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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.

Ask by Huff Hanson. in the United States
Mar 15,2025

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The polynomial \( f(x) = 8x^8 - 3x^5 + 8x \) has a total of 8 zeros, including their multiplicities.

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To find the number of zeros of the function \( f(x) = 8x^8 - 3x^5 + 8x \), we can factor out the common term, which is \( x \): \[ f(x) = x(8x^7 - 3x^4 + 8) \] This shows that \( x = 0 \) is one zero. Now, we need to find the zeros of the polynomial \( 8x^7 - 3x^4 + 8 \). Finding the roots of \( 8x^7 - 3x^4 + 8 = 0 \) can be complex, but we can analyze its degree and behavior. Since it's a 7th degree polynomial, it can have up to 7 zeros, considering their multiplicity (and potentially complex roots). To determine the real roots, we might need numerical methods or graphing tools. Nevertheless, without specific tools, we note that this polynomial does not change signs (as can be observed by evaluating it at various points), indicating that it may not have any real zeros apart from the one at \( x = 0 \). Thus, the total number of zeros of \( f(x) = 8x^8 - 3x^5 + 8x \), counting multiplicities, is: 1 (the zero at \( x = 0 \)) + 0 (potential real zeros from the polynomial) = **1**. So, the number of zeros is \( \boxed{1} \).

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