Problem

how many zeros and multiplicities does $f(x)=2 x^{5}-3 x^{3}+x^{2}+x-2$

Answer

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Answer

\(\boxed{\text{The function } f(x)=2 x^{5}-3 x^{3}+x^{2}+x-2 \text{ has 5 zeros, each with multiplicity 1.}}\)

Steps

Step 1 :The zeros of a polynomial are the values of x that make the polynomial equal to zero. The multiplicity of a zero is the number of times that zero appears as a root. To find the zeros and their multiplicities, we need to set the polynomial equal to zero and solve for x. However, this polynomial is not easily factorable, so we will need to use a numerical method to find the zeros.

Step 2 :The zeros of the function are complex numbers, which means they are not real zeros. The multiplicity of each zero is 1, as the derivative of the function at these points is not zero. Therefore, the function has 5 zeros, each with multiplicity 1.

Step 3 :\(\boxed{\text{The function } f(x)=2 x^{5}-3 x^{3}+x^{2}+x-2 \text{ has 5 zeros, each with multiplicity 1.}}\)

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