Factoring (x - 1)(x^2 - x - 2)Date: 11/04/97 at 23:22:03 From: Michelle Subject: Algebra 2 I just have a question about how to factor. I know I did it in Algebra but I'm not remembering how. The question is (X-1)(Xsquared-x-2). I have gotten to a point, but then I don't know how to finish. I have xcubed _ xsquared - 2 x - xsquared - x + 2 = 0 xcubed + 3x + 2 = 0 Then I can't figure out what to do next. Please help. Thanks, Michelle Date: 11/05/97 at 16:16:01 From: Doctor Keith Subject: Re: Algebra 2 Great question. I will go through the general ideas and path to solve the problem. If something does not seem clear write back and I will clarify it. For notational simplicity I will use x^2 for xsquared and sqrt(x) for the square root of x. First let's look at what we are trying to do. We are given a polynomial, which in your case is (x - 1)(x^2 - x - 2), and we want to find the factors. The factors of a polynomial are terms of the form (x - a) such that (x-a)(x-b)(x-c) = (x-1)(x^2 -x -2) i.e. the product of the factors equals the polynomial. Many people call this finding the zeros of a function because if we take x = a or x = b or x = c that term becomes zero, and thus the whole polynomial is zero at that point. Example: let x = a in the above factorization: f(x) = (x-a)(x-b)(x-c) f(a) = (a-a)(a-b)(a-c) = (0)(a-b)(a-c) = 0 Okay, what did that tell us? First we can see that we are given one of the factors of your problem, namely (x-1)! We thus only need to factor (x^2 - x - 2). As a general guideline I would not multiply through unless I had no other alternative. For instance (x^2 - x - 2)(x^2 - 2x - 4) is easier to factor by finding the factors of (x^2 - x - 2) and the factors of (x^2 - 2x - 4) than to find the factors of the product of the two. Note that if the factors of (x^2 - x - 2) = (x-a)(x-b) and the factors of (x^2 - 2x - 4) = (x-c)(x-d) then the factors of (x^2 - x - 2)(x^2 - 2x - 4) = (x-a)(x-b)(x-c)(x-d). The reason I mention zeros of a function is that this is often how factoring is referred to in later math courses and it is good to know when those come by that you are just doing something you already know. Now we need to find the factors of a quadratic, namely (x^2 -x -2). There are several ways to do this, and I will show you my favorites: 1) Use the quadratic equation: for ax^2 + bx + c the factors are (x-d)(x-e) with d, e given by -b + sqrt(b^2 -4ac) d = ------------------- 2a -b - sqrt(b^2 -4ac) e = ------------------- 2a This method works for all equations nicely and is really good for calculators, but can get messy for people. For your quadratic we have a = 1, b = -1, c = -2 thus 1 + sqrt(1 - 4(-2)) 1 + sqrt(1 +8) 1+3 d = -------------------- = ------------------- = ----- = 2 2 2 2 1 - sqrt(1 - 4(-2)) 1-3 e = -------------------- = ---- = -1 2 2 so: (x^2 - x - 2) = (x - 2)(x - -1) = (x-2)(x+1) 2) The second method is a structured guessing method. It could be solved explicitly (no guessing) but that would take as much work as the original problem. The advantage is that this can be good on tests if you are a quick guesser. Note that (x-a)(x-b)= x^2 - (a+b)x + ab = x^2 - x - 2 Thus we can use this by requiring: -(a+b) = -1 ab = -2 Now here is how we do the educated guessing. Since the order of the factors doesn't matter, let a >= b in general. From the second requirement we see that the product of a and b is negative; thus the two terms have opposite signs (assuming we are using real numbers, which is probably valid until college level math). Since a >= b we must have a > 0 and b < 0 and thus a > b. From the first requirement we see that a + b = 1, and since a > b and b = -|b| (|x| is the absolute magnitude, and just means dropping the sign for real numbers) we have a = 1 - b = 1 + |b| > |b| which tells us that the magnitude of a is bigger than the magnitude of b. Moreover, since b < 0 and a > |b| then -2 = -a|b| > -|b|^2 so -sqrt(2) < b < 0. This means that since a>|b| then a>sqrt(2). Now we have narrowed down our area of guessing. I usually try letting one of the terms be +/-1 and the other be whatever will satisfy the second requirement. We then check and see if the first requirement is satisfied. Since we know a can't be 1 but b could be -1, I will pick b = -1 and thus a(-1) = -2 so a = 2. Now we try this in our first requirement and get -(2-1) = -(1) = -1 and both conditions are satisfied on the first guess. (x - a)(x - b) = (x - 2)(x - -1) = (x - 2)(x + 1) In most classroom problems this will work within a couple guesses. What if it hadn't worked on the first try? Well, if the sum of our guesses were too big we would make b slightly more negative (know - sqrt(2)<b<-1 ) and find the a that works for this new guess in ab = -2 and try again. If the sum were too small, then we'd want to make b less negative (know -1<b<0) and again find the a that goes. As you can see, it converges quickly for simple problems, but for ab = 2.1345643433 this method is a lost cause. 3) Completing the square This is a particularly pretty way of solving the problem. What we do is note that (x-a)(x-a) = x^2 -2ax + a^2. Now we can write any quadratic as (x-a)^2 +b = x^2 -2ax +a^2 + b. So we require that in your problem -2a=-1 or a = 1/2 then we have that b + a^2 = b + 1/4 = -2 or b = -2.25 = -9/4 Now the factors are given by: (x-a+sqrt(-b))(x-a-sqrt(-b)) so for us, since sqrt(-b) = 3/2 we get (x-1/2+3/2)(x-1/2-3/2) = (x+1)(x-2) Okay, we have derived the same answer by three methods. Which is best? That depends on what is easiest for you. I will usually try a couple of guesses using method 2, because with a little practice you can get really good with it (even though it took more writing to explain because it is more intuitive), especially since I now know the range of expected values. If I don't get it quickly I will go to method 1 (quadratic equation), as this always works. When do I use method 3? When I want to prove something in my work, as the beauty and neatness of the expression really are impressive, and that is a big part of writing research papers. Our final answer is thus: (x-1)(x^2 -x -2) = (x-1)(x-2)(x+1) and we are done. I hope this has helped. If anything is unclear write back and I will explain it more. -Doctor Keith, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
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