Factoring a Quadratic with FractionsDate: 10/08/1999 at 23:08:27 From: Jill Fiebelkorn Subject: Algebra I am stuck on factoring expressions involving fractions. The problem is: x^2 + (1/2)x + (1/16) Date: 11/14/1999 at 20:45:02 From: Doctor Sandi Subject: Re: Algebra Hi Jill, Let's see if I can help you. When factoring ANY quadratic, you are looking for two numbers that will: (a) multiply together to give the end number in the expression (b) add/subtract together to give the middle number. This applies for any quadratic polynomial as long as the coefficient (the number in front) of x^2 is 1, which in this case it is. So, what numbers multiply to give (1/16) and add (in this case) to give (1/2)? As you do this, you have to keep your eye on the signs that you want to end up with as well. In this case you have only + so there are no problems with that. For now, let's think of common factors of 16 and 2 (the two denominators of our fractions.) What about the number 4? 4 * 4 = 16 and 2 * 4 = 8. If we take 1/4 so that the 4 is also a denominator, then we have: (1/4)*(1/4) = (1/16) and (1/4)+(1/4) = (2/4) = (1/2) Looks like we have our factors. As I said, look for something that when you multiply will become the last number in the expression, and when you add or subtract will be the middle number. You might ask, what about 8? 8 goes into 16 twice and 2 * 4 = 8. Try working out (1/8)+(1/8) and (1/8)*(1/8) and you will see that it is not the right answer. Sometimes it might take a while by trial and error to find the exact right answer, but it will be there. Here you will end up with (x+(1/4))(x+(1/4)) which can also be written as (x+(1/4))^2 When you are putting your factorization together, also think about how you want it to turn out, especially when dealing with negatives and positives. The way to tell if you have done it right is, of course, to multiply it out again. You should get the quadratic expression that you started out with. Just remember that the principle that I showed you is the same with any quadratic expression that has the coefficient of 1 for x^2. It can still be done when the coefficient of x^2 is greater than 1, but it's a bit harder. If you have one of those to solve and you can't get your teacher's help, just send it and we'll tackle it together, okay? I hope this has helped. If you would like to see some questions that other people have written in to Dr. Math and the answers that were sent to them, have a look at: http://mathforum.org/dr.math/tocs/factoring.middle.html If you would like to ask any other questions please don't hesitate to write to Dr. Math again and we will try to help you. - Doctor Sandi, The Math Forum http://mathforum.org/dr.math/ |
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