Factoring can be quite difficult for those who are new to the concept. There are many ways to go about it. The guess and check way seems to be the most common, and in my mind, it is the best, especially if one wants to go further into mathematics, than Calculus 1. But for those just getting through a required algebra course, here is another way to consider, that I picked up while tutoring some time ago: If you have heard of factor by grouping, then this concept will make some sense to you. Let's use an example to demenstrate how to do this operation: Ex| x2 + x - 2 With this guess and check method, we would use (x + 1)(x - 2) or (x + 2)(x - 1). When we "foil" this out, we see that the second choice is the correct factorization. But, instead of just using these guesses, why not have a concrete way to do this. Let's redo the example, with another method. Ex| x2 + x - 2 First... read more
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Whenever math gives you an easy topic, take it and run with it! I am sometimes surprised by how many students struggle (really struggle) with factoring. That said, I always struggled with quadratics when I was in high school. And forgetting to move negative signs and decimals when simplifying (whoops!). Fortunately, I've gotten past that. My point is that even tutors and teachers start somewhere, and common mistakes are called 'common mistakes,' for a reason. Each of us has an area where we are just not as strong. No shame in that! Factoring a number is nothing more than pulling apart its pieces. Glorified division, if you will. The difference with factoring is that you are dividing a given number by one of its smaller parts. For example: Let's use the number 20. We know from multiplication that we can multiply every number by 1, and come out with the same number. Our first two factors of 20 are 1 and 20. We also know... read more
I was working with a student today, and as we worked through the section in his book dealing with Trigonometric Identities and Pythagorean Identities, we stumbled across a problem that gave us a bit of trouble. The solution is not so complicated, but it sure had us stumped earlier. The problem was presented as such: Factor and simplify the following using Trigonometric and Pythagorean Identities: sec3(x) - sec2(x) - sec(x) + 1 We tried a couple of different approaches, such as factoring sec(x) from each term: sec(x) * [ sec2(x) - sec(x) - 1 + 1/sec(x) ] and factoring sec2(x) from each term: sec2(x) * [ sec(x) - 1 - 1/sec(x) + 1/sec2(x) ] We followed these approaches through a few steps, but nothing we were attempting led to the solution. After doing some reading online, I found that the solution required a... read more