Question

What are the possible number of positive, negative, and complex zeros of `f(x) = –3x^4 – 5x^3 – x^2 – 8x + 4`?

Answer 1

Look at changes of signs to find this has `1` positive zero, `1` or `3` negative zeros and `0` or `2` non-Real Complex zeros.

Then do some sums...

Explanation:

`f(x) = -3x^4-5x^3-x^2-8x+4`

Since there is one change of sign, `f(x)` has one positive zero.

`f(-x) = -3x^4+5x^3-x^2+8x+4`

Since there are three changes of sign `f(x)` has between `1` and `3` negative zeros.

Since `f(x)` has Real coefficients, any non-Real Complex zeros will occur in conjugate pairs, so `f(x)` has exactly `1` or `3` negative zeros counting multiplicity, and `0` or `2` non-Real Complex zeros.

`f'(x) = -12x^3-15x^2-2x-8`

Newton's method can be used to find approximate solutions.

Pick an initial approximation `a_0`.

Iterate using the formula:

`a_(i+1) = a_i - f(a_i)/(f'(a_i))`

Putting this into a spreadsheet and starting with `a_0 = 1` and `a_0 = -2`, we find the following approximations within a few steps:

`x ~~ 0.41998457522194`

`x ~~ -2.19460208831628`

We can then divide `f(x)` by `(x-0.42)` and `(x+2.195)` to get an approximate quadratic `-3x^2+0.325x-4.343` as follows:

Notice the remainder `0.013` of the second division. This indicates that the approximation is not too bad, but it is definitely an approximation.

Check the discriminant of the approximate quotient polynomial:

`-3x^2+0.325x-4.343`

`Delta = b^2-4ac = 0.325^2-(4*-3*-4.343) = 0.105625 - 52.116 = -52.010375`

Since this is negative, this quadratic has no Real zeros and we can be confident that our original quartic has exactly `2` non-Real Complex zeros, `1` positive zero and `1` negative one.