Click here for the most recent list of errors in Pre-Calculus Workbook for Dummies.
I’m not typically a reviewer of books, but I’m going to make an exception in this case. My shiny, new, first-edition copy of the Pre-Calculus Workbook for Dummies (Burger, Neal, Gilman; Wiley Publishing Inc.; 2009) showed up yesterday, and my experience has been less than positive. I already own Pre-Calculus for Dummies, but was getting weary from scouring the web for worksheets and quizzes to reinforce everything I was re-learning. (It’s amazing how much you can forget after 10 years outside of the classroom.) This workbook was going to allow me to be sure that I was grasping all of the material, so that I could finally crack open that Calculus I textbook and get a jump-start on the fall semester.
I was very impressed with the first few problems, as they didn’t waste time with simple examples. These were nitty-gritty, down-and-dirty problems that put my mettle to the test. So far, so good. Then we hit Chapter 2, and things went downhill fast. In the span of 3 pages, there were 4 errors in either examples or practice problems. The first time I went through, I assumed it had to be me that was in error. Granted, it’s a first edition book, but surely it was proofread well enough to avoid that many glaring mistakes. I went back and tried the problems again….still no dice. I read the step-by-step solutions to the problems, and was left more confused than when I started. Fortunately, I have a strong enough background in the basics, that I was ultimately able to determine that the mistakes weren’t on my end. I verified with a few of my mathematically inclined friends that, somehow, the folks at Wiley Publishing put out a remedial math book without bothering to check their own answers first. If I had been addressing the material for the first time, I would have given up in frustration by page 30.
I searched the internet for any errata for the book, but came up empty handed. The book is so new that there aren’t even any reviews on Amazon yet. So, instead of merely complaining about the book, I’ll try to be useful. I’ll use this blog post to list any errors I come across in the text, along with what I hope are understandable corrections. I’ll be the first to admit that I’m still very much in the process of relearning this material, so if you happen to notice something wrong with one of my corrections, please let me know. Also, if you’re using the book and find something that’s not quite right, send a comment, and I’ll look into it and add it to the page. Wiley Publishing doesn’t seem to have any documentation of errors in their books, so let’s all work together to fill in the gap.
I’m still convinced that the Pre-Calculus Workbook for Dummies could be one of the best resources for people interested in learning (or relearning) the math skills that will get them ready for college level calculus. It is shame that poor proofreading has turned it into such a source of frustration.
(Changes to the text are in yellow italics.)
Cheat Sheet (Tear-out page at beginning of book) – Sum-and-difference formulas
They got sin(x±y) right, but messed up cos(x±y) and tan(x±y).
These should read:

Page 27 – Problem #4
Solve for x in x3 – 5x > 4x2 . The answer is -1 < x < 0 or x > 5
The solution on page 36 says to subtract 4x2 from both sides of the equation*. It then proceeds to add 4x2 to the left of the inequality, instead of subtracting, as it just instructed. This, of course, will yield an incorrect answer.
The corrected Answer should read:
For this problem, start by gathering all your variables to one side of the inequality by subtracting 4x2 from each side: x3 - 4x2 – 5x > 0. Next, factor out x from each term: x(x2 - 4x – 5) > 0. Then factor the quadratic: x(x – 5)(x + 1) > 0. Setting your factors… Therefore your solution is -1 < x < 0 or x > 5.
*This problem is not an equation; it is an inequality. Remember: no “=”, no equation.
Page 29 – Example 1 (Top left corner of the page)
Write the solution for 5 – 2x > 4 in interval notation. The answer is (- ∞, 1/2)
It looks like they forgot to flip the inequality symbol when dividing both sides of the inequality by the negative 2. The way to get the correct answer is:
5 – 2x > 4
- 2x > -1
x < 1/2
Page 29 – Problem #6
Write the solution for the solution of x3 – 5x > 4x2 in interval notation, and graph the solution on a number line.
The solution on page 36 wittily points out that they’ve just given you this problem to solve in #4. It neglects, however, to correct the errors in the solution of #4. (Click here for a detailed solution to the inequality.) As for the correct answer, it should be (-1, 0)∪(5, ∞). The graph of the corrected solution looks like this:

Graphed at www.webgraphing.com. (Incredible math resource. Check it out!)
Page 29 – Problem #7
Graph the interval set (- ∞, -7)∪[-5, 2)∪(4, ∞) on a number line.
The solution on page 37 is incorrect, as they use an open circle instead of a solid circle at -5. Since the bracket before the 5 indicates a closed interval, that portion of the graph should look like this:

Graphed at www.webgraphing.com
Page 29 – Problem #8
Graph the solution of |2x -1| ≤ 3.
The solution on page 37 is correct, however the graph is incorrect. Because we are talking about closed intervals, the circles should be solid, not open as they are illustrated.
Page 31 – Example 2 (top right of page)
Solve for x in √2x -1 + 4 = x + 2
The answer is ultimately correct, but the way they got it contained an error. Towards the bottom of the solution, you are instructed to check for extraneous roots by plugging in your two possible solutions. In the right column, they replace both instances of x with 5, then solve down to 7=7. This part is correct. The substitution in the second column is wrong, however. Instead of replacing all instances x with 1, they replace the first x with 1, and the second x with 5. The solution should look like this:
√2(1) -1 + 4 = 1 + 2
√2-1 + 4 = 3
√1 + 4 = 3
1 + 4 = 3
5 ≠ 3
In the end, x=1 is still an extraneous root, but the substitution in the example was done incorrectly.
Page 36 – The explanation of the answer for problem 17
The answer is ultimately correct, but the explanation has a typo that might throw you off. The second sentence reads “Notice the 2 3√2 in both the numerator and the denominator.” You don’t notice this, because it’s not true. The sentence should read “Notice the 3√2 in both the numerator and the denominator.” From there on, the explanation will make sense.
Page 50 – Last paragraph (finding the oblique asymptote)
They leave out an important part of the description of how to find the oblique asymptote. The last sentence should read “The quotient you find (ignoring the remainder) is the equation you graph as an oblique asymptote.” Knowing that you should disregard the remainder is crucial, and will come in handy when you try to solve problem 14.
Page 65 – The graph of the answer for problem 12
I have no idea where they came up with the graph used in the answer. The asymptotes don’t correspond with the explanation, and the graph in the (-2,3) interval is drawn as a perfect parabola, when the points you get from solving the equation in that range indicate it is not. Here is the corrected graph, with vertical asymptotes placed at -2 and 3:

Notice the graph in the (-2,3) area is not a perfect parabola as was indicated on pg. 65.
Page 66 – The explanation and graph for problem 14
The explanation isn’t so much wrong, as it is incomplete. When describing the process of finding oblique asymptotes on page 50, they forgot to mention that you discard the remainder when finding the quotient of the numerator and denominator. This might leave you a little confused about what to do with the -13/(x + 1) you have as a remainder when you are done dividing. Just ignore it entirely and use x – 11 to graph your oblique asymptote. The second sentence should read, “Use long division (ignoring the remainder) to find the quotient x – 11…”
Also, the graph is inaccurate. When they graphed the oblique asymptote (y = x – 11), they gave it a y-intercept of -10. The asymptote should, obviously, cross the y-axis at -11. The values that they have graphed at x = -4 and x = 2 are off by 1 as well. The corrected graph should look like this:

The oblique asymptote has a y-intercept of -11 and and x-intercept of 11.
I’m still pretty early in the book, so there’s not much here yet. I’ll add more as I come across or am made aware of any other errors. In my wildest dreams, someone from Wiley will call up to offer me a proofreading credit in the next edition of the book. (Wouldn’t that look impressive on my resume?) More likely, I’ll be asked to pull the page, but until then, I’ll keep at it.