Quadratic TrinomialsDate: 14 Aug 1995 13:28:05 -0400 From: Anonymous Subject: algebra I'm reviewing some stuff for algebra 2, and I forgot QUADRATIC TRINOMIALS. It says "factor as the product linear binomials or write prime." Is there some trick to figuring this out? Help me out, please. * means squared 2x* + 5x + 2 25x* - 10x - 3 3x* - 10x + 3 Thanks Date: 14 Aug 1995 14:17:07 -0400 From: Dr. Ken Subject: Re: algebra Hello there! Well, the foolproof way that works every time is to use the Quadratic Formula and then multiply through by the leading coefficient. Oh, by the way, the standard notation for "squared" is ^2. For instance, in the first problem, you have 2x^2 + 5x + 2. Using the quadratic formula, we have x = (-b +- Sqrt{b^2 - 4ac})/2a = (-5 +- Sqrt{5^2 - 4*2*2})/2*2 = (-5 +- 3)/4 = -2 or -1/2 So following the pattern, we have the following factorization: (leading coefficient)(x - first root)(x - second root) = 2(x+2)(x+1/2) = (x+2)(2x+1), and that's the answer. If you use the quadratic formula and you don't get a perfect square inside the radical (what's inside the radical is called the discriminant) or you get a negative sign, then you can write "prime". I mean, you could try to show off and write 2x^2 + 4x + 1 = 2(x + 1 + Sqrt{2}/2)(x + 1 - Sqrt{2}/2), but somehow I don't think that's what your teacher had in mind. Good luck with it! -K Date: 15 Aug 1995 22:59:12 -0400 From: Anonymous Subject: Re: algebra Is there a shorter, easier, quicker way to do this? We didn't use the quadratic formula in my algebra class. We did everything else though. Date: 16 Aug 1995 15:57:47 -0400 From: Dr. Ethan Subject: Re: algebra Well, I must agree with Ken that the most reliable way is to learn the quadratic formula but if you want another way then you need to factor. And here is how you do it. It is a little more complicated then the quadratic formula but it is also a little easier to understand. Here is the idea, look at 2x^2 + 5x + 2 We need to find two binomials of the form (ax +b) which when multiplied together equals the desired trinomial. Well, let's assume they are (ax+b) and (cx + d) then (ax +b)(cx+d) = a*c x^2 + (bc + ad)x + b*d So a*c = 2 and b*d = 2 and (bc + ad) = 5 Well, to get this we realize that we can only choose 1 or 2 for any of the values. Now you just experiment till you find a combo that works. a=1 b=2 c=2 d=1 is one such combo because (x+2)(2x+1) = 2x^2 + 5x +2 This is just one example. Some are harder, but all factoring works on this principle. Write back if this isn't enough help. Ethan Doctor On Call |
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