Remainder and Factor TheoremDate: 01/29/98 at 08:09:46 From: Yanzhen Subject: Remainder and Factor Theorem Dear Dr. Maths, How do I answer the following questions? 1) 2x^4 - 3x^3 - 18x^2 - 7 divided by (x^2) -1 2) When the polynomial p(x) is divided by (x-1), the remainder is 5, and when p(x) is divided by (x-2), the remainder is 7. Find the remainder, when p(x) is divided by (x-1)(x-2). Thank you Dr Maths! Date: 02/24/98 at 14:30:13 From: Doctor Sonya Subject: Re: Remainder and Factor Theorem Dear Yanzhen, To solve this problem, there is an important theorem you need to know, called the Remainder and Factor theorem. As I see from the subject line, you've already heard of this, but I'll restate it anyway. The theorem says that when you divide a polynomial f(x) by another one g(x), there is a quotient q(x) and a remainder r(x). These are related by the equation f(x) = g(x)q(x) + r(x) The remainder r(x) either is zero, or has degree lower than g(x). For example, if g(x) is a linear polynomial, then r(x) should be a constant, and if g(x) is quadratic, then r(x) should be a linear polynomial of the form Ax+B. I'm not going to prove this theorem, but if you want more information on it, take a look in the Dr. Math archives, or write us back. All right, now that this theorem's been stated, here's how we'll use it in your problems. (1) Write f(x) = 2x^4 - 3x^3 - 18x^2 - 7. If the remainder is Ax+B when this is divided by x^2 - 1 = (x-1)(x+1), we have f(x) = (x-1)(x+1)q(x) + Ax + B. This may not look like we've simplified anything, but wait, what happens if you plug x = 1 into both sides of the equation? You get: f(1) = (1-1)(1+1)q(1) + 1A + B Now evaluate both sides and see what you get. Try the same thing with x = -1. Doing this will give you two equations for A and B, which you can then solve. Once you have A and B, plug their values back into the remember Ax + B, and you'll be done. (2) For your second problem, I'm going to write out what the problem says in terms of the remainder and factor theorem. p(x) = (x-1)q(x) + 5 p(x) = (x-2)Q(x) + 7 (Notice here that q(x) and Q(x) are different quotients.) Now, try plugging 1 into the first equation and 2 into the second equation. You'll get that: p(1) = 5 and that p(2) = 7 . Now, suppose the remainder is Ax+B when p(x) is divided by (x-1)(x-2). We write p(x) = (x-1)(x-2)q(x) + Ax + B. Here, q(x) is the quotient. Putting x = 1 and x = 2 respectively will give you two equations for A and B, which you can then solve simultaneously to find out what Ax + B is. When you plug in, remember that you already know what p(1) and p(2) are. -Doctors Sonya and Yiu, The Math Forum Check out our Web site http://mathforum.org/dr.math/ |
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