SOLUTION: Could someone please help? Use Descartes' Rule of Signs to find the number of possible number of positive zeros and the possible number of negative zeros of Q(x)=x^42x^3+4x^2-3x

Question 63740: Could someone please help?
Use Descartes' Rule of Signs to find the number of possible number of positive zeros and the possible number of negative zeros of Q(x)=x^42x^3+4x^2-3x-2.
Find the horizontal and vertical asymptotes of the function f(x)=x^2+3x-2/x-4
Answer by venugopalramana(3286) (Show Source): You can put this solution on YOUR website!
SEE THE FOLLOWING EXAMPLE AND TRY..
Using Descartes Rule of Signs to determine how many
positive and how many negative real zeros the
polynomial functions may have. Do not attempt to find
the zeros.
f(x)=x^6-4x^5-x^4-3x^3+3x^2+x+4
Thank You.
TO FIND NUMBER OF POSSIBLE POSITIVE ROOTS..
1.WRITE F(X) IN DESCENDING ORDER..
IT IS ALREADY WRITTEN HERE.
2.FIND NUMBER OF CHANGES IN SIGNS OF COEFFICIENTS OF
TERMS OF F(X)..
THERE ARE 2 CHANGES ...+1X^6 TO -4X^5...AND...-3X^3 TO
+3X^2.
HENCE THERE WILL BE A MAXIMUM OF 2 POSITIVE ROOTS.
NEXT TO FIND THE NUMBER OF POSSIBLE NEGATIVE ROOTS..
1.WRITE F(-X) IN DESCENDING ORDER..
IT IS ...
F(-X)= X^6+4X^5-X^4+3X^3+3X^2-X+4
2.FIND NUMBER OF CHANGES IN SIGNS OF COEFFICIENTS OF
TERMS OF F(-X)..
THERE ARE 3 CHANGES ...+4X^5 TO -X^4...AND...-X^4 TO
+3X^3....AND
+3X^2 TO -X
HENCE THERE WILL BE A MAXIMUM OF 3 NEGATIVE ROOTS.
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Find the horizontal and vertical asymptotes of the function f(x)=x^2+3x-2/x-4
HOPE YOU MEAN ... f(x)=(x^2+3x-2)/(x-4)..IF YOU DONT PUT BRACKETS IT MAY MEAN ANY THING LIKE.....f(x)=x^2+3x-[2/x]-4..ETC..ANY WAY TAKING AS MENTIONED
F(X) = Y = (X-2)(X-1)/(X-4)
VERTICAL ASYMPTOTE...
FIND WHEN Y WILL TEND TO INFINITY?
AS X TENDS TO 4 , Y TENDS TO INFINITY. HENCE X-4=0...OR...X=4 IS THE VERTICAL ASYMPTOTE.
HORIZONTAL ASYMPTOTE...
FIND WHETHER Y WILL TEND TO A FINITE LIMIT WHEN X TENDS TO + OR - INFINITY?
WE FIND THAT Y TENDS TO INFINITY AS X TENDS TO INFINITY ..HENCE THERE
ARE NO HORIZONTAL ASYMPTOTES

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