Factoring Articles - WyzAnt Tutor Blogshttp://www.wyzant.com/resources/blogs/factoringThis is an aggregate of all of the Factoring articles in WyzAnt.com's Tutors' Blogs. WyzAnt.com is your source for tutors and students.Fri, 26 Dec 2014 16:36:50 -0600http://www.wyzant.com/images/WyzAnt_white200.gifFactoring Articles - WyzAnt Tutor Blogshttp://www.wyzant.com/resources/blogs/factoringhttp://www.wyzant.com/resources/blogs/factoring275864http://www.wyzant.com/resources/blogs/275864/the_right_tool_for_the_jobJohn M.http://www.wyzant.com/resources/users/view/82890510The Right Tool for the Job<div>Back when I was still in middle school, I was sitting at my kitchen table during a family gathering, and my uncle posed the following puzzle for me to solve: A vendor is selling apples for 10 cents apiece, oranges for 5 cents apiece, and peanuts two for a penny. Someone comes along and buys exactly 100 items for exactly one dollar. How many apples, oranges and peanuts did that person buy?</div>
<div> </div>
<div>I took out a sheet of paper and a pencil and came up with the answer in a couple of minutes. This astonished my uncle because, it turns out, he had posed this problem to two adults, including a geometry teacher, and they couldn't solve it in less than a half hour. I had a bit of a reputation for mathematical cleverness, and he had posed this problem to stump me and test the extent of my cleverness. Decades later I still remember exactly how I solved it, probably because it was a boost to my ego to learn that I was apparently smarter than a geometry teacher!</div>
<div> </div>
<div>In this article I will explain exactly how I solved it, and then discuss why it was easier for me than for someone with more experience and education. If you don't already know the answer, spend a few minutes trying to solve it yourself before reading on. Pay special attention to the tools that you bring to bear on the problem.</div>
<div> </div>
<h2>Formulating the problem</h2>
<div>There are six things that we don't know: how many apples, oranges and peanuts were purchased, and how much money was spent on each type of item. However, since we know the price for each type of item, we can express the last three variables in terms of the first three. In other words, if <em>x</em> is the number of apples, <em>y </em>is the number of oranges, and <em>z</em> is the number of peanuts, then the buyer spent <em>10x </em>cents on apples, <em>5y</em> cents on oranges, and <em>z/2</em> cents on peanuts. The two things we do know can be expressed in two equations:</div>
<div> </div>
<div>1. <em>x + y + z = 100</em></div>
<div>2. <em>10x + 5y + z/2 = 100</em></div>
<div> </div>
<div>Equation (1) says the buyer bought exactly 100 items, and equation (2) expresses that the buyer spent exactly 100 cents, or one dollar.</div>
<div> </div>
<div>This is a system of linear equations in three variables. However, there are only two independent equations. This means that the system is <em>underconstrained</em>: there are infinitely many solutions in the real numbers. To solve the problem, we have to recognize that there are additional implicit constraints: first, that <em>x, y </em>and <em>z</em><em> </em>are all <em>non-negative </em> (you can't buy a negative number of items) and second, that they are all <em>natural numbers</em> (you can't buy half an apple or one third of an orange). Since we don't have ha'pennies, we also know that <em>z</em> must be an even number.</div>
<div> </div>
<div>The equations enable us to establish <em>bounds</em> on the values of the variables. Clearly, from equation (1) each of the three variables is less than or equal to 100. From equation (2) we can establish upper bounds for <em>x</em> and <em> y</em> by setting the other variables to zero:</div>
<div> </div>
<div>3. <em>10x ≤ 100 ⇒ x ≤ 10</em></div>
<div>4. <em>5y ≤ 100 ⇒ y ≤ 20 </em></div>
<div> </div>
<div>Since spending all the money on apples or oranges would result in far less than the requisite 100 items, these bounds are in fact strict. This means that the number of apples is one of the numbers {0,1,2,...,9}, and the number of oranges purchased is from the set {0,1,2,...,19}.</div>
<div> </div>
<div>The number of peanuts is likely to be large. If we have to resort to trial and error, it would be best to focus on the smaller numbers of apples and oranges. So, the first thing we can do is to eliminate the <em>z</em> variable by multiplying equation (2) by 2, and subtracting equation (1) from it:</div>
<div> </div>
<div>5. <em>20x + 10y + z = 200</em></div>
<div><em> - <span style="text-decoration: underline;">(x + y + z) = 100</span></em></div>
<div><em> 19x + 9y = 100</em></div>
<div> </div>
<div>We can then solve for <em>y</em> in terms of <em>x</em>:</div>
<div> </div>
<div>6. <em>y = (100 - 19x)/9 = 11 1/9 - (2 1/9)x</em></div>
<div> </div>
<div>It's at this point that can bring to bear the constraint that these numbers are all whole numbers. For <em>y</em> to be a whole number, the fractional part of the constant term, (1/9) must be cancelled by the fractional part of the term containing <em>x</em>. For this to happen, <em>x</em> must be congruent to 1 modulo 9. Among the possible values for <em>x</em>, only the number 1 itself is congruent to 1 modulo 9. Thus, <em>x</em> <em>= 1</em>, and from equation (6), <em>y = 9</em>. Substituting these values into equation (1) yields that <em>z = 90.</em></div>
<div> </div>
<h2>Hammers and Nails</h2>
<div>Why couldn't the adults arrive at this solution as quickly as I did? Was I really so much more clever? I think not. I believe that all three of us were victims of a peculiar human trait: we all have biases regarding what tools we bring to bear on a problem. There is a saying: "To a hammer, everything looks like a nail." We all have favorite tools in our toolbag, and we have specific ideas about when each tool is appropriate.</div>
<div> </div>
<div>Having a favorite tool leads us to apply that tool to just about every problem we confront (every problem looks like a nail for our hammer). Understanding the limitations of our tools can lead us to avoid using a tool because we don't believe it will get us anywhere.</div>
<div> </div>
<div>The fact that the system of linear equations is underconstrained likely lead the adults to abandon the use of algebra prematurely. They recognized that the problem involved the solution of diophantine equations (polynomial equations for which only integer solutions are acceptable), and so they likely sought to apply more sophisticated tools that are appropriate to the solution of diophantine equations (for example, computing the Smith Normal Form to determine two unimodular matrices). </div>
<div> </div>
<div>As a middle school student, I had just learned algebra, and knew nothing about diophantine equations, and so algebra was my shiny new hammer that I tried to apply to all problems. In fact, it could be argued that in choosing to eliminate the <em>z</em> variable, I had made a mistake. If I had continued my bounds analysis just a couple of steps further, I could have solved for <em>z</em>.</div>
<div> </div>
<div>Consider expressing <em>z</em> in terms of the other variables using equation (1):</div>
<div> </div>
<div>7. <em>z = 100 - (x + y)</em></div>
<div> </div>
<div>Since <em>x </em> is at most 9, and <em>y</em> is at most 19, </div>
<address> </address><address>8. (x<em> + y) ≤ 28.</em></address><address>9. z > 72</address><address> </address>
<div>In fact, we know that <em>x</em> and <em>y</em> cannot both have their maximum values at the same time. If the buyer buys 19 oranges, this costs 95 cents, and so there isn't enough money left to buy any apples. If the buyer buys 18 oranges, this costs 90 cents, and so exactly one apple can be purchased, and the total number of items is again 19. Each additional apple purchased requires that two fewer oranges are purchased, so increasing the number of apples reduces the sum. Thus, </div>
<div> </div>
<div>10. <em>(x + y) < 19</em></div>
<div>11. 81 < z < 100<em><br /></em></div>
<div> </div>
<div>Now, from equation (2), </div>
<div> </div>
<div>12. <em>z/2 = 100 - (10x + 5 y)</em></div>
<div>13. <em>z = 200 - 20x - 10y = 10 (20 - 2x - y)</em></div>
<div> </div>
<div>Equation (13) implies that <em>z</em> is an exact multiple of 10. Among the whole numbers between 81 and 100, 90 is the only multiple of 10. Thus, <em>z = 90.</em></div>
<div> </div>
<div>Substituting this value for <em>z</em> into equations (1) and (2) yields a system of two equations in two unknowns that can easily be solved for <em>x</em> and <em>y.</em></div>
<div> </div>
<div>By eliminating <em>z,</em> I had to resort to modular arithmetic to solve the problem. By focusing on <em>z</em>, I could have solved it just using bounds analysis and factoring. I was just lucky that my approach lead to a solution. The more educated mathematicians likely discounted it as a blind alley not worth going down.</div>
<div> </div>
<div>The moral of the story? Consider all tools that might get you to a solution, and don't discount anything until you have spent a little time seeing what it can get you. Sometimes the prima facie wrong path gets you to a solution faster.</div>
<div> </div>
<h2>John Von Neumann</h2>
<div>There is an old, likely apochryphal story that illustrates the penchant of even geniuses to inappropriately apply their favorite tools. A group of psychologists went around among mathematicians and physicists, posing the same problem to both: Two trains start out 100 miles apart heading towards one another on parallel tracks. One train travels at 40mph, the other at 60mph. Just as they begin to move (at the same time), a fly departs from the front of one train, flies at 100mph straight to the front of the other train, then immediately turns around and flies back to the first train. It keeps flying back and forth between the moving trains until the trains begin to pass one another. How far does the fly travel?</div>
<div> </div>
<div>The physicists solved the problem by noting that the trains approach each other at a relative 100mph, and so close the 100 mile distance between them in 1 hour. In that amount of time, the fly will travel 100 miles. This way of formulating the problem solves it in just two multiplications, and so the physicists all answered the question in less than a minute.</div>
<div> </div>
<div>The mathematicians formulated the problem as the sum of a series: they expressed the distance the fly would travel during each leg of its flight, then computed the sum. This approach requires a more complicated formulation, and a more complicated way to compute the solution, so the mathematicians all took several minutes to arrive at an answer.</div>
<div> </div>
<div>When the pyschologists posed the problem to John Von Neumann, a noted mathematician and one of</div>
<div>the founders of computer science, he solved it in under 30 seconds.</div>
<div> </div>
<div>The psychologists were surprised. They told him "for a mathematician, you certainly think like a physicist." He replied, "No way! No physicist could have summed that series as quickly as did!"</div>
<div> </div>Fri, 20 Jun 2014 17:24:15 -05002014-06-20T17:24:15-05:00260028http://www.wyzant.com/resources/blogs/260028/factoring_without_the_guess_and_check_methodShawil D.http://www.wyzant.com/resources/users/view/84848050Factoring without the "Guess and Check" method<div>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:</div>
<div> </div>
<div>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:</div>
<div> </div>
<div>Ex| x<sup>2</sup> + x - 2</div>
<div> </div>
<div>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.</div>
<div> </div>
<div>Let's redo the example, with another method.</div>
<div> </div>
<div>Ex| x<sup>2</sup> + x - 2</div>
<div> </div>
<div>First notice the -2, the negative shows us that the only way to this is (x - ?)(x + ?)</div>
<div>Now we look for the factors of 2. The only factors are 1,2 and 2, 1.</div>
<div>(In another case in which there are more factors, the difference or addition of these two factor must equal the middle term. That determines the correct pair of factors.</div>
<div>Now notice the middle term is positive. That means the larger of the two factors we choose must be positive also.</div>
<div>So, we can now write this as:</div>
<div> </div>
<div>x<sup>2</sup> + 2x - 1x - 2</div>
<div> </div>
<div>What we just did was replace the original "+x" with "2x - 1x". These two statements are equivalent, so this is fair game.</div>
<div> </div>
<div>Now, we group the left and right sides together. Note that when they are grouped, the negative stays with the 1, as shown below:</div>
<div> </div>
<div>(x<sup>2</sup> + 2x) + (-1x - 2)</div>
<div> </div>
<div>Now we factor out common terms from each:</div>
<div> </div>
<div>x is common on the left, and -1 is common on the right, so we factor out each of these:</div>
<div> </div>
<div>x(x + 2) - 1(x + 2)</div>
<div> </div>
<div>Now we check to see if our terms in parenthesis are equal. Since they are, we can group them into one term, and and the outer terms together. These two expressions are then multiplied:</div>
<div> </div>
<div>(x + -1)(x + 2)</div>
<div> </div>
<div>Which equals:</div>
<div> </div>
<div>(x - 1)(x + 2)</div>
<div> </div>
<div>If we foil this out, we get:</div>
<div> </div>
<div>x<sup>2</sup> + x - 2</div>
<div> </div>
<div>Which is our original problem. Therefore we have correctly factored this. And in so doing, we have learned a new way to factor that requires no guessing.</div>Mon, 10 Feb 2014 12:53:30 -06002014-02-10T12:53:30-06:00247050http://www.wyzant.com/resources/blogs/247050/factoring_why_use_itJessica S.http://www.wyzant.com/resources/users/view/84101090Factoring - Why use it?<div>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!</div>
<div> </div>
<div>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.</div>
<div> </div>
<div>For example:</div>
<div> </div>
<div>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.</div>
<div> </div>
<div>We also know that since 20 is an even number, we can divide it evenly by 2. Two times 10 equals 20. Those are our second factors of 20.</div>
<div> </div>
<div>We can also divide 20 by 4 evenly. We end up with 5. 5 x 4 = 20.</div>
<div> </div>
<div><strong>We know we're done factoring when we reach consecutive (numbers that are right after the other) numbers.</strong> Four and five are consecutive numbers.</div>
<div> </div>
<div>Our factors of 20 are 1 x 20, 2 x 10, and 4 x 5. And since mathematicians like things neat and orderly, we re-arrange these numbers from smallest to largest: 1, 2, 4, 5, 10, 20. Voila!</div>
<div> </div>
<div> </div>
<div> </div>
<div>But what about 3? or 6? I mean, theoretically, we can divide 20 by lots of different numbers.</div>
<div> </div>
<div> </div>
<div> </div>
<div>Well, yes and no. We CAN divide 20 by lots of different numbers, but they won't come out as a whole number. We'll end up with a whole bunch of decimals. (Or fractions - whichever you prefer.) And <strong>factors MUST be whole numbers, which means NO DECIMALS OR FRACTIONS.</strong></div>
<div> </div>
<div> </div>
<div>'Ok,' you might be saying, 'so, what? Why do I need to know this?'</div>
<div> </div>
<div> </div>
<div>Here's where factoring gets fun.</div>
<div> </div>
<div>Let's say you're taking a math test with no calculators allowed. Fun, yeah, I know. Stay with me, we're getting there! You're working on a proportions problem and come up with some big weird number like 176/24. You're running out of time on your test and REALLY don't want to divide that bad boy out to simplify, but you don't want to get the problem wrong simply from not simplifying OR make some stupid mistake that will cause you to get the answer wrong anyway. What now?</div>
<div> </div>
<div>FACTOR, FACTOR, FACTOR! Factoring to the rescue!</div>
<div> </div>
<div> </div>
<div>We have two even numbers, which means that they're both divisible by 2 evenly.</div>
<div> </div>
<div>You end up with:</div>
<div> </div>
<div><span style="text-decoration: underline;">2 x 88</span></div>
<div>2 x 12</div>
<div> </div>
<div>Ok, good start. Let's keep going and divide by 2 again.</div>
<div> </div>
<div><span style="text-decoration: underline;">2 x 2 x 44</span></div>
<div>2 x 2 x 6</div>
<div> </div>
<div>Now, do we HAVE to divide by the same number on both? Heck no, we're factoring! Your largest number, 44, has factors of 4 and 11. And, we know that 2 x 2 is 4 (make sure you keep track of all those 2's). When we have the whole thing factored all the way out, we end up with:</div>
<div> </div>
<div><span style="text-decoration: underline;">2 x 2 x 2 x 2 x 11</span></div>
<div>2 x 2 x 2 x 3</div>
<div> </div>
<div>So... how about all those 2's? Since we're working with a fraction, we can actually cancel some of those 2's out, since we know that dividing any number by itself equals 1. (2 ÷ 2 = 1)</div>
<div> </div>
<div><span style="text-decoration: underline;"><span style="text-decoration: line-through;">2 x 2 x 2</span> x 2 x 11</span><br /><span style="text-decoration: line-through;">2 x 2 x 2</span> x 3</div>
<div> </div>
<div>We come out with 22/3, or 7.33. That's a lot easier to divide out, now isn't it?</div>
<div> </div>
<div> </div>
<div>Obviously, you'll need some practice before using this as a time-saver on a test. This method does require knowledge on your part on how to factor as well as how to divide and multiply. It's a fabulous way to check your long division.</div>
<div> </div>
<div>Eventually, you will use this technique in more advanced math - like quadratics and square roots. And as a final note, you do NOT necessarily need to factor all the way down to prime numbers, as long as you're sure your math is correct. </div>Tue, 03 Dec 2013 00:45:51 -06002013-12-03T00:45:51-06:00243619http://www.wyzant.com/resources/blogs/243619/interesting_problem_from_a_section_on_trigonometric_identitiesSkyler H.http://www.wyzant.com/resources/users/view/82381760Interesting problem from a section on Trigonometric Identities<div>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.</div>
<div> </div>
<div>The problem was presented as such:</div>
<div> </div>
<div> </div>
<div> </div>
<div>Factor and simplify the following using Trigonometric and Pythagorean Identities:</div>
<div> </div>
<div>sec<sup>3</sup>(x) - sec<sup>2</sup>(x) - sec(x) + 1</div>
<div> </div>
<div> </div>
<div> </div>
<div>We tried a couple of different approaches, such as factoring sec(x) from each term:</div>
<div> </div>
<div>sec(x) * [ sec<sup>2</sup>(x) - sec(x) - 1 + 1/sec(x) ]</div>
<div> </div>
<div>and factoring sec<sup>2</sup>(x) from each term:</div>
<div> </div>
<div>sec<sup>2</sup>(x) * [ sec(x) - 1 - 1/sec(x) + 1/sec<sup>2</sup>(x) ]</div>
<div> </div>
<div>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 simple approach that I had overlooked.</div>
<div> </div>
<div>Instead of attempting to draw a factor from all four terms at once, we draw one factor from two of the terms and another factor from the other two terms. So from our original problem:</div>
<div> </div>
<div>sec<sup>3</sup>(x) - sec<sup>2</sup>(x) - sec(x) + 1</div>
<div> </div>
<div>We can rearrange the grouping of the terms, which is allowed by the Commutative Property of Addition:</div>
<div> </div>
<div>sec<sup>3</sup>(x) - sec(x) - sec<sup>2</sup>(x) + 1</div>
<div> </div>
<div>Looking at these two groups of terms, we can see that sec(x) can be factored from the first group. </div>
<div> </div>
<div>sec(x) * ( sec<sup>2</sup>(x) - 1 ) - sec<sup>2</sup>(x) + 1</div>
<div> </div>
<div>Now it becomes apparent that we can relate the second group of terms to the factoring we've just performed with a little trick: factoring -1 from the expression "-sec<sup>2</sup>(x) + 1":</div>
<div> </div>
<div>sec(x) * ( sec<sup>2</sup>(x) - 1 ) - 1 * ( sec<sup>2</sup>(x) - 1 )</div>
<div> </div>
<div>Now we have the same expression factored from both of the groups of two terms! Let's pull this factor out:</div>
<div> </div>
<div>( sec<sup>2</sup>(x) - 1 ) * ( sec(x) - 1 )</div>
<div> </div>
<div>We now have two approaches we can take to further simplify this problem. </div>
<div> </div>
<div> </div>
<div> </div>
<div>Method 1, Factoring the Difference of Squares:</div>
<div> </div>
<div>( sec<sup>2</sup>(x) - 1 ) * ( sec(x) - 1 )</div>
<div> </div>
<div>We can look at the first term (sec<sup>2</sup>(x) - 1) and see that it is the difference of squares; it is equivalent to (sec<sup>2</sup>(x) - 1<sup>2</sup>). Knowing this, we could factor the expression one step further to be left with:</div>
<div> </div>
<div>(sec(x) + 1)(sec(x) - 1)(sec(x) - 1)</div>
<div>(sec(x) + 1)(sec(x) - 1)<sup>2</sup></div>
<div> </div>
<div> </div>
<div> </div>
<div>Method 2, Relating to a Pythagorean Identity:</div>
<div> </div>
<div>( sec<sup>2</sup>(x) - 1 ) * ( sec(x) - 1 )</div>
<div> </div>
<div>We could alternatively take a look at this equation and recognize the presence of a Pythagorean Identity. The second Pythagorean Identity, tan<sup>2</sup>(θ) + 1 = sec<sup>2</sup>(θ), can easily be arranged to show that:</div>
<div> </div>
<div>tan<sup>2</sup>(θ) = sec<sup>2</sup>(θ) - 1</div>
<div> </div>
<div>We now see that tan<sup>2</sup>(x) can be substituted in for the first term in our equation:</div>
<div> </div>
<div>tan<sup>2</sup>(x) * ( sec(x) - 1 )</div>
<div> </div>
<div> </div>
<div> </div>
<div>While either of these methods is correct, it is my personal opinion that factoring the difference of squares is the more elegant expression. What do you think? Is there another approach we could have used? A more elegant solution?</div>
<div> </div>
<div>Thanks for reading and good luck with your studies,</div>
<div> </div>
<div>-Skyler</div>Fri, 08 Nov 2013 18:38:16 -06002013-11-08T18:38:16-06:00