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Mathematics 14 Online
OpenStudy (anonymous):

How do i Find the zeros for each equation and also how do i figure out the multiplicity of any zeros.

OpenStudy (mathstudent55):

The only way a product (multiplication) can equal zero is if at least one of the factors is zero. For example, let's say you're working with variables a, b, and c. If you are told that the product of a, b, and c equals zero, abc = 0 Then either a = 0 or b = 0 or c = 0 (or even two variables or even all three variables are zero, but at least one of them has to zero.) If none of those factors were zero, then you could not get them to multiply to zero. Do you undersrtand this so far?

OpenStudy (mathstudent55):

Using the above idea, when you are asked to solve an equation that looks like this: (x - 2)(x + 5) = 0, ask yourself, how can the product of x - 2 and x + 5 be equal to zero? The only way is if either x - 2 = 0 or x + 5 = 0. Now you need to find out what values of x would cause the factors to be zero, sol you solve two equations: x - 2 = 0 or x + 5 = 0 The solutions are: x = 2 or x = -5 This means that if x = 2, or x = -5, the produict of (x - 2)(x + 5) = 0 Since there was only one factor of x - 2 with solution x = 1 and one factor of x + 5 with solution x = -5, the multiplicity of each solution is 1. If you have a problem with factors raised to powers, such as (x - 2)^3(x + 5)^2 = 0, then you really have (x - 2)(x - 2)(x - 2)(x + 5)(x + 5) = 0 The next step should be x - 2 = 0 or x - 2 = 0 or x - 2 = 0 or x + 5 = 0 or x + 5 = 0 This will give the solutions x = 2 or x = 2 or x = 2 or x = -5 or x = -5 Since we are smart enough to realize that equal factors (such as the three factors x - 2) will give the same solution, we only need to solve them once, but to acknowlege that there are three factors x - 2, with three solutions x = -2, we state x = 2, multiplicity 3. Therefore when we se a problem such as (x - 2)^3(x + 5)^2 = 0 all we do to solve is x - 2 = 0 or x + 5 = 0 x = 2, multiplicity 3, or x = -5, multiplicity 2

OpenStudy (mathstudent55):

Often, these equations are not given to you as (x - 2)^2(x +6) = 0 They are really given as a polynomial equaling zero, and you first need to factor the polynomial before you have all the factors multiplied together equaling zero.

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