# How do you find the zeros of f(x) = 3x^4 +4x^3 +6x^2 -4?

May 28, 2016

Use a numerical method to find approximate zeros:

${x}_{1} \approx - 0.908845148702144$

${x}_{2} \approx 0.639396468497863$

${x}_{3 , 4} \approx - 0.53194232656453 \pm 1.41826916487156 i$

#### Explanation:

$f \left(x\right) = 3 {x}^{4} + 4 {x}^{3} + 6 {x}^{2} - 4$

By the rational root theorem, any rational zeros of $f \left(x\right)$ are expressible in the form $\frac{p}{q}$ for integers $p , q$ with $p$ a divisor of the constant term $- 4$ and $q$ a divisor of the coefficient $3$ of the leading term.

That means that the only possible rational zeros are:

$\pm \frac{1}{3}$, $\pm \frac{2}{3}$, $\pm 1$, $\pm \frac{4}{3}$, $\pm 2$, $\pm 4$

Evaluating $f \left(x\right)$ for each of these, we find that none is a zero.

So $f \left(x\right)$ has no rational zeros.

The coefficients of $f \left(x\right)$ have signs in the pattern: $+ + + -$

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

The coefficients of $f \left(- x\right)$ have signs in the pattern: $+ - + -$

Since this has three changes of sign and we already know about one positive zero, that means that $f \left(x\right)$ has $1$ or $3$ negative zeros.

We could use an algebraic method to find the zeros, but the method is lengthy and the resulting expressions in terms of square and cube roots are horribly messy.

So in this case I would recommend a numeric method.

We could use Durand-Kerner, which gives all four zeros in one go, but I will use Newton's method since we know we have at least $2$ Real zeros.

$f ' \left(x\right) = 12 {x}^{3} + 12 {x}^{2} + 6 x$

Newton's method starts by choosing an approximation ${a}_{0}$, then iterating using the formula:

${a}_{i + 1} = {a}_{i} - \frac{f \left({a}_{i}\right)}{f ' \left({a}_{i}\right)} = {a}_{i} - \frac{3 {x}^{4} + 4 {x}^{3} + 6 {x}^{2} - 4}{12 {x}^{3} + 12 {x}^{2} + 6 x}$

For Real zeros we can express this simply in a spreadsheet and find zeros:

${x}_{1} \approx - 0.908845148702144$

${x}_{2} \approx 0.639396468497863$

Then:

$\left(x - {x}_{1}\right) \left(x - {x}_{2}\right) \approx {x}^{2} + 0.26944868020428 x - 0.58111237849157$

Long dividing $f \frac{x}{\left(x - {x}_{1}\right) \left(x - {x}_{2}\right)}$ we get the approximate quadratic:

${x}^{2} + 1.0638846531291 x + 2.29445006281636$

Then we can use the quadratic formula to find zeros:

$x \approx - 0.53194232656453 \pm 1.41826916487156 i$

If you would rather not do the polynomial long division, then a good alternative is to make a more sophisticated spreadsheet with separate columns for Real and imaginary parts expressing the iteration formula above then try some Complex initial approximations.