Finding the Zeros of a Function
Date: 05/12/99 at 13:38:47 From: Toni Subject: Finding zeros Can you tell me how I would find the zeros of P(x) = (x^2-3)(2x-7)^2(x-1)^3 and what is meant by "state the multiplicity of each"?
Date: 05/13/99 at 17:11:53 From: Doctor Rick Subject: Re: finding zeros Hi, Toni. By definition, the zeros of P(x) are the values of x such that P(x) = 0. So finding the zeros of P(x) is the same as solving the equation (x^2-3)(2x-7)^2(x-1)^3 = 0 If a function is a product of expressions, P(x) = f(x)g(x)...h(x), then P(x) = 0 if and only if one (or more) of f(x), g(x), ...h(x) equals 0. (This is true because 0 * y = 0 for all y, and the product of any two non-zero numbers cannot be zero.) Since your P(x) is such a product, you need only look for zeros of the factors in the product. The problem is equivalent to the solution of x^2-3 = 0 or (2x-7)^2 = 0 or (x-1)^3 = 0 The first factor, (x^2-3), is not a product as it stands, but we can rewrite it as a product of two linear expressions: x^2 - 3 = (x + sqrt(3))(x - sqrt(3)) A difference of squares can always be written as the product of a sum and a difference in this way. Since a zero of either of these factors is a zero of (x^2 - 3), we have two new equations to solve: x + sqrt(3) = 0 or x - sqrt(3) = 0 Both these equations are linear - they only have the first power of the variable. A linear equation by definition has a zero of multiplicity 1; the zeros of these two linear equations are x = -sqrt(3) and x = sqrt(3). These are both zeros of the whole equation. The second and third factors are themselves products - the second is (2x-7)(2x-7) = 0 which is equivalent to 2x-7 = 0 or 2x-7 = 0 When x = 7/2, BOTH these equations are true. We say that x = 7/2 is a zero of multiplicity 2 because it is a zero of two linear factors of the function. Likewise with the third factor, (x-1)(x-1)(x-1) = 0 x-1 = 0 or x-1 = 0 or x-1 = 0 When x = 1, all 3 equations are true, so x = 1 is a zero of multiplicity 3. It's time to put it all together. The zeros of P(x) = (x^2-3)(2x-7)^2(x-1)^3 are: x = 7/2, multiplicity 2 x = 1, multiplicity 3 x = sqrt(3), multiplicity 1 x = - sqrt(3), multiplicity 1 Once you get the idea, you can practically read off the zeroes and their multiplicities directly from a function written in this form. Here is an extra tidbit. If you were to expand P(x) as a polynomial, the term with the highest exponent of x would be x^7 (actually 4x^7), making this a polynomial of degree 7. The sum of the multiplicities of the zeros of a polynomial is always the degree of the polynomial. In this way you can be sure that you have found all the zeros. I hope this helps. - Doctor Rick, The Math Forum http://mathforum.org/dr.math/
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