Finding the Sum of Solutions of a Polynomial EquationDate: 03/11/2004 at 21:04:54 From: Zahra Subject: x^3+2x^2-5x-6=0 What is the sum of the solutions of x^3 + 2x^2 - 5x - 6 = 0 given that x = 2 is a solution? I do not understand any part of this question. Date: 03/12/2004 at 14:01:18 From: Doctor Ian Subject: Re: x^3+2x^2-5x-6=0 Hi Zahra, Let's look at a simpler example: What is the sum of the solutions of x^2 - 5x + 6 = 0, given that x = 2 is a solution? If x = 2 is a solution, then x^2 - 5x + 6 = (x - 2)(x +/- something) So what we want to do is find the "something". (Do you see why this is true? Write back if you don't, and we can talk about that.) Now, how do we find it? Well, suppose I tell you that 12 = 4 * something How could you find the "something"? You'd divide: 12 / 4 = something = 3 Same thing here. We want to divide x^2 - 5x + 6 by (x - 2). The process works just like long division with numbers, except you deal with one term at a time instead of one digit at a time: x _____________ x - 2 ) x^2 - 5x + 6 x^2 - 2x <- This is x * (x - 2) -------- -3x <- This is -5x - (-2x), or -5x + 2x Now we bring the next term down, and keep going: x - 3 _____________ x - 2 ) x^2 - 5x + 6 x^2 - 2x -------- -3x + 6 -3x + 6 ------- 0 There's no remainder, so x^2 - 5x + 6 = (x - 2)(x - 3) If you multiply out the right side, you'll see that this is, in fact, correct. To finish our problem, the two solutions to the equation are 2 and 3, so their sum is 5. Now, with a quadratic, this is overkill. It's easier to use regular factoring methods. However, for higher-degree polynomials, it's a useful technique. If you have a cubic, and you know one of the roots is x = R, you can divide by (x - R) to get a quadratic, which you can then factor or solve normally. Or if you have a quartic polynomial, and you can find a root, you can divide to get a cubic; then if you can find another root, you can divide to get a quadratic. And so on. Does it make more sense now? - Doctor Ian, The Math Forum http://mathforum.org/dr.math/ Date: 03/12/2004 at 19:48:56 From: Zahra Subject: Thank you (x^3+2x^2-5x-6=0) Dear Dr. Ian: I just wanted to thank you for the great help. It was actually the first time I went on a math web site, and I loved it. I like the way you explained the question to me. I am 14 years old and I'm in 8th grade. This was my first year going to the Mu alpha competitions in Florida. I love it, and I will be there every year! Thanks again for your help. Sincerely, Zahra Date: 03/14/2004 at 19:34:29 From: Zahra Subject: x^3+2x^2-5x-6=0 Dear Dr. Ian - I understand the whole process you explained. But when I tried the original question I asked you it doesn't work. Here is what I did: x^2 + 4x + 3 _________________ x - 2 )x^3 + 2x^2 - 5x - 6 x^3 - 2x^2 ------------- 4x^2 - 5x - 6 4x^2 - 8x ---------- 3x - 6 3x - 6 ------ 0 Now, I don't think my answer is right. But I thought maybe I have to factor the answer I got, so I did. Here is what I got: x^2 + 4x + 3 (x + 1)(x + 3) But this can not be right because they say "given that x = 2". Thanks for the great help. Zahra Date: 03/14/2004 at 21:09:13 From: Doctor Ian Subject: Re: x^3+2x^2-5x-6=0 Hi Zahra, Actually, you've done all the steps perfectly! The only thing that remains is to see what the results are telling us. Recall that the first division tells us that x^3 + 2x^2 - 5x - 6 = (x - 2)(x^2 + 4x + 3) And you factored the quadratic, which shows that x^3 + 2x^2 - 5x - 6 = (x - 2)(x + 1)(x + 3) Note that for the expression on the left to be equal to zero, the expression on the right must also be equal to zero. But when is that true? The only way to get a product of zero is for one or more of the factors to be equal to zero. (Do you see why?) We've been told that x = 2 is a solution, and indeed, substituting 2 for x gives us a value of zero: (2 - 2)(2 + 1)(2 + 3) = 0 * 3 * 5 = 0 But there are two other solutions as well. One of them is x = -1, because adding -1 to 1 will make the second term zero: (-1 - 2)(-1 + 1)(-1 + 3) = -3 * 0 * 2 = 0 Can you see what the third solution has to be? - Doctor Ian, The Math Forum http://mathforum.org/dr.math/ Date: 03/15/2004 at 21:28:45 From: Zahra Subject: x^3+2x^2-5x-6=0 Dear Dr. Ian: Since (x + 1) gives a solution of -1 and (x - 2) gives 2, I would say that the third solution from (x + 3) is -3. Therefore, the sum of the solutions is (2 + -1 + -3) = -2 So I think I understand it now. Thank you for the great help! Sincerely, Zahra Date: 03/15/2004 at 21:54:26 From: Doctor Ian Subject: Re: x^3+2x^2-5x-6=0 You're most welcome! Let us know if you want help with anything else. - Doctor Ian, The Math Forum http://mathforum.org/dr.math/ |
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