\title{Self-Paced Study Guide in Algebra} \maketitle \section{How to Use This Guide} The \textit{Self-Paced Review} consists of review modules with exercises; problems and solutions; self-tests and solutions; and self-evaluations for the four topic areas Algebra, Geometry and Analytic Geometry, Trigonometry, and Exponentials \& Logarithms. In addition, previous \textit{Diagnostic Exams} with solutions are included. Each topic area is independent of the others. The \textit{Review Modules} are designed to introduce the core material for each topic area. A numbering system facilitates easy tracking of subject material. For example, in Algebra, the subtopic Linear Equations is numbered with \ref{alg:ex:two:three}. Problems and self-evaluations are categorized using this numbering system. When using the \textit{Self-Paced Review}, it is important to differentiate between concept learning and problem solving. The review modules are oriented toward refreshing concept understanding while the problems and self-tests are designed to develop problems solving ability. When reviewing the modules, exercises are liberally sprinkled throughout the modules: solve these exercises when working through the module. The problems should be attempted without looking at the solutions. If a problem cannot be solved after at least two honest efforts, then consult the solutions. Trying many times and then succeeding results in a better understanding than trying several times and reading the solution. The tests should be taken only when both understanding of the material and problem solving ability have been achieved. The self-evaluation is a useful tool to evaluate the mastery of the material. Finally, the previous Diagnostic Exams should provide the finishing touch. The review modules were written by Professor A\. P\. French (Physics Department) and Adeliada Moranescu (MIT Class of 1994). The problems and solutions were written by Professor Arthur Mattuck (Mathematics Department). This document was originally produced by the Undergraduate Academic Affairs Office, August, 1992, and transcribed to \LaTeX\ and edited for OCW by Tea Dorminy (MIT Class of 2013) in August, 2010. \newpage \tableofcontents \newpage \section{Algebra Review} One widely used algebra textbook\footnote{Hughes-Hallett, \textit{The Math Workshop: Algebra}, W\. W\. Norton Co., New York, 1980.} begins with the sentence: ``Algebra is really arithmetic in disguise.'' Arithmetic makes use of specific numbers; algebra develops general results that can be applied regardless of what the particular numbers are. That makes algebra much more powerful than arithmetic. But if you want to \textit{apply} a given algebraic result, you have to replace the $x$s and $y$s by specific numbers, and you have to know how to handle numerical quantities. So this algebra review begins with some stuff about numbers as such. \subsection{Scientific Notation} \label{alg:ex:two:one} Writing numbers in what is called `scientific notation' is an absolute necessity in science and engineering: you will be using it all the time. It is based on the fact that any positive number, however large or however small, can be written as a number between 1 and 10 multiplied by a power of 10. (And, for negative numbers, we simply put a $-{}$ sign in front.) Begin by recalling that: \begin{align*} 10^1&=10\\ 10^2&=10\cdot 10=100\\ 10^3&=10\cdot 10\cdot 10=1000\\ &\text{etc.} \end{align*} To \textit{multiply} together any two powers of 10, we simply \textit{add} the exponents: \[(10^m)(10^n)=10^{(m+n)}\] \textbf{These and related matters are discussed in more detail in the module on Exponentials \& Logarithms.} To \textit{divide} one power of 10 by another, we \textit{subtract} the exponents: \[\frac{10^m}{10^n}=10^{m-n}.\] Thus $10^9/10^4=10^{(9-4)}=10^5=100000$. It is implicit in this that the reciprocal of a positive power of 10 is an equal negative power:\[\frac{1}{10^n}=10^{-n}.\] The powers-of-10 notation also defines what is meant by $10^0$:\[10^0=\frac{10^n}{10^n}=1.\] This is all very simple, but you need to be careful when negative powers of 10 are involved. Give yourself some practice by doing the following exercises: \begin{exercise}\label{alg:ex:two:one:zero} Write the following numbers in scientific notation: \begin{enumerate} \item 80,516 \item 0.0751 \item 3,520,000 \item 0.000000081 \end{enumerate} \end{exercise} Note: The answers to the exercises are all collected together at the end of this module. We have tried to eliminate errors, but if you find anything that you think needs to be corrected, please write to us. \textit{A Note about Notation:} We are using the center dot, for instance, $1.5\cdot 10^4$, when we write a number as the product of a decimal with a power of 10. Perhaps you are accustomed to using the multiplication sign, ${}\times{}$, for this purpose, and you will find that both conventions are widely used. We prefer to use the center dot (especially in algebra and, later, in calculus) or, in some places, a parenthesis, to avoid any possible confusion with the variable $x$. \begin{exercise}\label{alg:ex:two:one:one} Evaluate: \begin{enumerate} \item $10^9\cdot 10^{-3}$ \item $10^9/10^{-4}$ \item $\frac{10^{-19}}{10^{-34}}$ \end{enumerate} Answers? You don't need \textit{us} to provide them! Just get the practice. \end{exercise} Now consider multiplying or dividing two numbers that are \textit{not} pure powers of 10. Take, for example, $2.6\cdot 10^3$ and $5.3\cdot 10^4$. Their \textit{product} is given by: \[(2.6\cdot 10^3)(5.3\cdot 10^4)=(2.6)(5.3)(10^3\cdot 10^4)=13.8\cdot 10^{3+4}=13.8\cdot 10^7=1.38\times 10^8.\] Their \textit{quotient} is given by: \[\frac{2.6\cdot 10^3}{5.3\cdot 10^4}=\left(\frac{2.6}{5.3}\right)10^{3-4}=0.68\cdot 10^{-1}=6.8\cdot 10^{-2}.\] Notice how we don't stop until we have converted the answer into a number between 1 and 10 multiplied by a power of 10. \begin{exercise}\label{alg:ex:two:one:two} Evaluate, in scientific notation: \begin{enumerate} \item $(1.5\cdot 10^4)(7.5\cdot 10^{-5})$ \item $(4.3\cdot 10^{-6})/(3.1\cdot 10^{-10}$ \item $\frac{(1.2\cdot 10^{-5})(1.5\cdot 10^{3})}{9\times 10^{-9}}$ \end{enumerate} \end{exercise} To \textit{add} or \textit{subtract} numbers in scientific notation, you first have to rearrange the numbers so that all the powers of 10 are the same; then you can add or subtract the decimal parts, leaving the power of 10 alone. It's usually best if you first express all the numbers in terms of the \textit{highest} power of 10. In this connection, remember that, with negative powers of 10, smaller exponents means bigger numbers: $10^{-3}$ is bigger than $10^{-5}$. You may, however, need to make a final adjustment if the combination of the decimal numbers is more than 10 or less than 1. %Page 3 Examples: \begin{align*} 2.1\cdot 10^3 + 3.5\cdot 10^5 &= (0.021 + 3.5)\cdot 10^5 = 3.521\cdot 10^5\\ 1.3\cdot 10^8 -8.4\cdot 10^7 &= (1.3-0.84)\cdot 10^8 = 0.46\cdot 10^8=4.6\cdot 10^7\\ 9.5\cdot 10^{-11} + 9.8\cdot 10^{-12} &= (9.5 + 0.98)\cdot 10^{-11} = 10.48\cdot 10^{-11}=1.048\cdot 10^{-10}\\ \end{align*} \begin{exercise}\label{alg:ex:two:one:three} Evaluate: \begin{enumerate} \item $9.76\cdot 10^9 + 7.5\cdot 10^8$ \item $1.25\cdot 10^6 - 7.85\cdot 10^5$ \item $4.21\cdot 10^{25} - 1.85\cdot 10^{26}$ \item $4.05\cdot 10^{-19} - 10^{-20}$ \item $(1.2\cdot 10^{-19})(5.2\cdot 10^{10} + 4\cdot 10^9)$ \end{enumerate} \end{exercise} \subsection{Significant Figures} \label{alg:ex:two:two} In the above examples and exercises in adding or subtracting numbers expressed in scientific notation, you will have noticed that you may end up with a large number of digits. But not all of these may be significant. For example, when we added $2.1\cdot 10^3$ and $3.5\cdot 10^5$, we got $3.521\cdot 10^5$. However, each of the numbers being combined were given with only two-digit accuracy. (We are assuming that 3.5 means simply that the number is closer to 3.5 than it is to 3.4 or 3.6. If it meant 3.500 then these extra zero digits should have been included.) This means that the $2.1\cdot 10^3$ did not add anything significant to the bigger number, and we were not justified in giving more than two digits in the final answer, which therefore should have been given as just $3.5\cdot 10^5$. Note, however, that if we were asked to add, say, $8.6\cdot 10^3$ to $3.5\cdot 10^5$, the smaller number would make a significant contribution to the final answer. The straight addition would give us $3.586\cdot 10^5$. Rounding this off to two digits would then give the answer $3.6\cdot 10^5$. The general rules governing significant figures are: \begin{enumerate} \item The final answer should not contain more digits than are justified by the least accurate of the numbers being combined; \textbf{but} \item Accuracy contained in the numbers being combined should not be sacrificed in the rounding-off process. \end{enumerate} A few examples will help spell out these conditions. (We'll ignore the powers-of-10 factors for this purpose.) \begin{tabular}{ll} Addition:&$1.63+2.1789+0.96432=4.77422$, rounded to $4.77$.\\ Subtraction:&$113.2-1.43=111.77$, rounded to 111.8.\\ Multiplication:&$(11.3)(0.43)=4.859$, rounded to $4.9$ (only two digits justified).\\ Division:&$\frac{1.30}{0.43}=3.02325814$, rounded to $3.0$ (two digits). \end{tabular} \textit{Beware of your calculator!} In the last examples, we used a pocket calculator to do the divisions. This automatically gave 10 digits in each answer. \textit{Most} of these digits are insignificant! Whenever you use your calculator for such a purpose, always ask yourself how many digits are justified and should be retained in the answer. Most of the calculations you will be doing will probably involve numbers with only a few significant digits. Get into the habit of cutting down your final answers to the proper size. As you can see, this situation will arise most importantly when you are doing divisions with your calculator. But watch out for surplus digits in multiplications as well. Now let's turn to algebra proper. \subsection{Linear Equations}\label{alg:linalg}\label{alg:ex:two:three} \subsubsection{Equations in one variable} These scarcely need any discussion. There is one unknown, say $x$, and an equation that relates this to given numbers or constants. The only job is to tidy things up so as to solve for $x$ explicitly. \begin{exercise}\label{alg:ex:two:three:zero} Solve for $x$: \begin{enumerate} \item $5\left(x+\frac 14\right)=2x-\frac 18$ \item $\frac 3x - \frac 45 = \frac 1x + \frac 13$ \item $3(ax+b)=5bx+c$ \end{enumerate} \end{exercise} \subsubsection{Simultaneous Equations in Two Variables} However two-variable equations are originally written, they can always be reduced to the following form: \begin{align*} a_1x + b_1 y &= c_1\\ a_2x + b_2 y &= c_2 \end{align*} where the coefficients $a_1,a_2,b_1,b_2,c_1,c_2$ may be positive or negative. Two essentially equivalent methods can be used to obtain the solutions for $x$ and $y$: \begin{enumerate} \item \textit{Substitution:} Use one of the equations (say the first) to get one of the unknowns (say $y$) in terms of $x$ and known quantities. Substitute into the other equation to solve for the remaining unknown ($x$). Plug this value of $x$ back into either of the initial equations to get $y$. \item \textit{Elimination:} Multiply the original equations by factors that make the coefficient of one unknown (say $y$) the same in both. By subtraction, eliminate $y$; this leads at once to $x$: Multiply 1st eq. by $b_2$: \[a_1b_2x + b_1b_2y = b_2c_1\] Multiply 2nd eq. by $b_1$: \[a_2b_1x + b_2b_1y = b_1c_2\] Subtract: \[(a_1b_2-a_2b_1)x=b_2c_1-b_1c_2\] Thus: \[x=\frac{b_2c_1-b_1c_2}{a_1b_2-a_2b_1}.\] Then solve for $y$ as before. But the last equation above points to a special situation: If the denominator $a_1b_2-a_2b_1$ is 0, then $x$ becomes indeterminate, and therefore so does $y$. \end{enumerate} Keep it neat! Good housekeeping in mathematics is very important. It will make it easier for you to check your work and it will help you to avoid errors. Notice how we put the equals signs underneath one another in the above analysis. Make a practice of doing this yourself. Concentrate on making this a habit, and don't throw it out the window when you are taking a quiz or exam. You're bound to benefit from being neat and orderly. \begin{exercise}\label{alg:ex:two:three:one} Solve for both unknowns: \begin{enumerate} \item \begin{align*} 2x-3y&=4\\ 3x-2y&=5 \end{align*} \item \begin{align*} 5a&=b-6\\ 2b&=a+4 \end{align*} \item \label{alg:exercise:iii:3} \begin{align*} \frac 3x + \frac 4y &= 5\\ \frac 4x - \frac 2y &= 3 \end{align*} \end{enumerate} [In \ref{alg:exercise:iii:3}, resist the temptation to multiply both equations throughout by $xy$ to clear the denominators. Just put $1/x=u$, $1/y=v$, and solve first for $(u,v)$, which are just as legitimate variables as $(x,y)$.] \end{exercise} \subsection{Polynomials} \label{alg:ex:two:four} Much of what you do in algebra (and later, in calculus) will be based on a familiarity with expressions made up of a sum of terms like $10x^3$ or $(3.2y^5)$ --- in other words, sums of products of numbers called coefficients and powers of variables such as $x$. Such an expression is called a \textit{polynomial} --- meaning simply something with many terms. Many important polynomials are made up of a set of terms each of which contains a different power of a single quantity $x$. We can then write \[P(x)=a_0+a_1x+a_2x^2+a_3x^3+\cdots+a_nx^n,\] where the quantities $a_0,a_1,a_2,a_3,\cdots,a_n$ are constant coefficients, labeled here to show which power of $x$ they are associated with. We have written the combination as $P(x)$, to indicate that this combination, $P$, is a certain \textit{function} of $x$ if $x$ is a mathematical variable. That is an aspect of polynomials which is important in calculus, but we won't expand on it in these algebra review notes. A polynomial has a highest power of $x$ in it; this is called the \textit{degree} of the polynomial. Thus if the highest term is $3x^4$, we say that the polynomial is ``of degree four'' or ``a quartic.'' For the most part, we shall not go beyond quadratics, of degree 2. Many polynomials are a binomial (two-term) combination of two variables, of the form $(x+y)$, raised to an arbitrary power. The simplest examples of this type are $(x+y)$ itself and $(x+y)^2=x^2+2xy+y^2$. A \textit{binomial expansion} is the result of multiplying out such an expression, like $(2x+y)(x+y)(x+3y)$, into a polynomial. If we want to add one polynomial to another, we simply add the terms with the same power of $x$. Multiplication of two polynomials is a bit more complicated, but again involves identifying all the terms that have the same power of $x$ and adding the coefficients to make a single term of the form $a_nx^n$. For instance: \begin{align*} (x+2)(2x^2-3x+4)&=x(2x^2-3x+4)+2(2x^2-3x+4)\\ &=(2x^3-3x^2+4x)+(4x^2-6x+8)\\ &=2x^3+(-3x^2+4x^2)+(4x-6x)+8\\ &=2x^3+x^2-2x+8 \end{align*} \begin{exercise}\label{alg:ex:two:four:zero} Multiply $3x^2+4x-5$ by $2x-1$. \end{exercise} \subsection{Quadratic Equations} \label{alg:ex:two:five} A quadratic equation is a polynomial equality that can be manipulated into the form \[ax^2+bx+c=0.\] We can solve the equation through the process of \textit{completing the square}. This technique is the basis of the general quadratic formula that you are likely familiar with already: \[x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\] We'll give a specific example and then derive the general formula. \subsubsection{Completing the Square: An Example} Suppose we are asked to solve the equation \[x^2-6x-4=0\] We recognize that the combination $x^2-6x$ is the first two terms of $(x-3)^2$. We just evaluate the complete square and see what is left over: \begin{align*} (x-3)^2&=x^2-6x+9\\ x^2-6x&=(x-3)^2-9\\ x^2-6x-4&=\left((x-3)^2-9\right)-4\\ &=(x-3)^2-13 \end{align*} But $x^2-6x-4=0\rightarrow (x-3)^2-13=0$, requiring $(x-3)=\pm\sqrt{13}$. Thus the solution is \[x=3\pm \sqrt{13}\] Note that, in this equation, our original quadratic expression reduces to a perfect square \textit{minus} a certain number, 13. When we set the whole expression equal to zero, this means that the perfect square is equal to a \textit{positive} number, and we can proceed to take the square root of both sides. But if the quadratic had been a perfect square \textit{plus} some number $n$ we would have arrived at an equation of the form \[(x+p)^2=-n\] We should then have been faced with taking the square root of a negative number. That would take us into the realm of \textit{imaginary} numbers, which is beyond the scope of anything you need consider for the present. \subsubsection{The General Quadratic Formula} It is possible to obtain the quadratic formula directly from the equation \[ax^2+bx+c=0\] Subtract the constant $c$ from both sides, to obtain \[ax^2+bx=-c\] Divide both sides by the coefficient $a$:\[ x^2+\frac{b}{a}x=-\frac{c}{a}\] Complete the square of the left-hand side by adding $(b/2a)^2$. Add the same quantity to the right side: \[x^2+\frac{b}{a}x+\left(\frac{b}{2a}\right)^2=\left(x+\frac{b}{2a}\right)^2=-\frac{c}{a}+\left(\frac{b}{2a}\right)^2\] Bring the right-hand side to a common denominator, and take the square root of both sides: \[x+\frac{b}{2a}=\pm\frac{\sqrt{b^2-4ac}}{2a}\] Subtract $\frac{b}{2a}$ from both sides, to obtain an expression for $x$: \[x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\] Now we have the general quadratic formula. Notice that, if the solutions are to be real numbers, we must have $b^2-4ac\geq 0$. \begin{exercise}\label{alg:ex:two:five:zero} Solve these quadratic equations by completing the square. If the coefficient of $x^2$ is not 1, you should follow the steps listed above in the derivation of the general formula. \begin{enumerate} \item $x^2-4x-12=0$ \item $9x^2-6x-1=0$ \item $x+2x^2=\frac 58$ \item $(x+2)(2x-1)+3(x+1)=4$ \end{enumerate} \end{exercise} \begin{exercise}\label{alg:ex:two:five:one} Solve these quadratics with the general formula: \begin{enumerate} \item $2x^2=9x-8$ \item $y=ut-\frac 12 gt^2$ (solving for $t$) \item $5x(x+2)=2(1-x)$ \item $0.2x^2-1.5x=3$ \item $x^2-2sx=1-2s^2$ (solving for $x$) \end{enumerate} \end{exercise} \subsection{Long Division \& Factoring} \label{alg:ex:two:six} Dividing one polynomial by another is very much like long division in arithmetic. We shall assume that the polynomial we are dividing into (the \textit{dividend}) is of a higher degree than the one we are dividing by (the \textit{divisor}). The reason for doing the division may be to find out if the higher-degree polynomial is exactly divisible by the lower-degree one; if it is, we have identified a way of \textit{factoring} the higher-degree polynomial. We write both dividend and divisor in decreasing powers of $x$, or whatever variable is being used. For example, we divide $4x^2+8x+3$ by $2x+1$: \polylongdiv{4x^2+8x+3}{2x+1} It works! $2x+1$ divides exactly into $4x^2+8x+3$; the result is $2x+3$, and so we have factored the quadratic into two linear factors. \begin{exercise}\label{alg:ex:two:six:zero} Divide $5x^2-6x-8$ by $x-2$. \end{exercise} Let's now change the above example of long division slightly, by making our dividend $4x^2+8x+5$. There will now be a remainder of 2. The remainder 2 means that the quotient still has a fraction in it. \[\frac{4x^2+8x+5}{2x+1}=2x+3+\frac{2}{2x+1}\] This is analogous to the similar process with numbers; for instance, \[\frac{423}{10}=42+\frac 3{10}\] With numbers we know that the remainder is always a number less than the divisor when dividing by 10; the possible remainders are $0,1,\cdots,9$. For polynomial, the remainder is always of lower degree than the divisor. For instance, \[\frac{a_5x^5+a_4x^4+a_3x^3+a_2x^2+a_1x+a_0}{x^2+1}=Q(x)+\frac{R(x)}{x^2+1}\] The quotient $Q(x)$ has degree $3=5-2$ and the remainder $R(x)$ has degree at most 1. In other words, there are constants $b_0,b_1,b_2,b_3,c_0,c_1$ for which \[Q(x)=b_3x^3+b_2x^2+b_1x+b_0 \text{ and } R(x)=c_1x+c_0\] \begin{exercise}\label{alg:ex:two:six:one} Divide $2x^3+2x^2-10x+4$ by $x+3$. What is the degree of the remainder? \end{exercise} \begin{exercise}\label{alg:ex:two:six:two} If the polynomial $x^5+4x^4-6x^3+5x^2-2x+3$ were divided by $x^2+2$, what would be the degree of the quotient and the degree of the remainder? Don't actually do the division! \end{exercise} \begin{exercise}\label{alg:ex:two:six:three} What happens if you divide a polynomial of lower degree by one of higher degree? \end{exercise} Factoring is an important part of many calculations. But it is often hard to see which polynomial evenly (exactly) divides another. In fact, factoring polynomials and numbers is an important problem for both mathematicians and computer scientists. Here is the only tool you need to know for now: \begin{center} If $P(r)=0$, then $(x-r)$ divides $P(x)$ evenly. \end{center} This is what we mean by saying that $r$ is a \textit{root} of $P(x)$. Example: $P(x)=x^2+x-2$. $P(x)$ has the root $-2$, since $P(-2)=(-2)^2+(-2)-2=0$. Then $P(x)$ is exactly divisible by $(x-(-2))=x+2$: \polylongdiv{x^2+x-2}{x+2} The quotient is $x-1$. Therefore, $x^2+x-2=(x+2)(x-1)$. The actual process of division here is of course just like that in the example at the beginning of the section. But there we gave the value of the divisor, instead of looking for it from scratch. When dividing by a first order factor that divides evenly, you may save time by solving for coefficients using the following scheme: \begin{align*} x^2+x-2&=(x+2)(ax+b)\\ &=ax^2+(2a+b)x+2b \end{align*} Equating coefficients gives $a=1, b=-1$. (But you can see in this case that $a=1$ without writing down anything; then you can solve for $b$ in $x^2+x-2=(x+2)(x+b)$.) Here is another approach to factoring: For polynomials $x^n+a_{n-1}x^{n-1}+\cdots+a_0$ with leading coefficient 1, and integers for the other coefficients, any integer root must be a divisor of the constant term $a_0$. Thus in the polynomial $x^2+x-2$ we have $a_0=-2$, and there are four integers that might work: $\pm 1$ and $\pm 2$. Finally, for quadratic equations the roots are always obtainable by the quadratic formula: \[x^2+x-2=0\rightarrow x=\frac{-1\pm\sqrt{1^2-4(-2)}}2=\frac{-1\pm 3}{2}=-2\text{ or } 1\] In general, $ax^2+bx+c=a(x-r_1)(x-r_2)$ where $r_1$ and $r_2$ are the roots. If the quadratic formula leads to the square root of a negative number than the quadratic has no real roots and no real factors. If you allow the use of complex numbers as coefficients, then the quadratic factors as usual. Look out for easily factorable quadratics! Be on the alert for the following: \begin{itemize} \item Quadratics whose coefficients are small or simple enough for you to make a shrewd guess at the factorization \item Quadratics that are perfect squares \item Expressions of the form $a^2x^2-b^2$, which can be immediately factored into $(ax+b)(ax-b)$ \end{itemize} Don't overuse the quadratic formula when factoring a polynomial like $x^2+x-2$. Your first instinct should be to check $\pm 1$ and $\pm 2$ as roots. If you do use the quadratic formula, you should be prepared to \textit{doublecheck} your answer. Arithmetic errors can be as significant in a physical problem as the difference between going the right way and the wrong way down a one-way street. \begin{exercise}\label{alg:ex:two:six:four} Test whether the following quadratics can be factored, and find the factors if they exist: \begin{enumerate} \item $8x^2+14x-15$ \item $2x^2-3x+10$ \item $9x^2-24ax+16a^2$ \item $2x^2+10x-56$ \item $100x^4-10^{10}$ (This is of 4th degree, but first put $x^2=u$.) \end{enumerate} \end{exercise} \begin{exercise}\label{alg:ex:two:six:five} Solve these quadratic equations by factoring: \begin{enumerate} \item $x^2-5x+6=0$ \item $2x^2-5x-12=0$ \item $3x^2-5x=0$ \item $6x^2-7x-20=0$ \end{enumerate} \end{exercise} \subsection{Some Tricks of the Trade} \label{alg:ex:two:seven} \subsubsection{Getting Rid of Radicals}(Mathematical, not political) Radicals are a nuisance if one is trying to solve an algebraic equation, and one usually wants to get rid of them. The way to get rid of a radical is of course to square it (or cube for a cube root, etc.); but this only works if the radical stands by itself on one side of an equation. Thus if you have $\sqrt{x-1}+a=b$, you don't gain anything by squaring both sides. That simply gives you $x-1+2a\sqrt{x-1}+a^2=b^2$. But, if you first isolate the radical on the left-hand side, you have \[\sqrt{x-1}=b-a\] Then when you square, you get $x-1=(b-a)^2$, and you are in business. You may have to go through this routine more than once. Example: Suppose you have $\sqrt{2x-1}-1=\sqrt{x-1}$. Now you can't isolate both radicals, and it doesn't help to put them together on one side of the equation. So you do the best you can, with one of the radicals isolated in the equation as it stands. Squaring, you get \[2x-1-2\sqrt{2x-1}+1=x-1\] Now we can isolate the remaining radical: \[2\sqrt{2x-1}=x+1\] Squaring again, and rearranging, gives us a quadratic equation: \[x^2-6x+5=0\] with roots $x=1$ or $x=5$. \textbf{Warning!} Squaring may introduce so-called \textit{extraneous} roots, because, for example, if we started with $x=3$ and squared it, we would have $x^2=9$. Taking the square root of this would appear to allow $x=-3$ as well as $x=3$. So always check your final results to see if they fit the equation in its original form. \begin{exercise}\label{alg:ex:two:seven:zero} Solve for $x$: \begin{enumerate} \item $\sqrt{2x-1}=x-2$ (Be careful!) \item $\sqrt{x-3}=\sqrt{2x-5}-1$ \end{enumerate} \end{exercise} \subsubsection{Combining Fractions} \begin{enumerate} \item Multiplying: $\frac ab \frac cd = \frac{ac}{bd}$ \item Dividing: $\frac{\frac ab}{\frac cd}=\frac ab \frac dc = \frac{ad}{bc}$ \item Adding: $\frac ab + \frac cd = \frac{ad+bc}{bd}$ \item Subtracting: $\frac ab - \frac cd = \frac{ad-bc}{bd}$ \end{enumerate} The quantities $a,b,c,d$ may be ordinary numbers, but they may also be algebraic expressions. In either case, look first to see if they can be factored. In (a) and (b) this may lead to simple cancellation; in (c) and (d) it may enable you to find a least common denominator that is simpler than the product $bd$. In the latter case, convert each fraction to one with this common denominator and then add and subtract the numerators as required. This may involve less complexity than if you mechanically apply the above expressions. \begin{exercise}\label{alg:ex:two:seven:one} Evaluate the following combinations of fractions, bringing the result to a single denominator in each case and canceling where possible: \begin{enumerate} \item $\displaystyle \left(\frac{x-y}{x^2+y^2}\right)\left(\frac{y}{y-x}\right)$ \item $\displaystyle \frac{\left(\frac{x^2y+y^3}{2x}\right)}{\left(\frac{2xy+y^2}{4x^3}\right)}$ \item $\displaystyle \frac{x}{4y^2x}+\frac{z^2}{8xy}$ \item $\displaystyle \frac{4z}{xy^2}-\frac{2x}{y^3x}+\frac{z}{x^2y}$ \end{enumerate} \end{exercise} \section{Answers to Exercises} Answers to exercise \ref{alg:ex:two:one:one} not provided. Exercise \ref{alg:ex:two:one:zero}: (a) $8.0516\cdot 10^4$; (b) $7.51\cdot 10^{-2}$; (c) $3.52\cdot 10^6$; (d) $8.1\cdot 10^{-8}$ Exercise \ref{alg:ex:two:one:two}: (a) $1.125$; (b) $1.39\cdot 10^{4}$; (c) $2\cdot 10^6$ Exercise \ref{alg:ex:two:one:three}: (a) $1.051\cdot 10^{10}$; (b) $4.65\cdot 10^{5}$; (c) $-1.429\cdot 10^{26}$; (d) $3.95\cdot 10^{-19}$; (e) $6.72\cdot 10^{-9}$ Exercise \ref{alg:ex:two:three:zero}: (a) $x=\frac{-11}{24}$; (b) $x=\frac{30}{17}$; (c) $x=\frac{c-3b}{3a-5b}$ Exercise \ref{alg:ex:two:three:one}: (a) $x=\frac 75$, $y=-\frac 25$; (b) $a=-\frac 89$, $b=\frac{14}{9}$; (c) $x=1$, $y=2$ Exercise \ref{alg:ex:two:four:zero}: $6x^3+5x^2-14x+5$ Exercise \ref{alg:ex:two:five:zero}: (a) $6,-2$; (b) $\frac{1\pm\sqrt{2}}3$; (c) $\frac{-1\pm \sqrt{6}}{4}$; (d) $\frac{-3\pm\sqrt{15}}2$ Exercise \ref{alg:ex:two:five:one}: (a) $\frac{9\pm\sqrt{17}}4$; (b) $\frac{u\pm\sqrt{u^2-2gy}}g$; (c) $\frac{-6\pm\sqrt{46}}5$; (d) $\frac{15\pm\sqrt{465}}4$; (e) $s\pm\sqrt{1-s^2}$ Exercise \ref{alg:ex:two:six:zero}: $5x+4$ Exercise \ref{alg:ex:two:six:one}: The degree of the remainder is zero. (The remainder is $-2=-2x^0$.) Exercise \ref{alg:ex:two:six:two}: Degree of quotient: 3; degree of remainder: 1 (at most). Exercise \ref{alg:ex:two:six:three}: You can't do it by long division. We assumed that the divider had a lower degree than the dividend, in order for the long division to work. It's like trying to divide 2 by 7. So the answer is (again, without using long division!): quotient ${}=0$; remainder = dividend. Exercise \ref{alg:ex:two:six:four}: (a) $(4x-3)(2x+5)$; (b) No (complex roots); (c) $(3x-4a)^2$; (d) $2\left(x+\frac 52 - \frac{\sqrt{137}}2\right)\left(x+\frac 52 + \frac{\sqrt{137}}2\right)$; (e) $100\left(x^2+10^4\right)\left(x+100\right)\left(x-100\right)$ Exercise \ref{alg:ex:two:six:five}: (a) $(x-2)(x-3)=0\rightarrow 2,3$; (b) $(2x-3)(x+4)=0\rightarrow \frac 32, -4$; (c) $x(3x-5)=0\rightarrow \frac 53, 0$; (d) $(2x-5)(3x+4)=0\rightarrow \frac 52, -\frac 43$ Exercise \ref{alg:ex:two:seven:zero}: (a) $x=5$ ($x=1$ doesn't work, since in the original equation the LHS is a radical, and thus greater than 0; and the RHS is $1-2=-1<0$); (b) $x=3$ or $7$ Exercise \ref{alg:ex:two:seven:one}: (a) $\frac{-y}{x^2+y^2}$; (b) $\frac{2x^2\left(x^2+y^2\right)}{2x+y}$; (c) $\frac{2x^2+yz^3}{8xy^2z}$; (d) $\frac{4xyz^2-2x^3+y^2z^2}{x^2y^3z}$ \textbf{This module is based in large part on an earlier module prepared by the Department of Mathematics.} \section{Algebra Review Problems} \ref{alg:ex:two:one}, \ref{alg:ex:two:two} \textbf{Calculations} \begin{problem} Express as a single number, in scientific notation $a\times 10^k$, $1\leq a<10$: \begin{enumerate} \item $\frac{(.024)(3\times 10^{-2})}{8\times 10^{12}}$ \item $\frac{1600}{32\times 10^{-4}}$ \end{enumerate} \end{problem} \begin{problem} $6.25\times 10^{24}$ molecules of water fill a .2 liter glass. Approximately how much of this volume (in liters) does one molecule account for? Give your answer in scientific notation with the correct number of significant figures. \end{problem} \begin{problem} Using $E=mc^2$, where $c=3\times 10^{8}$ meters/sec, in $km\cdot sec$ units how much energy $E$ is mass-equivalent to $3\times 10^{-18}$ kg of water? \end{problem} \begin{problem} Looking at a representative $.60$ ml sample under her microscope, Michelle counts exactly 120 bacteria. If the sample is drawn from a flask containing 2000. ml of water, how many bacteria are expected to be in the flask? \end{problem} \ref{alg:ex:two:three} \textbf{Linear equations} \begin{problem} Solve for $x$ in terms of $a$ and $b$: $\frac{x+a}{x-a}=b$. \end{problem} \begin{problem} Solve simultaneously for $x$ and $y$: \begin{enumerate} \item\begin{align*} 2x-3y&=-1\\3x-2y&=6\end{align*} \item\begin{align*} 4x-3y&=9\\5y-3x-7&=0\end{align*} \item\begin{align*} x+2y&=a\\x-y&=b\end{align*} \end{enumerate} \end{problem} \begin{problem} Brown rice comes in 5 lb bags costing $\$2$ a bag; wild rice comes in 2 lb. bags costing $\$10$ a bag. Egbert has just spent $\$32$ buying 34 lbs. of rice for his commune. How many bags of each did he buy? \end{problem} \begin{problem} How many liters of liquid endersol should be added to 20 liters of water to get a solution that is 40\% endersol? \end{problem} \ref{alg:ex:two:four} \textbf{Polynomials, binomial theorem} The general \textit{binomial theorem} is something you should know. It states that \[(a+b)^n=a^n+na^{n-1}b+\cdots+\binom{n}{k}a^{n-k}b^k+\cdots+nab^{n-1}+b^n\] where the coefficient of the general term is \[\binom{n}{k}=\frac{n!}{k!(n-k)!}=\frac{n(n-1)(n-2)\cdots(n-k+1)}{k!}\] The cases $n=2$ and $n=3$ are given in the review modules. \begin{problem} Write $(x+1)^3$ as a polynomial in $x$ using the binomial theorem. \end{problem} \begin{problem} Simplify $\frac{(x+h)^3-x^3}{h}$, writing it as a polynomial in $x$ and $h$. \end{problem} \begin{problem} What are the coefficients $a,b,c$ in $(x+1)^12=ax^12+bx^11+cx^10+\cdots$? \end{problem} \begin{problem} Write $(x+y)^4$ as a polynomial in $x$ and $y$. \end{problem} \ref{alg:ex:two:five} \textbf{Quadratic equations} \textit{Try a selection of these problems; similar problems are grouped together. If the coefficient of $x^2$ is not 1, be especially careful in using the quadratic formula.} \begin{problem} For each of the following, first try to find the roots by factoring the left hand side. If successful, then use the quadratic formula to check your answer; if not, find the roots by the quadratic formula. Try a, c, and d, b for more practice. \begin{enumerate} \item $x^2-7x+12=0$ \item $x^2+2x-35=0$ \item $x^2-6x+5=0$ \item $2x^2-5x+3=0$ \end{enumerate} \end{problem} \begin{problem} \begin{enumerate} \item Solve for $t$ in terms of the other variables: $y=vt-\frac 12 gt^2$. \item Express $x$ in terms of $a$, $b$, and $h$, if $h=2ax-bx^2$ and $a$, $b$, $h$, $x>0$. \end{enumerate} \end{problem} \begin{problem} Solve $x^2+3x-1=0$, and tell whether both roots are positive, negative, or of opposite signs. How could you have answered this last question without first solving the equation? \end{problem} \begin{problem} Let $a$, $b$ be the distinct roots of $x^2-4x+2$, with $a0$. \end{problem} \begin{problem} Combine the two terms: \begin{enumerate} \item $\frac{a}{a+h}+\frac{h}{a-h}$ \item $\frac{x}{x+1}+\frac{2}{x-2}$ \end{enumerate} \end{problem} \begin{problem} Solve: $3(x+1)^{-1}+(x-3)^{-1}=1$. \end{problem} \begin{problem} For each positive integer $n$, $a_n$ is given: evaluate and simplify $a_{n+1}/a_n$: \begin{enumerate} \item $a_n=\frac{x^n}{n!}$ \item $a_n=\frac{x^{2n}}{n(n+1)}$ \end{enumerate} \end{problem} \textbf{Geometric series and geometric progressions} The formula for the sum of the \textit{infinite geometric series} is: \[1+r+r^2+r^3+\cdots+r^n+\cdots=\frac{1}{1-r}\] if $-10$, while $-3-\sqrt{13}<0$. Therefore, the roots are of opposite signs. Alternately, since $x^2+3x-1=(x-r_1)(x-r_2)$, we have $r_1r_2=-1$, which shows that roots have opposite signs without actually calculating them. \end{solution} \begin{solution} $x^2-4x+2=0$, by formula: $x=\frac{-4\pm\sqrt{16-4\cdot 2}}2=2\pm\sqrt{2}$. Since $\sqrt{2}\approx 1.4$, $2-\sqrt{2}<1<2+\sqrt{2}$. \end{solution} \begin{solution} \begin{enumerate} \item $x^2+4x-b=0$ by formula gives $x=\frac{-4\pm\sqrt{16+4b}}2$, so roots are real iff (if and only if) $16+4b\geq 0$ or $\boxed{b\geq -4}$. \item $x^2+bx+1=0$ by formula gives $x=\frac{-b\pm\sqrt{b^2-4}}2$, so roots are real iff $b^2-4\geq 0\rightarrow b^2\geq 4$, so iff $b\geq 2$ or $b\leq -2$. \end{enumerate} \end{solution} \begin{solution} Let $x$ be the width, $y$ be the height. We have \begin{align*} y-2&=x\\xy&=7\end{align*} Substitution into the second gives \begin{align*} (y-2)y&=7\\ y^2-2y-7&=0\\ y&=\frac{2\pm\sqrt{4+28}}2=2\sqrt{2}+1\\ x&=2\sqrt{2}-1 \end{align*} \end{solution} \begin{solution} Let $x$ be the number. Then $3x+\frac 2x=5\rightarrow 3x^2-5x+2=0$. By formula, \[x=\frac{5\pm\sqrt{26-4\cdot 2\cdot 3}}{6}=\frac{5\pm 1}{6}=\boxed{1,\frac 23}\] \end{solution} \begin{solution} Solve $16t^2-40t-100=0$, or $4t^2-10t-25=0$: this gives $t=\frac{10\pm\sqrt{100+4\cdot 25\cdot 4}}{2\cdot 4}=\frac{10\pm 10\sqrt{5}}{8}=\boxed{\frac 54\left(1+\sqrt{5}\right)}$. We reject the other solution because it makes no physical sense. \end{solution} \begin{solution} \begin{enumerate} \item $f(-1)=0$ means that $x+1$ is a factor. Thus, $x^3-2x^2-x+2=(x+1)(x^2-3x+2)=(x+1)(x-2)(x-1)$. (We get the factor $x^2-3x+2$ by long division.) \item $f(2)=0$ means that $x-2$ is a factor; $x^3-2x-4=(x-2)(x^2+2x+2)$. \item $x^4-16=(x^2-4)(x^2+4)=(x-2)(x+2)(x^2+4)$. \item $x^3-x^2-9x+9$ --- We try as roots the factors of 9, and see that 1 is a root, so $x-1$ is a factor. $x^3-x^2-9x+9=(x-1)(x^2-9)=(x-1)(x-3)(x+3)$. \end{enumerate} \end{solution} \begin{solution} $x-1$ divides $f(x)$ means that $f(1)=0$. We calculate therefore $f(1): 5\cdot 1^8-3\cdot 1^5+1^4-2\cdot 1-1=0$, so $x-1$ is a factor. \end{solution} \begin{solution} \begin{enumerate} \item $x-1$ is a factor implies $f(1)=0$; $1^3-3\cdot 1^2+c\cdot 1-2=0$ so $c-4=0\rightarrow \boxed{c=4}$. \item $x+2$ is a factor implies $f(-2)=0$; $(-2)^4+4(-2)^3+c=0\rightarrow 16-32+c=0\rightarrow\boxed{c=16}$. \end{enumerate} \end{solution} \begin{solution} \begin{enumerate} \item \begin{align*} \sqrt{3x+10}&=x+2\\ 3x+10&=x^2+4x+4\\ x^2+x-6&=0\\ (x+3)(x-2)&=0 \end{align*} We reject $x=-3$, since it doesn't satisfy the equation we started with; thus the only solution is $\boxed{x=2}$. \item \begin{align*} \sqrt{1-x^2}&=\frac{x}{\sqrt{3}}\\ 1-x^2&=\frac{x^2}{3}\\ 1&=\frac 43 x^2\\ x&=\sqrt{\frac 34}=\boxed{\frac{\sqrt{3}}2} \end{align*} Again, we reject $-\sqrt{3}/2$ since it doesn't work in the original equation. \end{enumerate} \end{solution} \begin{solution} $\frac{1+x^2}{1-x^2}=y$ -- solve for $x^2$: $1+x^2=y-x^2y\rightarrow x^2(1+y)=y-1\rightarrow x^2=\frac{y-1}{y+1}$, so \[\boxed{x=\pm\sqrt{\frac{y-1}{y+1}}}\] \end{solution} \begin{solution} \begin{align*} \frac{1}{\sqrt{m^2-c^2}}&=a\\ \frac{1}{m^2-c^2}&=a^2\\ \frac{1}{a^2}&=m^2-c^2\\ m^2&=c^2+\frac{1}{a^2}\\ m&=\sqrt{c^2+\frac{1}{a^2}} \end{align*} \end{solution} \begin{solution} \begin{enumerate} \item \[\frac{a}{a+h}+\frac{h}{a-h}=\frac{a(a-h)+h(a+h)}{(a+h)(a-h)}=\frac{a^2+h^2}{a^2-h^2}\] \item \[\frac{x}{x+1}+\frac{2}{x-2}=\frac{x^2+2}{(x+1)(x-2)}\] \end{enumerate} \end{solution} \begin{solution} \[\frac{3}{x+1}+\frac{1}{x-3}=1\rightarrow \frac{4x-8}{x^2-2x-3}=1\rightarrow 4x-8=x^2-2x-3\rightarrow x^2-6x+5=0\] This last factors to $(x-5)(x-1)=0$, so $\boxed{x=1,5}$ are the solutions. \end{solution} \begin{solution} \begin{enumerate} \item $\frac{a_{n+1}}{a_n}=\frac{x^{n+1}}{(n+1)!}\cdot \frac{n!}{x^n}=\boxed{\frac{x}{n+1}}$ \item $\frac{a_{n+1}}{a_n}=\frac{x^{2(n+1)}}{(n+1)(n+2)}\cdot \frac{n(n+1)}{x^{2n}}=\boxed{x^2\cdot \frac{n}{n+2}}$ \end{enumerate} \end{solution} \begin{solution} \begin{enumerate} \item $1-\frac 12+\frac 14-\frac 18=\frac{1}{1-\left(-\frac 12\right)}=\frac{1}{3/2}=\boxed{\frac 23}$ ($r=-\frac 12$) \item $2+\frac{2}{10}+\frac{2}{100}+\cdots=2\cdot\frac{1}{1-\frac{1}{10}}=\boxed{\frac{20}9}$ ($r=\frac{1}{10}$) \end{enumerate} \end{solution} \begin{solution} \begin{enumerate} \item $.444\cdots=\frac{4}{10}+\frac{4}{100}+\frac{4}{1000}+\cdots=\frac{4}{10}\left(1+\frac{1}{10}+\frac{1}{100}+\cdots\right)=\frac{4}{10}\left(\frac{1}{1-\frac{1}{10}}\right)=\boxed{\frac{4}{9}}$ \item $.121212\cdots=\frac{12}{100}+\frac{12}{10^4}+\frac{12}{10^6}+\cdots=\frac{12}{100}\cdot\frac{1}{1-1/100}=\boxed{\frac{12}{99}}$ \end{enumerate} \end{solution} \begin{solution} \[100+\frac 23 100 + \frac 23 \cdot \frac 23 \cdot 100 +\cdots=100\left(1+\frac 23 +\left(\frac 23\right)^2+\cdots\right)=100\cdot\frac{1}{1-2/3}=\boxed{300}\] \end{solution} \begin{solution} $(1+r+\cdots+r^n)(1-r)=(1+r+\cdots+r^n)-(r+r^2+\cdots+r^{n+1})=1-r^{n+1}$. \end{solution} \begin{solution} \begin{align*} 1+r+\cdots+r^n&=(1+r+r^2+\cdots)-r^{n+1}(1+r+\cdots)\\ &=\frac{1}{1-r}-\frac{r^{n+1}}{1-r}\\ &=\frac{1-r^{n+1}}{1-r} \end{align*} \end{solution} \section{Algebra Diagnostic Test \#1} \begin{problem} Evaluate and express in scientific notation with the correct number of significant figures $\frac{1}{8\cdot 10^{12}}$ and $\frac{160}{32\cdot 10^{-4}}$. \end{problem} \begin{problem} Jack and Janet are buying an assortment of fruit. Jack wants the number of apples to be three less than twice the number of oranges. Janet wants to buy two more apples than oranges. How many oranges and apples should they get? \end{problem} \begin{problem} Expand the following polynomial and determine its degree: \[x^4-3x^3+3x^2+(2-x^2)(x^2+2)\] \end{problem} \begin{problem} A man throws a rock off the ledge of a 100 foot high cliff at time $t=0$. The height of the rock above ground level is then given in terms of the time $t$ in seconds by $40t-16t^2+100$. After how many seconds does the rock hit the ground? \end{problem} \begin{problem} Factor $x^3-x^2-16x-16$. \end{problem} \begin{problem} Simplify \[\frac{(z+h)^3-z^3}{h}\] \end{problem} \section{Algebra Diagnostic Test Solutions} \begin{sproblem} Evaluate and express in scientific notation with the correct number of significant figures $\frac{1}{8\cdot 10^{12}}$ and $\frac{160}{.32\cdot 10^{-4}}$. \end{sproblem} \begin{solution} \[\frac{1}{8\cdot 10^{12}}=\frac 18 \times 10^{-12}=.125\times 10^{-12}=\boxed{1.3\times 10^{-13}}\] \[\frac{160}{.32\times 10^{-4}}=\frac{16\times 10^{1}}{32\times 10^{-6}}=.5\times 10^7=\boxed{5\times 10^{6}}\] \end{solution} \begin{sproblem} Jack and Janet are buying an assortment of fruit. Jack wants the number of apples to be three less than twice the number of oranges. Janet wants to buy two more apples than oranges. How many oranges and apples should they get? \end{sproblem} \begin{solution} Let $x$ be the number of apples, $y$ be the number of oranges. Then we have the system of equations \begin{align*} x&=2y-3\\ x&=y+2 \end{align*} Solve simultaneously. Subtracting, $0=y-5$ so $y=5$, $x=7$. Answer: \boxed{5 oranges, 7 apples}. \end{solution} \begin{sproblem} Expand the following polynomial and determine its degree: \[x^4-3x^3+3x^2+(2-x^2)(x^2+2)\] \end{sproblem} \begin{solution} \begin{align*} {}&=x^4-3x^3+3x^2+(2x^2+4-x^4-2x^2)\\ {}&=-3x^3+3x^2+4 \end{align*} \boxed{degree 3} \end{solution} \begin{sproblem} A man throws a rock of the ledge of a 100 foot high cliff at time $t=0$. The height of the rock above ground level is then given in terms of the time $t$ (seconds) by $40t-16t^2+100$. After how many seconds does the rock hit the ground? \end{sproblem} \begin{solution} The problem states the height at time $t$ is $40t-16t^2+100$, and asks what is $t$ when the height is 0? Thus, we solve $-16t^2+40t+100=0\rightarrow 4t^2-10t-25=0$. This gives \[t=\frac{10\pm\sqrt{100-4\cdot 4\cdot (-25)}}{2\cdot 4}=\frac{10\pm\sqrt{500}}{8}=\frac{10}{8}\left(1+\sqrt{5}\right)\] as we reject the negative root. Thus, $\boxed{t=\frac 54\left(1+\sqrt{5}\right)}$. \end{solution} \begin{sproblem} Factor $x^3-x^2-16x+16$. \end{sproblem} \begin{solution} Let $f(x)=x^3-x^2-16x+16$. By inspection, $f(1)=0$, so $x-1$ is a factor. Now $f(x)=(x-1)(x^2-16)$ (determine $x^2-16$ by division or inspection); this factors further to $\boxed{(x-1)(x-4)(x+4)}$. \end{solution} \begin{sproblem} Simplify $\frac{(x+h)^3-x^3}{h}$. \end{sproblem} \begin{solution} \begin{align*} \frac{(x+h)^3-x^3}{h}&=\frac{x^3+3x^2h+3xh^2+h^3-x^3}{h}\\ &=\boxed{3x^2+3xh+h^2} \end{align*} \end{solution} \section{Algebra Diagnostic Test \#2} \begin{problem} $6.25\times 10^{24}$ molecules of water fill a $.2$ liter glass. Approximately how much of this volume in liters does one molecule account for? \end{problem} \begin{problem} Solve for $x$ and $y$ in terms of $a$ and $b$, simplifying your answer as much as possible: \begin{align*} x+2y&=a\\ x-y&=b \end{align*} \end{problem} \begin{problem} Write $(x+1)^3$ as a polynomial in $x$. \end{problem} \begin{problem} Suppose $a$ and $b$ are the distinct roots of $x^2-4x-2$, with $a