Finding a Single Pair of FactorsDate: 02/27/2002 at 20:00:02 From: Lauren Subject: ax squared - bx - c = The problem is: 34x^2 - 41x - 15 I have tried all of the factors of 34, and everything else, but I can't get the answer because the only factors of 15 are 1 and 15, and 3 and 5. Please help! Date: 02/27/2002 at 23:17:31 From: Doctor Peterson Subject: Re: ax squared - bx - c = Hi, Lauren. I would just use the quadratic formula to find the zeroes, from which I would build the factors. That's the easy way. But there is a better way to factor this than the method you are using, one which I only learned recently. Rather than look at combinations of factors of 34 and factors of 15, just look for a pair of factors of 34*-15 = -510 whose sum is -41. You can find that easily. I'll demonstrate the work by doing a different problem as a model. We'll factor 21x^2 - 23x - 20 We want to factor ac = 21*-20 = -420, so that the sum of the factors is -23. (Note that the factors will have opposite signs, so their sum will be the _difference_ of two factors of +420.) We can try 42 and 10, but their difference is too big; 21 and 20 have too small a difference; but 35 and 12 are just right. Writing 23 as 35-12, our quadratic is 21x^2 - 35x + 12x - 20 Taking this in pairs and factoring each pair, we get 7x(3x - 5) + 4(3x - 5) Aha! They have a common factor, so we can combine them: (7x + 4)(3x - 5) Notice that if we had reversed the order of the middle terms, it would still work: 21x^2 + 12x - 35x - 20 3x(7x + 4) - 5(7x + 4) (3x - 5)(7x + 4) Here is where I learned the method: Two Methods of Factoring Quadratics http://mathforum.org/library/drmath/view/52878.html Looking for a single pair of factors this way is much easier than the way I learned, which is probably what you are trying to do! Have fun finishing your problem. - Doctor Peterson, The Math Forum http://mathforum.org/dr.math/ Date: 04/23/2009 at 12:49:59 From: Li Subject: Factoring Trinomials Dear Dr. Math, In refereeing for a journal article I came up with the following: If y = ax then ax^2 + bx + c = (1/a)(y^2 + by + ac) which gives a quick and easy method of factoring ax^2 + bx + c. I thought it was new until I Googled it and found your post: Factoring Trinomials http://mathforum.org/library/drmath/view/56442.html My explanation above may be shorter and easier for students to understand. Also, for an experienced user, the method can be presented in a very quick and direct form: ax^2 + bx + c = (1/a)(ax + ?)(ax + ?) where the two ?'s should multiply into ac and sum into b. Best regards, Li Date: 04/23/2009 at 23:25:20 From: Doctor Peterson Subject: Re: Factoring Trinomials Hi, Li. I've never until now heard an explanation of this method that actually justifies it and doesn't just tell the student to magically replace one trinomial with another; both of your approaches here make a lot of sense, and may change it from a method I avoid into one I may end up teaching as my preferred method. (My math department officially forbids teaching the "slip and slide" method, as they call it, because it does not involve transforming the expression into a series of equivalent expressions, and so is mathematically unjustified.) I currently teach the "ac" method, which I learned for the first time in the course of answering this question; that method is taught in the texts I've used since then, along with trial and error. Let me fill in the details in what you wrote, to see how I would explain it to students. I generally start by showing an example, and then generalize; so I'll work with the trinomial 6x^2 - x - 12. First, for students able to handle a change of variables: We choose (for no obvious reason, except that it turns out to work) to rewrite the equation in terms of a new variable u = 6x. This means we are replacing x with u/6: 6x^2 - x - 12 = 6(u/6)^2 - (u/6) - 12 = 1/6 u^2 - 1/6 u - 12 Now we factor out 1/6, which means we see 12 as 72/6: u^2 - u - 72 = ------------ 6 Now we just have to factor a trinomial of the easy kind; we get (u - 9)(u + 8) = -------------- 6 We have to put the answer in terms of x, so we replace u with 6x: (6x - 9)(6x + 8) = ---------------- 6 Now we factor out the common factor from each factor here, and cancel those with the 6: 3(2x - 3) 2(3x + 4) = ------------------- 6 = (2x - 3)(3x + 4) This, as you say, is the same method as on the page you refer to, except that every step is justified and we aren't just jumping from one trinomial to one that is not equal to it. Your second approach doesn't require the second variable, which beginning students would have trouble with; here's how I'd explain it: We try putting 6x in each of the prospective factors, just as we put x in each factor in the easy case; but we realize we have to divide that by 6 to make it come out right, so we write (6x + _)(6x + _) 6x^2 - x - 12 = ---------------- 6 What's written so far makes sense, since we have 36x^2 / 6 = 6x^2. Now, just as in the simple case, we look at how we can fill in the blanks. Call them p and q for now. The last term of the trinomial will be pq/6, so pq has to equal -12. The middle term will be (6x*q + p*6x) / 6 = (p + q)x so p + q has to equal -1. As usual we find that p and q are -9 and 8, so we have (6x - 9)(6x + 8) 6x^2 - x - 12 = ---------------- 6 and from here we do the same as in the other method, factoring out the common factor. The nice thing here is that, like the ac method I usually teach, it centers around the same process as in the first case (finding a pair of numbers whose sum and product are known), but it doesn't require the process of factoring by grouping, and it's easy to see why it works (though not, perhaps, how you would think of it!). So I'm going to try this last approach with my next class and see if they like it. Thanks for your contribution! If you have any more thoughts (and especially if I've missed a good way to explain it) please let me know. - Doctor Peterson, The Math Forum http://mathforum.org/dr.math/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994- The Math Forum at NCTM. All rights reserved.
http://mathforum.org/dr.math/