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

Jan 7, 2016

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:

$f \left(x\right) = - 3 {x}^{4} - 5 {x}^{3} - {x}^{2} - 8 x + 4$

Since there is one change of sign, $f \left(x\right)$ has one positive zero.

$f \left(- x\right) = - 3 {x}^{4} + 5 {x}^{3} - {x}^{2} + 8 x + 4$

Since there are three changes of sign $f \left(x\right)$ has between $1$ and $3$ negative zeros.

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

$f ' \left(x\right) = - 12 {x}^{3} - 15 {x}^{2} - 2 x - 8$

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

Pick an initial approximation ${a}_{0}$.

Iterate using the formula:

${a}_{i + 1} = {a}_{i} - f \frac{{a}_{i}}{f ' \left({a}_{i}\right)}$

Putting this into a spreadsheet and starting with ${a}_{0} = 1$ and ${a}_{0} = - 2$, we find the following approximations within a few steps:

$x \approx 0.41998457522194$

$x \approx - 2.19460208831628$

We can then divide $f \left(x\right)$ by $\left(x - 0.42\right)$ and $\left(x + 2.195\right)$ to get an approximate quadratic $- 3 {x}^{2} + 0.325 x - 4.343$ 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:

$- 3 {x}^{2} + 0.325 x - 4.343$

$\Delta = {b}^{2} - 4 a c = {0.325}^{2} - \left(4 \cdot - 3 \cdot - 4.343\right) = 0.105625 - 52.116 = - 52.010375$

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.