Multiply first and last: $3 * -8 = -24$

What are the factors of -24 that add up to 10? 12 and -2

Here's where it gets weird. We multiplied the first and last so the product is too large by a factor of 3 so we need to cancel that extra 3. We'll take care of it in this next step.

**The factors of 3x² are going to be 3x and 3x,**which is where that extra 3 is included.

Create a big fraction: $\dfrac{(3x+12)(3x-2)}{3}$

What's cool is that the two factors, 12 and -2 are placed right into the big fraction without consideration for where they go. They are guaranteed to reduce. What's incredibly cool is that you have no big list of factors to run a trial and error method against.

Reduce $\dfrac{3(x+4)(3x-2)}{3} = (x+4)(3x-2)$

It works if you forget to factor out a common factor. It works for all trinomials of this type.

Factor $24x^2+4x-20$

Multiply first and last: $24 * -20 = -480$

What are the factors of -480 that add up to 4? 24 and -20

Create a big fraction: $\dfrac{(24x+24)(24x-20)}{24}$

Reduce $\dfrac{24(x+1)4(6x-5)}{24} = 4(x+1)(6x-5)$

It would be easier if you did factor out that 4:

Factor $24x^2+4x-20 = 4(6x^2+x-5)$

Multiply first and last: $6 * -5 = -30$

What are the factors of -30 that add up to 1? 6 and -5

Create a big fraction: $4*\dfrac{(6x+6)(6x-5)}{6}$

Reduce $4*\dfrac{6(x+1)(6x-5)}{6} = 4(x+1)(6x-5)$

Thank you for posting this. The method looks good. It is not as simple when either the constant or the leading coefficient has two or more digits. However, it is still effective. Here are a few examples:

ReplyDelete1. 18z^2 + 19z - 12

2. 6 + 7a - 20a^2

3. 32n^2 - 4n - 15

Ah, but #1 gives you a great opportunity to talk about 18*12 = 2*3*3 * 2*2*3 and using that info to find the two factors (27 & 8) more quickly. Beats the hell out of the 72 combinations of the guess and check method.

ReplyDelete#2 just needs to have -1 factored out first.

#3, see note 1. 2*2*2*2*2 * 3*5 ... 20 and 24.

I agree, it is a great opportunity.

ReplyDeleteI personally never had a problem with the guess and check method, but perhaps I had an above average aptitude for arithmetic. It was never a random search. For example, given 32n^2 - 4n - 15, it was obvious that the factors were a positive and negative with the absolute value of the negative larger than the positive. Also, since the linear coefficient -4 was relatively small, it also seemed obvious to begin by checking factor pairs of relatively equal size; that is, 4 and 8 with 3 and 5. It usually took only three or four quick trials to produce the binomial factors.

As a teacher, I like to develop this kind of tactical thinking in my students. This should in no way imply that the method you shared does not involve the same kind of thinking, as your comment clearly points out. The student must think "tactically" to find factors that yield a sum of -4. (By the way, it was not necessary to factor out -1 in example #2. Just lead each binomial with and have a divisor of 6.)

Thank you again for posting these great ideas.