CHAPTER
2

Polish Up Your Algebra Skills

In This Chapter

• Creating linear equations
• The properties of exponents
• Factoring polynomials
• Solving quadratic equations

If you are an aspiring calculus student, somewhere in your past you probably had to do battle with the beast called algebra. Not many people have positive memories associated with their algebraic experiences, and I am no different. Forget the fact that I was a math major, a calculus teacher, and even took my calculator to bed with me when I was young (a true but very sad story). I hated algebra for many reasons, not the least of which was that I felt I could never keep up with it. Every time I seemed to understand algebra, we’d be moving on to a new topic much harder than the last.

Being an algebra student is sort of like battling a famous boxer. Here is this champion of mathematical reasoning that has stood unchallenged for hundreds of years, and you’re in the ring going toe-to-toe with it. You never really reach back for that knockout punch because you’re too busy fending off your opponent’s blows. When the bell rings to signal the end of the fight, all you can think is “I survived!” and hope that someone can carry you out of the ring.

Perhaps you didn’t hate algebra as much as I did. You might be one of those lucky people who understood algebra easily. You are very lucky. For the rest of us, however, there is hope. Algebra is much easier in retrospect than when you were first being pummeled by it. As calculus is a grand extension of algebra, you will, of course, need a large repertoire of algebra skills. So it’s time to slip those old boxing gloves back on and go a few rounds with your old sparring partner. The good news is you’ve undoubtedly gotten stronger since the last bout. If, however, a brief algebra review is not enough for you, pick up this book’s prequel, The Complete Idiot’s Guide to Algebra, by yours truly.

Walk the Line: Linear Equations

Graphs play a large role in calculus, and the simplest of graphs, the line, surprisingly pops up all the time. As such, it is important that you can recognize, write, and analyze graphs and equations of lines. To begin, remember that a line’s equation always has three components: two variable terms and a constant (numeric) term. One of the most common ways to write an equation is in standard form.

Common Forms of Linear Equations

A line in standard form looks like this: Ax + By = C. In other words, the variable terms are on the left side and the number is on the right side of the equal sign. Also, to officially be in standard form, the coefficients (A, B, and C) must be integers, and A is supposed to be positive. What’s the purpose of standard form? A linear equation can have many different forms (for example, x + y = 2 is the same line as x = 2 – y). However, once in standard form, all lines with the same graph have the exact same equation. Therefore, standard form is especially handy for instructors; they’ll often ask that answers be put into standard form to avoid alternate correct answers.

There are two major ways to create the equation of a line. One requires that you have the slope and the y-intercept of the line. Appropriately enough, it is called slope-intercept form: y = mx + b. In this equation, m represents the slope and b the y-intercept. Notice the major characteristic of an equation in slope-intercept form: it is solved for y. In other words, y appears by itself on the left side of the equation.

Example 1: Write the equation of a line with slope –3 and y-intercept 5.

Solution: In slope intercept form, m = –3 and b = 5, so plug those into the slope-intercept formula:

y = mx + b
y = –3x + 5

Another way to create a linear equation requires a little less information—only a point and the slope (the point doesn’t have to be the y-intercept). This (thanks to the vast creativity of mathematicians) is called point-slope form. Given the point (x₁, y₁) and slope m, the equation of the resulting line will be y – y₁ = m(x – x₁).

You will find this form extremely handy throughout the rest of your travels with calculus, so make sure you understand it. Don’t get confused between the x’s and x₁’s or the y’s and the y₁’s. The variables with the subscript represent the coordinates of the point you’re given. Don’t replace the other x and y with anything—these variables are left in your final answer. Watch how easy this is.

Example 2: If a line g contains the point (–5,2) and has slope , what is the equation of g in standard form?

Solution: Because you are given a slope and a point (which is not the y-intercept), you should use point-slope form to create the equation of the line. Therefore, , x₁ = –5, and y₁ = 2. Plug these values into point-slope form and get:

If this equation is supposed to be in standard form, you’re not allowed to have any fractions. Remember that the coefficients have to be integers, so to get rid of the fractions, multiply the entire equation by 5:

Now, move the variables to the left and the constants to the right and make sure the x term is positive; this puts everything in standard form:

x + 5y = 5

Calculating Slope

You might have noticed that both of the ways we use to create lines absolutely require that you know the slope of the line. The slope of the line is that important (almost as important as wearing both shoes and a shirt if you want to buy a Slurpee at 7-Eleven). The slope of a line is a number that describes precisely how “slanty” that line is—the larger the value of the slope, the steeper the line. Furthermore, the sign of the slope (in most cases Capricorn) will tell you whether or not the line rises or falls as it travels.

Figure 2.1

Calculating the slope of a line.

As shown in Figure 2.1, lines with shallower inclines have smaller slopes. If the line rises (from left to right), the slope is positive; if, however, it falls from left to right, the slope is negative. Horizontal lines have 0 slope (neither positive nor negative), and vertical lines are said to have an undefined slope, or no slope at all.

It is very easy to calculate the slope of any line: find any two points on the line, (a,b) and (c,d), and plug them into this formula:

In essence, you are finding the difference in the y’s and dividing by the difference in the x’s. If the numerator is larger, the y’s are changing faster, and the line is getting steeper. On the other hand, if the denominator is larger, the line is moving more quickly to the left or right than up and down, creating a shallow incline.

You should also remember that parallel lines have equal slopes, whereas perpendicular lines have slopes that are negative reciprocals of one another. Therefore, if line g has slope , then a parallel line h would have slope also; a perpendicular line k would have slope .

Example 3: Find the equation of line j given that it is parallel to the line 2x – y = 6 and contains the point (–1,1); write j in slope-intercept form.

Solution: This problem requires you to create the equation of a line, and you’ll find that the best way to do this every time is via point-slope form. So you need a point and a slope. Well, you already have the point: (–1,1). Using your keen sense of deduction, you know that only the slope is left to find and that’ll be that. But how to find the slope?

If j is parallel to 2x – y = 6, then the lines must have the same slope, so what’s the slope of 2x – y = 6? Here’s the key: if you solve it for y, it will be in slope-intercept form, and the slope, m, is simply the coefficient of x. When you do so, you get y = 2x – 6. Therefore, the slope of both lines is 2, and you can use point-slope form to write the equation of j:

Solve for y to put the equation in slope-intercept form:

Interpreting Linear Graphs

Calculus has undergone a renaissance over the last few decades, as researchers have gained more insight about the most effective way to present and learn mathematical material. Without climbing onto a soapbox, allow me to present the “bottom line”: you need to learn the concepts behind the math, not just memorize a series of steps to reach a solution.

Understanding based on pure memorization is fragile—without constant practice, it shatters. Therefore, throughout this book (and most likely throughout your calculus course) you will be presented with nontraditional problems, including problems presented graphically. If these types of problems feel strange, don’t worry. They are meant to stretch your understanding of the topic at hand, and upon wrestling with them for a bit, they provide you with a deeper and longer-lasting mastery of mathematics.

In the next example, you’re not given a specific slope, intercept, or point to make a line. Instead, you’re given a graph. All the information you need is in there—you just have to harvest it yourself.

Example 4: The graph of line p is presented in Figure 2.2. Express the equation of the line perpendicular to p with the same x-intercept in standard form.

Figure 2.2

The graph of line p.

Solution: Before you jump into the solution, analyze the graph of the line and collect all the information you can, even if it later proves unnecessary. Here are my observations:

• The slope of p must be negative, because the line travels down as you move from left to right.
• The x-intercept of line p is 1, because it passes through the x-axis at point (1,0).
• The y-intercept of line p is 3, because it passes through the y-axis at point (0,3).

None of those are earth-shattering observations by any means, but that is all you need to solve the problem. You’re asked to find the equation of the line perpendicular to p, which means the slope of that line is the opposite reciprocal of the slope of p. You are given two points through which p passes, so apply the slope formula with (a,b) = (1,0) and (c,d) = (0,3):

The slope of p is –3. My assumption that the slope was going to be negative was correct, but you didn’t doubt me for a moment, did you? The slope of the line perpendicular to p must be the opposite reciprocal of –3, which is .

According to the problem, the new line shares the same x-intercept as p, so the new line must also pass through point (1,0). You now know the slope of the line you are creating (m = ) and a point on the line (x₁,y₁) = (1,0). Apply the point-slope formula.

The line needs to be in standard form, so no fractions are allowed. Multiply both sides of the equation by 3 to eliminate the fractions.

A linear equation in standard form has the variable terms on the left side of the equation and the constant (number term) on the right. Subtract x from both sides.

–x + 3y = –1

Almost finished! A line in standard form must have a positive x-coefficient, so multiply everything by –1:

The equation of the line perpendicular to line p that passes through the same x-intercept is x – 3y = 1.

You’ve Got the Power: Exponential Rules

I find that exponents are the bane of many calculus students. Whether they never learned exponents well in the first place or simply make careless mistakes, exponential errors are a treasure trove of frustration. Therefore, it’s worth your while to spend a few minutes and refresh yourself on the major exponential rules. You may find this exercise empowering.

• Rule one:

Explanation: If you multiply two terms with the same base (here it’s x), add the powers and keep the base. For example, .

• Rule two:

Explanation: This is the opposite of rule one. If you divide (instead of multiply) two terms with the same base, then you subtract (instead of add) the powers and keep the base. For example, .

• Rule three:

Explanation: A negative exponent indicates that a variable is in the wrong spot, and belongs in the opposite part of the fraction, but it only affects the variable it’s touching. For example, in the expression , only the y is raised to a negative power, so it needs to be in the opposite part of the fraction. Correctly simplified, that fraction looks like this: . Note that the exponent becomes positive when it moves to the right place. Remember that a happy (positive) exponent is where it belongs in a fraction.

• Rule four: (x^a)^b = x^ab

Explanation: If an exponential expression is raised to a power, you should multiply the exponents and keep the base. For example, (h⁷)³ = h²¹.

• Rule five:

Explanation: The numerator of the fractional power remains the exponent. The denominator of the power tells you what sort of radical (square root, cube root, etc.). For example, 4^3/2 can be simplified as either or . Either way, the answer is 8.

Example 5: Simplify xy^1/3(x²y)³.

Solution: Your first step should be to raise (x²y) to the third power. You have to use rule four twice (the current exponent of y is understood to be 1 if it is not written). This gives you . The problem now looks like this: .

To finish, you have to multiply the x’s and y’s together using rule one:

Breaking Up Is Hard to Do: Factoring Polynomials

Factoring is one of those things you see over and over in algebra. I have found that even among my students who disliked math, factoring was popular; it’s something that some people just “got,” even when most everything else escaped them. This is not the case, however, in many European schools, a fact that surprised my colleagues and me when I was a high school teacher.

Canadian exchange students gave me blank stares when we discussed factoring in class. This is not to say that these students were not extremely intelligent (they were); they just used other methods. However, factoring comes in handy throughout calculus, so I deem it important enough to cover here. Call it patriotism.

Factoring is basically reverse multiplying—undoing the process of multiplication to see what was there to begin with. For example, you can break down the number 6 into factors of 3 and 2, since 3 ∙ 2 = 6. There can be more than one correct way to factor something.

Greatest Common Factor

Use the greatest common factor method of factoring if you have terms with elements in common. It’s easier than it sounds. Take the expression 4x + 8.

Notice that both terms can be divided by 4, making 4 a common factor. Therefore, you can write the expression in the factored form of 4(x + 2).

In effect, I have “pulled out” the common factor of 4, and what’s left behind are the terms once 4 has been divided out of each. In this type of problem, ask yourself, “What do each of the terms have in common?” and then pull that greatest common factor out of each to write your answer in factored form.

Special Factoring Patterns

You should feel comfortable factoring trinomials such as x² + 5x + 4 using whatever method suits you. Most people play with binomial pairs until they stumble across something that works, in this case (x + 4)(x + 1), whereas others undertake more complicated means. Regardless of your personal “flair,” there are some patterns you should have memorized:

• Difference of perfect squares: a² – b² = (a + b)(a – b)

Explanation: A perfect square is a number like 16, which can be created by multiplying something times itself. In the case of 16, that something is 4, since 4 times itself is 16. If you see one perfect square being subtracted from another, you can automatically factor it using the pattern above. For example, x² – 25 is a difference of x² and 25, and both are perfect squares. Thus, it can be factored as (x + 5)(x – 5).

• Sum of perfect cubes: a³ + b³ = (a + b)(a² – ab + b²)

Explanation: Perfect cubes are similar to perfect squares. The number 125 is a perfect cube because 5 ∙ 5 ∙ 5 = 125. This pattern is a little clumsier to memorize, but it can come in handy occasionally. This formula can be altered just slightly to factor the difference of perfect cubes, as illustrated in the next bullet. Other than a couple of sign changes, the process is the same.

• Difference of perfect cubes: a³ – b³ = (a – b)(a² + ab + b²)

Explanation: Enough with the symbols for these formulas—let’s do an example.

Example 6: Factor x³ – 27 using the difference of perfect cubes factoring pattern.

Solution: Note that x is a perfect cube since x ∙ x ∙ x = x³, and 27 is also, since 3 ∙ 3 ∙ 3 = 27. Therefore, x³ – 27 corresponds to a³ – b³ in the formula, making a = x and b = 3. Now, all that’s left to do is plug a and b into the formula:

You cannot factor (x² + 3x + 9) any further, so you are finished.

Solving Quadratic Equations

Before you put algebra in the rearview mirror, there’s one last stop. Sure, you’ve been able to solve equations like x + 9 = 12 forever, but when the equations get a little trickier, maybe you get a little panicky. Forgetting how to solve quadratic equations (equations whose highest exponent is a 2) has distinct symptoms: dizziness, shortness of breath, nausea, and loss of appetite. To fight this ailment, take the following 3 tablespoons of quadratic problem solving and call me in the morning.

Every quadratic equation can be solved with the quadratic formula (method three, which follows), but it’s important that you know the other two methods as well. Factoring is undoubtedly the fastest of the three methods, so you should try it first. Few people choose completing the square as their first option, but it (like the quadratic formula) works every time, though it requires a few more steps than its counterpart. However, you have to learn completing the square, because it pops up later in calculus, when you least expect it.

Method One: Factoring

To begin, set your quadratic equation equal to 0; this means add and subtract the terms as necessary to get them all to one side of the equation. If the resulting equation is factorable, factor it and set each individual term equal to 0. These little baby equations will give you the solutions to the equation. That’s all there is to it.

Example 7: Solve the equation 3x² + 4x = –1 by factoring.

Solution: Always start the factoring method by setting the equation equal to 0. In this case, start by adding 1 to each side of the equation: 3x² + 4x + 1 = 0.

Now, factor the equation and set each factor equal to 0. This creates two cute little mini-equations that need to be solved, giving you the final answer:

This equation has two solutions: or x = –1. You can check them by plugging each separately into the original equation, and you’ll find that the result is true.

Method Two: Completing the Square

As I mentioned earlier, this method is a little trickier than the other two, but you really do need to learn it now, or you’ll be coming back to figure it out later. I’ve discovered that it’s best to learn this method in the context of an example, so let’s go to it.

Example 8: Solve the equation 2x² + 12x – 18 = 0 by completing the square.

Solution: In this method, unlike factoring, you want the constant separate from the variable terms, so move the constant to the right side of the equation by adding 18 to both sides:

2x² + 12x = 18

This is important: For completing the square to work, the coefficient of x² must be 1. In this case, it is 2, so to eliminate that pesky coefficient, divide every term in the equation by 2:

x² + 6x = 9

Here’s the key to completing the square: take half of the coefficient of the x term, square it, and add it to both sides. In this problem, the x coefficient is 6, so take half of it (3) and square that (3² = 9). Add the result (9) to both sides of the equation:

At this point, if you’ve done everything correctly, the left side of the equation will be factorable. In fact, it will be a perfect square!

To solve the equation, take the square root of both sides. That will cancel out the exponent. Whenever you do this, you have to add a ± sign in front of the right side of the equation. This is always done when square rooting both sides of any equation:

To solve for x, subtract 3 from each side, and that’s it. It would also be good form to simplify into :

Method Three: The Quadratic Formula

The quadratic formula is one-stop shopping for all your quadratic equation needs. All you have to do is make sure your equation is set equal to 0, and you’re halfway there. Your equation will then look like this: ax² + bx + c = 0, where a, b, and c are the coefficients as indicated. Take those numbers and plug them straight into this formula (which you should definitely memorize):

You’ll get the same answer you would achieve by completing the square. Just to convince you that the answer’s the same, we’ll do the problem in Example 8 again, but this time with the quadratic formula.

Example 9: Solve the equation 2x² + 12x – 18 = 0, this time using the quadratic formula.

Solution: Because the equation is already set equal to 0, it is in form ax² + bx + c = 0, and a = 2, b = 12, and c = –18. Plug these values into the quadratic formula and simplify:

So although there are fewer steps to the quadratic formula, there is some room for error during computation. You should practice both methods, but primarily use the one that feels more comfortable to you.

Synthesizing the Quadratic Solution Methods

Whenever you are learning (or reviewing) mathematical techniques, it’s easy to get lost in the details. While it’s true that many solution methods require you to understand and follow a series of steps, math is more than a process to follow. It is not merely a numbered list of commands to execute robotically.

You might be asking yourself, “When do I know which technique to use?” Well, you should always try factoring first, because it’s usually the easiest method. If factoring doesn’t work, then which should you choose: the quadratic formula or completing the square?

Honestly, the choice is yours. Select the method that feels most comfortable to you, because although the methods are very different in process, they are actually more related than they may at first appear. In fact, did you know that if you ever forget the quadratic formula, you can generate it from scratch? Just complete the square on the generic quadratic equation ax² + bx + c = 0.

Example 10: Generate the quadratic formula by completing the square for ax² + bx + c = 0.

Solution: The coefficient of x² must be 1 in order to complete the square, so divide each term by the current coefficient (a) and then move the constant (the term with no x-part) to the right side of the equation by subtracting:

According to the technique described in Example 8, you must take half of the x-coefficient , square it , and add the result to both sides of the equation.

The left side of the equation is a perfect square. Notice that the term inside the squared quantity is , the value that you squared just a moment ago.

In order to add the fractions on the right side, you will need common denominators.

You may have noticed the familiar “b² – 4ac” from the quadratic formula. You’re almost done! To solve for x, take the square root of both sides of the equation.

Subtract from both sides to solve for x, and you’ve generated the quadratic formula as if by magic!

The moral of this story: although the solution methods for quadratic equations may look quite different from one another, they have a lot in common.

Before wrapping up our discussion on quadratic equations, let’s review the relationship between the factors and x-intercepts of a quadratic equation.

Example 11: Create a quadratic equation that has x-intercepts x = –2 and x = 5.

Solution: First, a warning: there are many possible correct answers, but this is the easiest solution. You may want to look back at Example 7 for a moment, because you’re going to follow that process in reverse.

If x = –2 is a solution for the quadratic equation, then you could add 2 to both sides of the equation to get an equivalent equation:

Similarly, you could subtract 5 from both sides of the equation x = 5:

Notice that (x + 2) and (x – 5) could be factors of the quadratic equation you’re looking for, because they both equal zero. If one of the factors equals 0, then multiplying anything by that factor also gives you zero.

Why is this good news? If the entire equation equals 0, then you’ve found a root, an x-intercept. Therefore, the simplest quadratic equation with x-intercepts –2 and 5 would be the product of (x + 2) and (x – 5):

Quadratic equation y = x² – 3x – 10 has x-intercepts x = –2 and x = 5. If you’re skeptical, substitute them back into the equation to verify. You’ll get y = 0, which means points (–2,0) and (5,0) lie on the graph (and also on the x-axis).

The Least You Need to Know

• Basic equation solving is an important skill in calculus.
• Reviewing the five exponential rules will prevent arithmetic mistakes in the long run.
• You can create the equation of a line with just a little information using point-slope form.
• There are three major ways to solve quadratic equations, each important for different reasons.