The roots of a polynomial written in the form
[tex] (x-x_1)(x-x_2)(x-x_3)\ldots(x-x_n) [/tex]
Are exactly
[tex] x_1,\ x_2,\ x_3,\ldots x_n [/tex]
The multiplicity of a solution [tex] x_i [/tex] is how many times the parenthesis [tex] (x-x_i) [/tex] appears in the factorization of the polynomial.
So, if we write the powers explicitly, your polynomial is written as
[tex] (x + 5)^3(x - 9)^2(x + 1) = (x + 5)(x + 5)(x + 5)(x - 9)(x - 9)(x + 1) [/tex]
Which means that the solutions are:
The roots of [tex]f(x) = (x + 5)^3(x-9)^2(x+1)[/tex] are:
-5 with multiplicity 3
9 with multiplicity 2
-1 with multiplicity 1
The given function is:
[tex]f(x) = (x + 5)^3(x-9)^2(x+1)[/tex]
The roots of the function [tex]f(x) = (x + 5)^3(x-9)^2(x+1)[/tex] are the values of x that makes f(x) to be zero
To find the roots of the function [tex]f(x) = (x + 5)^3(x-9)^2(x+1)[/tex], equate each of the terms to zero
The multiplicity of the roots is the power of the term from which the root is gotten
Equate each of the terms to zero
[tex](x+5)^3 = 0\\x + 5 = 0\\x = -5[/tex]
Since the power is 3, the multiplicity is 3
One root is -5 with a multiplicity of 3
[tex](x-9)^2 = 0\\x - 9 = 0\\x = 9[/tex]
Since the power is 2, the multiplicity is 2
One root is 9 with a multiplicity of 2
[tex](x+1)^2 = 0\\x +1 = 0\\x = -1[/tex]
Since the power is 1, the multiplicity is 1
One root is -1 with a multiplicity of 1
Therefore, the roots of [tex]f(x) = (x + 5)^3(x-9)^2(x+1)[/tex] are:
-5 with multiplicity 3
9 with multiplicity 2
-1 with multiplicity 1
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