Factoring Trinomials With Prime FactorsDate: 05/16/2002 at 21:39:26 From: Scott Subject: Factoring Trinomials Factor each trinomial, if possible. If the trinomial cannot be factored using integers, write PRIME. 1) 48x^2 + 22x - 15 I've multiplied -15 and 48 and I usually get -720 and I can't find a number that goes into -720 and adds up to 22. 2) -4y^2 + 19y - 21 I've multiplied -21 and -4 and I get 84, but I can't find any factors of 84, that equal 19. 3) 6a^2 - 7a + 18 I've multiplied 18 and 6 and I get 108, but I can't find any factors of 108 that equal -7. Help! Date: 05/17/2002 at 12:30:32 From: Doctor Peterson Subject: Re: Factoring Trinomials Hi, Scott. If I don't see the appropriate pair of factors on sight, I make a systematic search by first finding the prime factorization. I'll demonstrate on the first, 720 = 2*2*2*2*3*3*5. Since the product is -720, we will have a positive and a negative factor, and it is the _difference_ of two factors of 720 that has to equal 22. Taking a first guess, let's just divide the prime factors evenly (so the factors will be fairly close together), and try 2*2*2*2 * 3*3*5 16 * 45 The difference is 45-16 = 29. That's a bit too big, so we have to make the factors closer. Maybe if we swap a 2 for a 3 it will make the 16 big enough: 2*2*2*3 * 2*3*5 24 * 30 The difference is now 30-24 = 6. That's too small. How about instead swapping a 4 for a 5: 2*2*5 * 2*2*3*3 20 * 36 The difference is now 36-20 = 16. We want to make the smaller one smaller, so let's start here and swap a 10 for a 9: 2*3*3 * 2*2*2*5 18 * 40 The difference is now 40-18 = 22. We found it! It would be a lot harder if this were in fact prime. To show that there is no way to factor it, you would have to make a list of ALL pairs of factors and show that none work. In this case, there are 30 factors, in 15 pairs, so that wouldn't be terribly hard, if you know how to make such a list. In some cases, however, it's obvious that you can't factor it: If the desired sum of two positive factors is less than twice the square of the product, then no pair of factors can give a sum so low. Can you see why? But ultimately, the most efficient way to see whether a quadratic can be factored is to use the quadratic formula to find its zeros. Factoring is a waste of time in hard cases. - Doctor Peterson, The Math Forum http://mathforum.org/dr.math/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994-2015 The Math Forum
http://mathforum.org/dr.math/