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.

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