RicardoCa

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Answer:

Let `f(x)=x^7-7x^6+x^5-3x^4+x^2-2x+3`

(1) First we note that the degree is odd with a positive leading coefficient. So as `x -> -oo` the function goes to `-oo` and as `x-> oo` the function goes to `oo` .

(2) Since the degree is odd, there must be at least 1 real zero.

(3) The only possible rational zeros are `+-1,+-3` . These are ruled out by checking with substitution or synthetic division.

(4) We note that f(-1)=-6,f(0)=3,f(1)=-6,f(2)=-333,...,f(6)=-42741,f(7)=9642. f(x) is a polynomial, so it is continuous everywhere. We can apply the Intermediate Value Theorem on the intervals [-1,0],[0,1], and [6,7] to conclude that there are real zeros in each interval.

** The Intermediate Value Theorem says that if f(x) is continuous on an interval [a,b], and either f(a)>f(b) or vice versa, then for any real number k between f(a) and f(b) there exists a point c in [a,b] such that f(c)=k. In this case, if f(a)>0 and f(b)<0, there is a c in [a,b] such that f(c)=0 **

(5) Using synthetic division with -1 we get 1,-8,9,-12,12,-11,9,-6; since these numbers are alternating in sign there is no zero less than -1. The function tends to negative infinity for x<-1.

(6) Using synthetic division with 7 we get 1,0,1,4,28,197,1377,9642; Since these numbers are all non-negative, there can be no real zero greater than 7, so the function tends to infinity for x>7.

From (3) we know there are no rational roots. From (4) we know there are at least 3 real roots. Further investigation suggests these are the only real roots as any other roots must be in [-1,7] and there don't appear to be any other roots in this interval.

**Thus there are 3 real roots, and 4 complex roots.**

** The possibilities were 1 real, 6 complex;3 real and 4 complex; 5 real and 2 complex; or 7 real zeros. Complex roots must appear as conjugate pairs, and the odd power guarantees at least one real root. **

20

Chirkash

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