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9 votes
According to Descartes’s rule of sign, how many possible positive and negative roots are there for the equation

0 = -4x^6 - 3x^5 + 2x^2 - 4x + 1

Number of possible positive roots -

Number of possible negative roots -

0
1, only
2, only
0 or 2
3, only
1 or 3

User Hangon
by
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1 Answer

29 votes
29 votes

Answer:

answer below

Explanation:

-4x⁶ - 3x⁵ + 2x² - 4x + 1

↑ ↑ ↑

changes signs 3 times: possible positive roots 3 or 3-2=1

replace x with -x

-4x⁶ + 3x⁵ + 2x² +↓ 4x + 1

possible negative roots: 1

User BitRulez
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