Reflections
… there are programs that can perform mathematics at the symbolic level, and some of them are free. This fact leads to an objection, heard with increasing frequency, that people will use programs like Maxima to avoid learning anything about mathematics at all, instead depending on symbolic math software to conceal their ignorance. - More
“Determine which of (x – 3), (x – 1), (x + 2) are factors of P(x) = 2x^3 + 7x^2 + 7x + 2, and hence factorise P(x) completely.”
This was the first question on a recent test I gave. The students could have asked Maxima:
(%i1) factor (2*x^3 + 7*x^2 + 7*x + 2);
(%o1) (x + 1) (x + 2) (2 x + 1)
or asked the web site QuickMath which uses WebMathematica:
http://www.calc101.com/ is a similar site.
Most of the problems we deal with in class can be instantly solved by free symbolic math software. So why don’t we allow the students access? Why do we insist they should learn what machines are far better at than humans and instead use the time to what the machines are poor at, namely problem discovery and problem solving?
There is a risk that we will use math software to become intellectually lazy. But the possibility exists that, in partnership with computer math software, people will learn much more about mathematics than they would be likely to do while using the older methods. Because we have computers to perform low-level computations, we can spend our time acquiring mathematical knowledge at a higher level. - More
I believe several universities now allow this kind of software, but at the pre-university level it is unheard of. At IB there is even an exam where calculators are banned. I fail to grasp why.
February 14th, 2010 at 3:02 am
Strictly speaking, the Maxima and QuickMath solutions do not answer the question as posed. They do not first determine which of the three candidates is a factor of P(x), and *hence* factorise P(x) completely. Instead, they present all three factors simultaneously.
The fastest solution to the question might be to notice that P(1) and P(3) must be positive, and so, by the Factor Theorem, the only possible factor is (x + 2). Then ask QuickMath to factor P(x)/(x + 2) — or first simplify it and then factor the result.
February 14th, 2010 at 6:25 am
I agree, but if you just factorise straight away the answers to the first three questions can be deduced, and should be mentioned in the solution.
The test I referred to has been completely solved by Maxima here: http://easyquestion.net/maxima/maxima%20does%20the%20test.html. Can you find better ways to solve some of the problems.
A forum for Maxima is here: http://sourceforge.net/projects/wxmaxima/forums/forum/435775.
Finally, I wrote about the same topic here: http://www.easyquestion.net/learninginadigitalworld/2010/02/12/when-will-we-train-students-to-be-more-than-second-hand-calculators/.
February 14th, 2010 at 5:59 pm
Waking up this morning I realised that, assuming that one of the polynomials listed is actually a factor, the whole question can be solved by inspection.
Firstly, (x + 2) must be the first factor, as noted above.
Next, because P(x) is a symmetric polynomial, if r is a root then so is 1/r. (Set z = 1/x and simplify.) Thus 1/2 is a root and we see that (2x + 1) is the corresponding factor.
Finally, because P(x) is a symmetric polynomial of *odd* degree, P(1) = 0 and (x + 1) is the factor.
Try getting Maxima to do that!
I didn’t see any slick solutions for the other questions. I agree with your point about students and calculators.
February 14th, 2010 at 8:06 pm
Excuse my ignorance, but what is a symmetric polynomial?
February 14th, 2010 at 10:20 pm
Oops, sorry, I meant palindromic polynomial: http://en.wikipedia.org/wiki/Palindromic_polynomial
February 14th, 2010 at 10:28 pm
And I meant to say -2 is a root -1/2 is a root! (Not 1/2.) And P(-1) = 0.
February 14th, 2010 at 10:30 pm
WordPress deleted my double implication sign between “root” and “-1/2″ in the above post.
February 14th, 2010 at 11:12 pm
Very nice! I had no idea that the P(x) given was so interesting.