Question

1. use descartess rule of signs to determine the possible numbers of positive and negative real zeros of $f(x)=5x^4 - 2x^3 - 8x^2 - 6x + 4$.

what is the possible number of positive real zeros?
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what is the possible number of negative real zeros?
\boxed{} (use a comma to separate answers as needed.)

Explanation:

Step1: Analyze \( f(x) \) for positive real zeros

For \( f(x) = 5x^4 - 2x^3 - 8x^2 - 6x + 4 \), count the sign changes:

• From \( 5x^4 \) to \( -2x^3 \): sign change (1st)
• From \( -6x \) to \( +4 \): sign change (2nd)

Total sign changes: 2. By Descartes's Rule, positive real zeros: 2 or \( 2 - 2 = 0 \).

Step2: Analyze \( f(-x) \) for negative real zeros

Compute \( f(-x) = 5(-x)^4 - 2(-x)^3 - 8(-x)^2 - 6(-x) + 4 = 5x^4 + 2x^3 - 8x^2 + 6x + 4 \).
Count sign changes:

• From \( +2x^3 \) to \( -8x^2 \): sign change (1st)
• From \( -8x^2 \) to \( +6x \): sign change (2nd)

Total sign changes: 2. By Descartes's Rule, negative real zeros: 2 or \( 2 - 2 = 0 \).

Answer:

Possible number of positive real zeros: \( 2, 0 \)
Possible number of negative real zeros: \( 2, 0 \)