Register

According to the fundamental theorem of algebra, how many zeros does the function f(x) = 15x17 + 41x12 + 13x3 − 10 have?

According to the fundamental theorem of algebra, how many zeros does the function f(x) = 15x17 + 41x12 + 13x3 − 10 have?
214k views
1 vote
According to the fundamental theorem of algebra, how many zeros does the function f(x) = 15x17 + 41x12 + 13x3 − 10 have?

User Pratik Khadka
by
8.2k points

2 Answers

2 votes

Answer:

17 zeros!

Explanation:

User Josh Bush
by
8.4k points
5 votes
The number of zeroes = the number of degrees of the equation ( highest exponent.

So the answer to this one is 17 zeroes.
User Elias Prado
by
9.0k points
← Prev Question Next Question →

Related questions

User Oyarzun
asked Dec 27, 2022 74.4k views
distributive law to simplify 15x13 + 15x17
Oyarzun asked Dec 27, 2022
by Oyarzun
8.9k points
2 answers
0 votes
74.4k views
Ask a Question
Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.

9.4m questions

12.2m answers

Categories

Other Questions

How do you can you solve this problem 37 + y = 87; y =
What is .725 as a fraction
How do you estimate of 4 5/8 X 1/3
Link Copied!