Raymond B. answered 12/18/25
Math, microeconomics or criminal justice
(x+2)^3(x-1) has zeroes -2 with multiplicity 3 and 1 with multiplicity 1
zeroes = the x intercepts = roots
set each factor = 0 and solve for x to find the zeroes'
(x+2)(x+2)(x+2)(x-1) has zeroes -2,-2,-2,1
Nikolai O.
I believe that the multiplicity for -2 is 3, not 1. Multiplicity refers to the number of times a zero appears in the factorization of the f(x). The polynomial here is: (x+2)^3 * (x-1) Which is equal to: (x+2) * (x+2) * (x+2) * (x+1) Thus, the zero x = -2 "appears" in the factored form of f(x) three times. Thus, x = -2 has a multiplicity of three, not one. Similarly, the zero x = -1 "appears" in the factored form of f(x) once. Thus, x = -1 has a multiplicity of one. Side note: The multiplicity of all zeros in a polynomial must add up to its degree. f(x) in expanded form is: (x+2)^3 * (x+1) = x^4 + 7x^3 + 18x^2 + 20x + 8 Since the highest term is x^4, f(x) is a fourth-degree polynomial. Notice that the zeros have a multiplicity of the zeros of f(x) are: x = -2 <- Multiplicity of three x = 1 <- Multiplicity of one Therefore, the multiplicity "add up" to four. This is a great way to check whether or not your answer is correct.09/01/23