndExpansion We start with the equation $pq=6$. \ We list all the pairs of negative numbers with a product of $6.$\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} & $6$ & \\ $-1$ & & $-6$ \\ $-2$ & & $-3$% \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 sum of the first pair is $-7$ and the sum of the second pair is $-5$. \ There are no pairs with sum $-4$. %TCIMACRO{\TeXButton{\end{minipage}}{\end{minipage}}}% %BeginExpansion \end{minipage}% %EndExpansion %TCIMACRO{\TeXButton{\solinside}{\solinside}}% %BeginExpansion \solinside% %EndExpansion This indicates that the trinomial cannot be factored. \ Therefore, our answer is \fbox{$3x^{2}-4x+2$} for the factored form.\vspace{0.08in} \pagebreak %TCIMACRO{\TeXButton{\vissza}{\vissza}}% %BeginExpansion \vissza% %EndExpansion Sometimes the product $ac$ is too rich in divisors to list all pairs. \ If the second term has a small coefficient, there is a neat shortcut to find the righ $p$ and $q$. \ Th following example illustrates this trick. \ %TCIMACRO{\TeXButton{\exstart}{\exstart}}% %BeginExpansion \exstart% %EndExpansion \textbf{Example 4.} \ Completely factor the expression $24x^{2}+x-10$.% \vspace{0.08in} %TCIMACRO{\TeXButton{\sol}{\sol}}% %BeginExpansion \sol% %EndExpansion \textbf{Solution: }\ In this case, $ac=-240$. \ We are looking for $p$ and $% q $ with $pq=-240$ and $p+q=1$. \ The number $-240$ has quite a number of divisors, so we will find the right pair without listing all the pairs. \ This trick is not possible for all cases, but it will work here. \ The negative product $pq$ indicates that one of $p$ and $q$ is positive and the other is negative. \ The positive sum $1$ indicates that between the two numbers $p$ and $q$, the positive one has a greater absolute value.\vspace{% 0.08in} %TCIMACRO{\TeXButton{\solinside}{\solinside}}% %BeginExpansion \solinside% %EndExpansion When we add a positive and a negative number, we subtract the absolute values. \ Considering the absolute values $\left\vert p\right\vert $ and $% \left\vert q\right\vert $, their product is $240$ and their difference is $1$% . \ That means that the right pair of factors are very close to each other. \ Consequently, both numbers must be fairly close to the square root of $240$% . \ We enter $\sqrt{240}$ into our calculator. \ This is not an integer, the calculator shows $\sqrt{240}=\allowbreak 15.\,\allowbreak 491\,933...$. \ If two numbers are \ close to each other \ and their product is $240$, then they both are fairly close to $15$. \ So we round up to $16$ and start looking for divisors counting backward: \ $16,$ $15$, $14$, etc. \ Except, we won't even reach $14$ because we immediately bump into $15$ and $16$. \ Our $p$ and $q$ is $16$ and $-15$. \ We can now easily factor by grouping. \begin{eqnarray*} 24x^{2}+x-10 &=&24x^{2}+16x-15x-10 \\ &=&8x\left( 3x+2\right) -5\left( 2x+2\right) =\left( 8x-5\right) \left( 3x+2\right) \end{eqnarray*} %TCIMACRO{\TeXButton{\exstart}{\exstart}}% %BeginExpansion \exstart% %EndExpansion \textbf{Example 5.} \ One side of a rectangle is $2$ feet shorter than three times another side. \ Find the sides of the rectangle if we also know that its area is $176\unit{ft}^{2}$. %TCIMACRO{\TeXButton{\sol}{\sol}}% %BeginExpansion \sol% %EndExpansion \textbf{Solution: }\ If we label one side as $x$, then the other side can be written as $3x-2$. \ The equation will express the area of the rectangle. \ \begin{eqnarray*} x\left( 3x-2\right) &=&176 \\ 3x^{2}-2x &=&176 \\ 3x^{2}-2x-176 &=&0 \end{eqnarray*}% We will factor this trinomial by the AC-method. \ In this case, $ac=3\left( -176\right) =\allowbreak -528$. \ To factor $3x^{2}-2x-176$, we need to find two integers $p$ and $q$ with product $-528$ and sum $-2$. The negative product indicates that one of $p$ and $q$ is positive, the other one is negative. \ Therefore, the difference between their absolute values is $2$. \ That is fairly small for a large product such as $-528$. \ If the two numbers multiplying each other to $528$ are close to each other, they also must be close to $\sqrt{528}=\allowbreak 22.\,\allowbreak 978\,251....$ \ \ We roll up to $23$ and start looking for pairs of integers multiplying to $% 528$. \ We will try $23$, $22$, $21$, and so on. \ We almost immediately find $22$ and $24$. \ Therefore $p$ and $q$ are $-24$ and $22$. \ We can now easily factor the trinomial. \begin{eqnarray*} 3x^{2}-2x-176 &=&3x^{2}-24x+22x-176 \\ &=&3x\left( x-8\right) +11\left( 2x-8\right) \\ &=&\left( 3x+11\right) \left( x-8\right) \end{eqnarray*}% Applying the zero product rule, we solve $3x+11=0$ and $x-8=0$ \ ad obtain $% x=-\dfrac{11}{3}$ \ or $x=8$. \ Since $x$ represents a distance and distances cannot be negative, we rule out $-\dfrac{11}{3}$ and are left with $x=8$. \ If $x$ is $8$, then $3x-2=3\cdot 8-2=22$. \ Thus the sides of the rectangle are \fbox{$8\unit{ft}$ and $22\unit{ft}$} long. \ %TCIMACRO{\TeXButton{\solinside}{\solinside}}% %BeginExpansion \solinside% %EndExpansion We check: \ $22$ is indeed two less than three times $8,$ and the area of the rectangle is $8\unit{ft}\left( 22\unit{ft}\right) =\allowbreak 176\unit{% ft}^{2}$, and so our solution is correct.\vspace{0.08in} %TCIMACRO{\TeXButton{\vissza}{\vissza}}% %BeginExpansion \vissza% %EndExpansion Quadratic equations often have two solutions. \ In the previous example, one solution of the equation was easily ruled out, but that is not always the case. \ Often times both solutions of the equation result in a meaningful solution. \ The next example illustrates this.\vspace{0.08in} %TCIMACRO{\TeXButton{\exstart}{\exstart}}% %BeginExpansion \exstart% %EndExpansion \textbf{Example 6.} \ Twice the square of a number is $35$ greater than three times the number. \ \ Find all such numbers.\vspace{0.08in} %TCIMACRO{\TeXButton{\sol}{\sol}}% %BeginExpansion \sol% %EndExpansion \textbf{Solution: }\ If we label this number by $x$, then the square of this number is $x^{2}$ and twice the square is $2x^{2}$.\ The equation will compare twice the square of the number and three times the number.% \begin{eqnarray*} 2x^{2} &=&3x+35 \\ 2x^{2}-3x-35 &=&0 \end{eqnarray*}% We will factor $2x^{2}-3x-35$. \ The product $ac$ is $-70$. \ Therefore, \ we are lookin for two integers $p$ and $q$ with product $-70$ and sum $-3$. \ These are easily found: $-10$ and $7$. \ We now know how to take apart the middle term before grouping. \ \begin{eqnarray*} 2x^{2}-3x-35 &=&0 \\ 2x^{2}-10x+7x-35 &=&0 \\ 2x\left( x-5\right) +7\left( x-5\right) &=&0 \\ \left( 2x+7\right) \left( x-5\right) &=&0 \end{eqnarray*}% We solve the linear equations $2x+7=0$ and $x-5=0$ and obtain $x=-\dfrac{7}{2% }$ and $x=5$. \ We check both: %TCIMACRO{\TeXButton{\solinside}{\solinside}}% %BeginExpansion \solinside% %EndExpansion If the number is $5$, then twice its square is $2\cdot 5^{2}=50$, and three times the number is $15$. \ Indeed, $50$ is $35$ greater than $15.$ \ So $5$ works. %TCIMACRO{\TeXButton{\solinside}{\solinside}}% %BeginExpansion \solinside% %EndExpansion If the number is $-\dfrac{7}{2}$, then twice its square is $2\left( -\dfrac{7% }{2}\right) ^{2}=2\cdot \dfrac{49}{4}=\dfrac{49}{2}$. \ Three times the number is \newline $3\left( -\dfrac{7}{2}\right) =\allowbreak -\dfrac{21}{2}$. \ The difference between $\dfrac{49}{2}$ and $-\dfrac{21}{2}$ is $\dfrac{49}{2}-\left( -% \dfrac{21}{2}\right) =\dfrac{49+21}{2}=\dfrac{70}{2}=35$. \ %TCIMACRO{\TeXButton{\solinside}{\solinside}}% %BeginExpansion \solinside% %EndExpansion We found that both \fbox{$-\dfrac{7}{2}$ and $5$} works. %TCIMACRO{\TeXButton{\vissza}{\vissza}}% %BeginExpansion \vissza% %EndExpansion \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. a) \ $30x-15y+6ax-3ay$ \ \ \ \ \ \ \ \ \ f) $b^{2}-a+ab^{2}-1$ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ k) \ $14x-12x^{2}+10$ b) \ $xy-y-x+1$ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ g) \ $% 2m^{2}-18n^{4}+2m^{2}p^{2}-18n^{4}p^{2}$ \ \ \ \ \ \ \ \ \ l) \ $% 5m^{2}-7mn+2n^{2}$\ \ c) \ $6a^{2}b^{2}-4a^{2}bc-10a^{2}c^{2}$ \ \ \ \ \ \ \ \ \ h) \ $% a^{2}x^{2}-a^{2}y^{2}+b^{2}x^{2}-b^{2}y^{2}$ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ m) \ $29px-21p^{2}+10x^{2}$ d) \ $a^{2}m+2a^{2}n-b^{2}m-2b^{2}n$ \ \ \ \ \ i) \ $3x^{2}-2x-1$ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ n) \ $% 2x^{4}-3y^{4}+x^{2}y^{2}$ e) \ $x^{2}-4y^{2}+m^{2}x^{2}-4m^{2}y^{2}$ \ \ \ \ \ \ j) \ $6x^{2}-5x+1$ \ \ \ \ \ \ \ \ \ \ \ \ \ \item Solve each of the following equations. a) \ $6x^{4}-x^{3}=2x^{2}$ \ \ \ \ \ \ \ \ b) \ $5a^{2}+5=26a$ \ \ \ \ \ \ \ \ \ $\ $c) \ $11p+35p^{2}=6$ \item One side of a rectangle is $4$ $\unit{in}$ shorter than $3$ times the other side. Find the sides of the rectangle if its area is $319$ $\unit{in}% ^{2}$. \end{enumerate} \pagebreak\ \ \ \ \FRAME{itbpF}{0.8129in}{0.8129in}{0.2006in}{}{}{answers.jpg}{\special% {language "Scientific Word";type "GRAPHIC";maintain-aspect-ratio TRUE;display "USEDEF";valid_file "F";width 0.8129in;height 0.8129in;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 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) \ $3\left( a+5\right) \left( 2x-y\right) $ \ \ \ \ \ b) \ $\left( x-1\right) \left( y-1\right) $ \ \ \ \ \ \ c) \ $2a^{2}\left( b+c\right) \left( 3b-5c\right) $ \ \ \ \ \ d) \ $\left( a-b\right) \left( a+b\right) \left( m+2n\right) $ e) \ $\left( x-2y\right) \left( x+2y\right) \left( m^{2}+1\right) $ \ \ \ \ \ f) $\left( b-1\right) \left( b+1\right) \left( a+1\right) $ \ \ \ \ \ g) \ $-2\left( 3n^{2}-m\right) \left( 3n^{2}+m\right) \left( p^{2}+1\right) $ h) \ $\left( x-y\right) \left( x+y\right) \left( a^{2}+b^{2}\right) $ \ \ \ \ i) \ $\left( 3x+1\right) \left( x-1\right) $ \ \ \ \ \ \ \ j) \ $\left( 3x-1\right) \left( 2x-1\right) $ k) \ $-2\left( 2x+1\right) \left( 3x-5\right) $ \ \ \ \ \ l) \ $\left( 5m-2n\right) \left( m-n\right) $ \ \ \ \ \ m) \ $\left( 5x-3p\right) \left( 2x+7p\right) $ n) \ $\left( x-y\right) \left( x+y\right) \left( 2x^{2}+3y^{2}\right) $ \ \ \item a) \ $-\dfrac{1}{2},0,\dfrac{2}{3}$ \ \ \ \ \ \ \ \ b) \ $\dfrac{1}{5}% ,5$ \ \ \ \ \ \ \ \ c) \ $-\dfrac{3}{5},\dfrac{2}{7}$ \item $11$ $\unit{in}$ by $29$ $\unit{in}\vspace{0.7in}\vspace{5in}$ \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}