# Determine the different possibilities as to the kind of roots that x^9-4x^8+7x^6+x^5-3x^4+x^2-2x+2 will have.Show answers in a table columns: number of positive roots, number of negative roots,...

Determine the different possibilities as to the kind of roots that x^9-4x^8+7x^6+x^5-3x^4+x^2-2x+2 will have.

Show answers in a table

columns: number of positive roots, number of negative roots, number of imaginary roots

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Given `x^9-4x^8+7x^6+x^5-3x^4+x^2-2x+2` , find the number and type of possible roots:

(1) From a corollary of the fundamental theorem of algebra we know that there are 9 roots, some of which could be imaginary. Further, we know that all imaginary roots appear as conjugate pairs.

**Therefore, there is at least 1 real root.**

(2) We can apply Descartes' rule of signs to determine the possible number of positive and negative real roots. (See links below)

(a) The number of sign changes in the expression is 6. **So the possible number of positive roots are 6,4, and 2**

(b) Replacing x with -x, we reapply the rule of signs:

`-x^9-4x^8+7x^6-x^5-3x^4+x^2+2x+2` has 3 sign changes, **so the number of possible negative roots is 3 or 1**

(4) Putting it all together:

**We list the possible roots as positive, negative, imaginary:**

**6,3,0 6 positive, 3 negative, no imaginary****6,1,2** etc...**4,3,2****4,1,4****2,3,4****2,1,6****0,3,6****0,1,8**

There are 8 possible combinations of positive, negative, and imaginary roots. The actual roots are 2 positive, 1 negative, and 6 imaginary. Note that we are guaranteed at least 1 negative root.

**Sources:**