Question

how do you find the mutiplicity of an equation?

Answer 1

what do you mean?

Answer 2

for the equation f(x)=(x-3)^2(x-5)^3(2x+1) I'm asked to find the zeros and the multiplicity. The zeros are 3,5, and -1/2

Answer 3

The multiplicity is the multiple of times that there is that zero. For example,\[(x-3) ^ {2}\]
has a multiplicity of 2, because it has the exponent of 2 there.
\[(x-5) ^{3}\]
and the multiplicity of x-5 is 3, because the exponent is 3.
The rest would be only a multiplicity of 1 because they are only there once.

Answer 4

\(\bf f(x)=(x-3)^\color{red}{2}(x-5)^\color{blue}{3}(2x+1)\\ \quad \\
\implies f(x)=\color{red}{(x-3)(x-3)}\color{blue}{(x-5)(x-5)(x-5)}(2x+1) \)

see the "multiplicity" of those roots?

Answer 5

so the multiplicity is the degree?

Answer 6

Yes.

Answer 7

more or less, yes, is how many times it REPEATS or MULTIPLIES itself

Answer 8

so for f(x)=x^3-x^2-4x+4 my multiplicity would be 3,2, and 1?

Answer 9

no because that polynomial is not in "factored form" first, once you have the factors and thus the roots, then you can see who is REPEATING itself

Answer 10

Not necessarily. That is not in factored form.

Answer 11

This is the graph of the first function. There is a double root at 3 (multiplicity of 2) and a triple root at 5 (multiplicity of 3.)

Answer 12

so how do get the multiplicity of the second problem?

Answer 13

x^3-x^2-4x+4
(x^2-4)(x-1)
x^3-x^2-4x+4
\[(x^{2}-4)(x-1)\]
You can factor the \[(x^{2}-4)\]
down to (\[(x+2)(x-2)\]
So, your final factored form would be \[(x-1)(x-2)(x+2)\]
Because there is no repeated factors (exponent) the multiplicity of all of them would be 1.

Answer 14

so the multiplicity *

Answer 15

got it, thanks!