he number of zeros of $$ f(x)=8 x^{8}-3 x^{5}+8 x $$ is $$ \square $$, provided that each zero is counted according to its multiplicity.

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1. Factor the given polynomial:
   $$
   f(x)=8x^8-3x^5+8x = x\left(8x^7-3x^4+8\right)
   $$

2. Notice that $$f(x)$$ is written as a product:
   - The factor $$x$$ contributes a zero (namely, $$x=0$$) with multiplicity 1.
   - The other factor is a degree 7 polynomial, $$8x^7-3x^4+8$$.

3. By the Fundamental Theorem of Algebra, a polynomial of degree 7 has 7 zeros (counting multiplicities) in the complex numbers.

4. Therefore, the total number of zeros for $$f(x)$$ is:
   $$
   1 + 7 = 8
   $$

Thus, the number of zeros of $$f(x)=8x^8-3x^5+8x$$ is $$8$$.
The polynomial $$ f(x) = 8x^8 - 3x^5 + 8x $$ has a total of 8 zeros, including their multiplicities.

he number of zeros of is , provided that each zero is counted according to its multiplicity.

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