It is often difficult to factor a trinomial in the form ax^2 + bx + c when lead coefficient is not one. It is especially difficult to factor when the lead coefficient is not prime. The following is a method that makes this much easier.
We multiply ax^2 + bx + c by a to get ax^2 + bx + c and try to factor this realizing that after we are done we must divide out the factor of a we put in originally. Suppose this can be factored into
(ax + r)(ax + s). This new expression multiplies out as a2x2 + axs + axr + rs or a^2x^2 + x(as + ar) + rs. Thus, we only need to find an r and s such that r + s = b (as + ar = ab) and rs = ac. We only need to find r and r.
Example: Factor: 6x^2 + 11x – 10
1. Multiply 6 by the -10 to create the trinomial x^2 + 11x – 60 (I call this the dummy trinomial. The silly name helps the kids remember that it is only temporary.)
2. (6x + )(6x - ) and then only need to find the numbers that multiply to – 60 and add to 11.
3. (6x + 15)(6x – 4) (still the dummy)
4. And finally: (6x + 15) /3(6x - 4)/2 = (2x + 5)(3x - 2)
We multiply ax^2 + bx + c by a to get ax^2 + bx + c and try to factor this realizing that after we are done we must divide out the factor of a we put in originally. Suppose this can be factored into
(ax + r)(ax + s). This new expression multiplies out as a2x2 + axs + axr + rs or a^2x^2 + x(as + ar) + rs. Thus, we only need to find an r and s such that r + s = b (as + ar = ab) and rs = ac. We only need to find r and r.
Example: Factor: 6x^2 + 11x – 10
1. Multiply 6 by the -10 to create the trinomial x^2 + 11x – 60 (I call this the dummy trinomial. The silly name helps the kids remember that it is only temporary.)
2. (6x + )(6x - ) and then only need to find the numbers that multiply to – 60 and add to 11.
3. (6x + 15)(6x – 4) (still the dummy)
4. And finally: (6x + 15) /3(6x - 4)/2 = (2x + 5)(3x - 2)
4 comments:
That is a really interesting way to do it. Now I need a free hour to sit down and figure out why it works.
I teach my kids something similar, but with a slight twist....(no r and s to divide by!)
Look for things that multiply to ac and add up to b. Then split up the middle term into those two numbers and factor by grouping. I'll do the same example you did.
6x^2+11x-10 (So numbers that multiply to -60, add to 11 are 15 and -4.)
Rewrite: 6x^2+15x-4x-10
First two terms have 3x in common and terms 3 and 4 have -2 in common.
Factoring those out we get:
3x(2x+5)-2(2x+5)
Both terms have (2x+5) so factor that out, leaving you with:
(2x+5)(3x-2) (!)
Note: It doesn't matter which order you split the middle term....it would also work to do the rewrite step as 6x^2-4x+15x-10.
Great- now I'm going to be factoring trinomials for the rest of the day, lol.
For the record, "anonymous" wasn't me, but I do the same thing...
here
Jonathan
Post a Comment