asked 76.0k views
1 vote
What are the possible numbers of positive, negative, and complex zeros of f(x) = -3x4 + 5x3 - x2 + 8x + 4?

2 Answers

6 votes
Positive Roots: 3 or 1
Negative Roots: 1
answered
User Yangfan
by
7.9k points
3 votes

Answer:

3 positive real root and 1 negative real root and no complex root.

Explanation:

Here, the given function,


f(x) = -3x^4 + 5x^3 - x^2 + 8x + 4

Since, the coefficient of variables are,

-3, 5, -1, 8, 4,

The sign of variables goes from Negative(-3) to positive (5) , positive(5) to negative(-1) and negative (-1) to positive (8),

So, the total changes in sign = 3,

By the Descartes's rule of sign,

Hence, the number of real positive roots = 3,

Also,


f(-x) = -3x^4 - 5x^3 - x^2 - 8x + 4

Since, the coefficient of variables are,

-3, -5, -1, -8, 4

The sign of varibles goes to negative (-8) to positive (4),

So, the total changes in sign = 1,

Hence, the number of real negative roots = 1,

Now, the degree of the function f(x) is 4,

Thus, the highest number of roots of f(x) is 4,

So, it does not have any complex root.

answered
User Tyronen
by
8.6k points
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