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Month: November 2015

AMATYC 2015 – First Impressions

AMATYC 2015 – First Impressions

You can always see which way the pendulum is swinging at AMATYC. Lots of sessions on tablet/smart phone apps, getting online students more engaged, and various pathways approaches. So different from my first AMATYC.

I have seen, firsthand, how mobile technology can impact the classroom. A couple of weeks ago I was introducing my students to the graph of a parabola. I wanted to graph a parabola using Desmos, have my students observe the results, and tell me which points they felt were important. Of course, the projector failed. Not a problem for my students – they whipped out their smart phones and graphed it themselves.

I’m learning about so many new (to me) aps – and I will share things as the week progresses.

I’m afraid we have always treated online students as being on their own, even when we try to make our classroom more collaborative. It probably has something to do with how easy it is to administer an online class and how powerful today’s learning management systems are, including online homework. But we need to teach the online classes, not just administer them. Community is important whether you are a face-to-face student or an online student. The easier it is to get disconnected, the harder it is to learn and persevere.

My thoughts on the new pathways are complicated, and I will address those in a future blog. I think that redesigning our courses to meet the true needs of our students is a good idea. We cannot continue to teach 21st century students using 20th century approaches. But we still need to teach important skills and develop critical thinking.

More to follow …

George

Factoring Trinomials (x^2+bx+c): Sometimes You Should Start With b, Not c

Factoring Trinomials (x^2+bx+c): Sometimes You Should Start With b, Not c

In today’s class we ran into the trinomial

x^2+3x-208

This can be challenging for students just learning to factor because they are not familiar with the factor set of 208. Traditionally I recommend that students follow the 10-second rule:

If you cannot find the correct factor pairing within 10 seconds, you should move on to listing all factor pairs of c.

This would result in the following pairs:

1\cdot208, 2\cdot104, 4\cdot52, 8\cdot26, 13\cdot16That means that students need to try 13 potential factors (1 through 13) before they find the correct pair. Here’s an alternate approach: focus on the term 3x. Since c is negative, we know that m and n will have opposite signs. If we ignore those signs, we know that the two values must be 3 apart from each other or, in other words, have a difference of 3. I told my students to start multiplying numbers that were 3 apart from each other. We started with 10 & 13, but that product was too small. We bumped them up to 11 & 14, 12 & 15, and finally 13 & 16. Bingo!

Now consider the trinomial

x^2+29x+208

Focusing on the middle term (29x) might not seem to be much help here. However, we could start with two numbers that have a sum of 29 like 14 & 15 and work our way back.

Give it a try and let me know what you think.

George

Factoring – A New Approach

Factoring – A New Approach

This semester I have been trying to help my students discover mathematics, rather than having everything coming straight from me. My goal is to get them to think more while also gaining confidence in their own abilities. One topic that has always given my students trouble is factoring trinomials that have a leading coefficient other than 1. Here’s what I tried.

We had spent the day before factoring trinomials that did have a leading coefficient of 1, and my students understood the overall goal of rewriting the trinomial as a product of binomials. I started with one of the new trinomials, along with one of the factors.

2x^2+13x+15=(2x+3)( ? )

I asked them to come up with the other factor. They figured out that the missing factor had to contain x in order to get to 2x^2, and it also had to contain 5 in order to get to 15.

2x^2+13x+15=(2x+3)(x+5)

Now I had them multiply to check that their product would produce the middle term of 13x.

I then ran them through a series of problems where I provided parts of the factored form.

6x^2+23x-18=(3x-2)(?)

10x^2+23x+12=(2x+?)(5x+?)

8x^2-2x-21=(?+3)(?-7)

I then turned them loose on factoring

2x^2+15x+7

I gave them no prompts or hints, and waited to see what they could come up with. Many of my students were able to put it together and explain what they did to their classmates. I then went into a formal presentation of how to factor these by trial and error.

Why did this work? For starters, students developed their own understanding of how the variable terms in the factors have to multiply to be ax^2 and how the constant terms in the factors have to have a product of c. They learned how to check to see if the factored form produced a middle term of bx. They developed their own intuition rather than me robbing them of that opportunity.

A week has passed and they still seem to be on track. The exam is next Tuesday – I will let you know how it goes. Fingers confidently crossed.