what is the number of complex roots (maximum) : f(x)=10x^5-15x^4+12x^3-18x^2+2x-3

Answer 1

Decartes rule of signs tells us that we have a maximum of 5 positive roots as there are 5 sign changes and no negative roots as the are no sign changes when we evaluate f(-x) Going down by twos we could also have 3 or 1 positive roots

+ roots -roots imaginary roots total roots

5 0 0 5

3 0 2 5

1 0 4 5

So maximum number of imaginary roots is 4. That does not means that there are necessarily 4 imaginary root; this is just the maximum possible

Answer 2

Note how, looking at coefficients in pairs, the second coefficient is 3/2 * the first. This, in combination with the fact that all odd coefficients are positive, guarantees that there is one real coefficient, 3/2. (if the value is greater than 3/2, each pair of terms will add to a positive, and if it is less than 3/2, each pair will be negative - we could factor the successive terms as x^even number * (x - 3/2).

1 real and 4 complexes.

Now, to show this result algebraically.

3/2 | 10 -15 12 -18 2 -3

15 0 18 0 3

10 0 12 0 2 0

Now, the factoring is then (2x - 3)(5x^4 + 6^2 + 1)

(We take half of the synthetic division bottom to change x - 3/2 into 2x - 3

5x^4 + 6x^2 + 1 >= 1 for all x.5

Thus, this factor has four complex roots, and the polynomial has four complex roots

We can easily find the complex roots.

5x^4 + 6x^2 + 1 = (5x^2 + 1)(x^2 + 1) = 5(x + 1/√5 i)(x - 1/√5 i)(x + i)(x - i)

Thus, the original polynomial factors as 5(2x - 3)(x + 1/√5 i)(x - 1/√5 i)(x + i)(x - i)

If you prefer, we can write this as (2x - 3)(√5x + i)(√5x - i)(x + i)(x - i)

Our roots are (3/2, -1/√5 i, - i, i, 1/√5 i)

If we had simply answered the question as to the maximum possible number of complex roots without factoring, our answer would have also been four, as the maximum number if the degree of the polynomial if even and the degree - 1 if odd.