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How many possible positive and negative real roots does the polynomial x^5 - 4x^4 + 5x^3 - 9x^2 + 2x - 8 have?

a) 0
b) 1
c) 2
d) 3

User CodyK
by
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1 Answer

5 votes

Final answer:

The polynomial x^5 - 4x^4 + 5x^3 - 9x^2 + 2x - 8 can have up to 3 possible positive real roots and 1 possible negative real root, as determined by Descartes' Rule of Signs. The polynomial has up to 3 positive and 1 negative real roots.

Step-by-step explanation:

To determine how many possible positive and negative real roots the polynomial x^5 - 4x^4 + 5x^3 - 9x^2 + 2x - 8 has, we can use Descartes' Rule of Signs. First, we count the number of sign changes in the polynomial to find the number of possible positive roots. The original polynomial has 3 sign changes (from x^5 to -4x^4, from 5x^3 to -9x^2, and from 2x to -8), indicating up to 3 possible positive real roots.

Next, we replace x with -x to find the number of possible negative roots: (-x)^5 + 4(-x)^4 + 5(-x)^3 + 9(-x)^2 - 2(-x) - 8 = -x^5 - 4x^4 - 5x^3 + 9x^2 + 2x - 8, which has 1 sign change (from 9x^2 to 2x), indicating there is 1 possible negative real root. Therefore, the polynomial has up to 3 positive and 1 negative real roots.

User Antwane
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