Reducing Fractions and its Application to Rational Expressions
There is a variety of reasons today why a given student doesn’t learn or master a presented method or technique. As teachers , we are aware of diverse learning styles and conditions in the classroom. And, despite hard work and willingness to learn on the part of the student, the set objective is not met. Logically then, what do we do? Alternative approaches to a problem are often sought and this is where “thinking outside the box” may come in handy; especially when a novel idea works and appeals to others.
Imagine the plight of the Math 10 or Math 11 student whose factoring skills are less than adequate. That person may find factoring a chore due to a lack of success with a conventional method. In order to increase their comprehension, a student needs another approach to make the task more suitable. Consider reducing the fraction 39/65 to lowest terms. If we know that 13 is a factor of both 39 and 65, then we can write it as
The educator knows that 13 is the greatest common factor but the student may not. Similarly, how would it be apparent to a student in an expression like
For reference and clarity I would now like to refer to my publication in the article, “Reducing Fractions”, published in the Scientific Journal of Junior Math and Science, London, England 1990, which demonstrates the premise that the only possible factors available to reduce a fraction to lowest terms, come from the difference between the numerator and the denominator. Numerically, it looks like this:
(1) 65 – 39 = 26 demonstrates the difference between the numerator and the denominator
(2) Factors of 26: 1, 2, 13, 26
(3) Disregard 1, 2, and 26 because they are all even
(4) Try 13. If 13 doesn’t work, then nothing else will.
When transferring this concept to rational expressions, examine the following two examples:
e.g. 1) Reduce
Step: (2) Disregard 5 and consider (x-3) because 5 doesn’t divide evenly into the numerator or the denominator but (x-3) might.
At this stage, we can factor more (3) and easily because we know that (x-3)
is one of the desired factors; failing that, use long division
e.g. 2) Reduce Steps: (1)
(2) Disregard x and (x+1) but consider (x-1) because neither x nor (x+1) divide evenly into neither the numerator nor the denominator; the only remaining factor to consider is (x-1).
What I also like about this method is that we can discover which factors will not work in a given situation:
Immediately we can see that there cannot be a common factor of the form (x+a). (Assuming that the algebra is done correctly of course)
As we can see, there is no point in factoring and looking far a common term if indeed none exists to begin with.
Now it’s no secret that a method for finding the g.c.f. of two polynomials does exist, but it does involve long division, and therefore, it would look something like this:
Find the g.c.f. for and , or g.c.f. .
Divide one into the other, and keep track of the remainder. Now divide the remainder into the previous divisor, and again, keep track of the remainder. Continue this last step until the remainder is zero. The divisor, which gives zero as a remainder, is our g.c.f.
This means we would have:
Take only (x-3) as –5 is not a factor of the form (x+a) and because –5 doesn’t divide evenly into either the numerator or the denominator.
Actually, finding the g.c.f. in this manner is part of the reason why the above method of subtraction works. The teacher now has more than one way of presenting this material to various types of learners, and can provide alternatives for the reluctant student.
A welcome application of this approach is the calculation of limits for the calculus student. In general, we have:
Instead of evaluating directly, and giving the indeterminate form , we can subtract
the two polynomials, factor this difference, and then try to reduce it to its lowest terms.
This would create the following:
This reveals that (1) the expression can be reduced, and that (2) (x+a) is the common factor. A numerical example would look like:
= Disregard x and consider (x+2) because x doesn’t divide evenly into the numerator or the denominator.
In summary, some students see math as a necessary evil. However, now they can “get into it” a bit more because someone has found a method which makes sense to them.