Hi Lesya P
If by zero you mean the zeroes or x intercepts these can be found by simply setting the factors equal to zero. Since your function is in factored form you can do that directly
zeroes -2 and 1
multiplicity is 1 x =1
UPDATED multiplicity is 3 for x =-2 per the associated exponent
You can graph your function on graphing calculator, or at desmos.com to see the zeroes.
Doug C.
And since the multiplicity of each zero is odd, the graph "passes through" the x-axis at those zeros. Try changing the exponent to an even integer on one or both factors to see that the graph "bounces off" the x-axis at zeros with even multiplicity. Then try raising the exponents to higher multiplicity (both odd and even) to see the graph flatten in the vicinity of the zeros: desmos.com/calculator/6lhf6bnf7e
Report
09/01/23
Brenda D.
tutor
Thanks and yes the multiplicity of x=-2 is 3 it appears once in the graph but per the exponent it is still (x+2)(x+2)(x+2). I will update
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09/02/23
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