How many real roots does this polynomial have? f (x) = 4x^5 + x^3 + 7x - 2 Options: a) 1 b) 3 c) 4 d) 5

I know total number of roots is 5, but how do we find how many of those are real? Please help, thanks.

1 Answer
George C.
Dec 3, 2017

a) #1#

Explanation:

Given:

#f(x) = 4x^5+x^3+7x-2#

Note that the pattern of signs of the coefficients is #+ + + -#. With one change of signs, Descartes' Rule of Signs tells us that #f(x)# has exactly one positive real zero.

Note that:

#f(-x) = -4x^5-x^3-7x-2#

The pattern of the signs of the coefficients of #f(-x)# is #- - - -#. With no changes of signs, Descartes' Rule of Signs tells us that #f(x)# has no negative real zeros.

So #f(x)# has exactly one real zero, which is positive.

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