How do you find all the zeros of -9x^4 + 3x^3 - 7x^2 + 7x - 9 = 0?

May 8, 2016

See explanation...

Explanation:

$f \left(x\right) = - 9 {x}^{4} + 3 {x}^{3} - 7 {x}^{2} + 7 x - 9$

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 $- 9$ and $q$ a divisor of the coefficient $- 9$ of the leading term.

That means that the only possible rational zeros are:

$\pm \frac{1}{9}$, $\pm \frac{1}{3}$, $\pm 1$, $\pm 3$, $\pm 9$

In addition, note that $f \left(- x\right) = - 9 {x}^{4} - 3 {x}^{3} - 7 {x}^{2} - 7 x - 9$ has all negative coefficients, so there are no negative zeros.

So the only possible rational zeros are:

$\frac{1}{9} , \frac{1}{3} , 1 , 3 , 9$

Evaluating $f \left(x\right)$ for these values we find that none of them are zeros. So $f \left(x\right)$ has only irrational and/or Complex zeros.

At this stage we have two main choices. We can seek an algebraic solution, which is doable, but lengthy and complicated. Alternatively, we can seek a numerical solution.

One simple method to seek numerical values for the zeros is Newton's:

• Start with an approximation ${a}_{0}$ for the zero of $f \left(x\right)$.

• Refine the approximation using the formula:

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

• Repeat until you get the desired numerical accuracy.

In our example:

$f ' \left(x\right) = - 36 {x}^{3} + 9 {x}^{2} - 14 x + 7$

So our iterative step is:

${a}_{i + 1} = {a}_{i} - \frac{- 9 {x}^{4} + 3 {x}^{3} - 7 {x}^{2} + 7 x - 9}{- 36 {x}^{3} + 9 {x}^{2} - 14 x + 7}$

$= {a}_{i} - \frac{9 {x}^{4} - 3 {x}^{3} + 7 {x}^{2} - 7 x + 9}{36 x - 9 {x}^{2} + 14 x - 7}$

The slight complication is that the zeros we are looking for are all non-Real Complex ones, so to use this formula in a spreadsheet (which typically do not understand Complex numbers), you need to split the Real and imaginary parts into separate columns and express the Complex arithmetic in terms of those parts.

This takes a little patience to set up, but once you have it, you can try different initial approximations, ${a}_{0}$ easily to find zeros:

$x \approx - 0.48389 \pm 0.98869 i$

$x \approx 0.65056 \pm 0.63411 i$