To some extent what you can say about the location of the roots of

polynomials depends on how much work you do. But I will assume that

you are expected to use Descartes rule of signs for these.

I will do the first one.

Look at the signs of the coefficients:

+ - - + - +

These change from + to - or - to + 4 times, so by DesCartes rule of

signs this can have either 4, 2 or 0 positive real roots.

Now replace x by -x in this to obtain:

-x^5 - 6x^4 + 3x^3 + 7x^2 + 8x + 1

so now we have:

- - + + + +

so there is 1 change from + to - or - to +, so in this case there must be

exactly 1 negative real root, (because if there are k changes in the signs

there can be k, k-2, ... , q, where q is zero if k is even and 1 if k is odd).

Now complex roots of a polynomial with real coefficients occur in conjugate

pairs, and the total number of roots is equal to the order of the polynomial

so here the order is 5 so there are a total of 5 roots altogether.

So here we have 1 negative root, 4, 2 or 0 positive real roots, with 0, 2 or 4

complex roots depending on the number positive roots. So in list form we

have the following possibilities:

1 negative, 4 positive 0 complex

1 negative, 2 positive 2 complex

1 negative, 0 positive 4 complex

The other problems are similar

RonL