according to the fundamental theorem of algebra, how many zeros does the function f (x) = 17x^15+41x^12+13x^3-10 have?

Question

Mathematics

Isaiah Pugh

15 December, 02:56

according to the fundamental theorem of algebra, how many zeros does the function f (x) = 17x^15+41x^12+13x^3-10 have?

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Answers (1)

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Nola Craig · Answer 1

Answer: 15

Step-by-step explanation:

The total number of zeros (which includes real & complex roots) is the degree of the function. The degree of a function is the biggest exponent.

f (x) = 17x¹⁵ + 41x¹² + 13x³ - 10

has a degree of 15, so the function has 15 zeros.