Factoring - The Lizzie Method
The "Lizzie Method" which is also known as the "fraction method" or "Vieta's Theorem" gives an easy-to-teach and use method for factoring quadratics in the form ax2+bx+c. This method has been taught previously, but was independently rediscovered by a Michigan high school student in 2003. I like the name "The Lizzie Method" for this technique because it helps students realize that the mathematical techniques that we use in our classes are created by people and the informality helps emphasize that in this instance, it was someone like them who came up with this technique on her own.
The big drawback to this method as opposed to factoring ax2+bx+c by grouping (also known as the ac method, the split method and the British method) is that the procedure seems a bit arbitrary. At the Algebra I level, I think that the arbitrarinesss is inescapable, but when this method is taught in an Algebra II class or above, the lesson can incorporate a proof showing why this method works.
Prerequisites
Students should have a solid grounding in factoring quadratics in the form x2+bx+c by finding factors of c whose sum or difference is b. This ability is essential to using this method.
It's also worth reviewing simplifying fractions, especially with students in a 2-year Algebra I sequence who might not have spent significant time on this in well over a year.
The method
1. Our first step is to identify what sort of factors that we have, using a procedure similar to how we identify the type of factors of x2+bx+c. If c is positive, then we have either the product of two sums (b is positive) or the product of two differences (b is negative). If c is negative, then we have the product of a sum and a difference.
2. Next we take the product of the leading coefficient a and the final constant c and look for two factors that either sum to b (when c is positive) or whose difference is b (when c is negative). Note that this is very much like what we did with factors of c for factoring x2+bx+c.
For example, if we wish to factor 6x2-11x-10, we first note that c=-10 so we have the product of a sum and a difference. Then we take the product 6×10=60 and find two factors of 60 whose difference is -11. In this case we have have 4×15=60 and 4-15=-11.
3. Next we set up a pair of factors by with the constants being each of the two numbers that we found in the previous step divided by our leading coefficient. So in our example we have:
(x+4/6)(x-15/6)
4. We simplify the fractions that appear above. Now our example will look like:
(x+2/3)(x-5/2)
5. If we're looking for the solutions of x2+bx+c=0, we can use these factors as they are. The solutions will be -2/3 and 5/2.
6. But we should still find the factored form so that we can confirm that our factorization is correct by multiplying the two factors together. We need to get rid of any fractions that exist by multiplying each factor by the denominator of the fraction. On paper, we can do this multiplication quickly by moving the denominator from the bottom of the fraction to next to the x. Our factorization now looks like:
(3x+2)(2x-5)
Points to ponder
Once we've completely simplified the two fractions (step 4), the product of the denominators will always be a.
When teaching this, it's a good idea to review multiplying polynomials and take a close look at the terms that we get along the way to help make it clear where some of the numbers in this process come from.
I'd be curious to know when and why Vieta's theorem fell out of the US mathematics curriculum. Doing some research, this is apparently standard issue in many (most?) European algebra texts.
Last updated 26 Feb 2006. Send me an e-mail.