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$x^3-5x^2+x+8=0$

I know that the zeros are approx. $-1.07$, $1.72$, and $4.34$ by looking by using a graphing calculator, but how do I find the zeros without? Rational roots theorem does not work here!

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    $\begingroup$ Is this an exercise or a part of another problem? In general, you either have to use the general formula or modify the polynomial to another form using some trick (e.g. knowing one answer beforehand!). Sometimes other things in the problem may help in finding such a trick. $\endgroup$ – JiK Jan 14 '14 at 20:34
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Use the bisection-method or the newton-method. Also possible is the regula falsi.

Idea of the bisection-method :

Let a and b be numbers, such that f(a) < 0 and f(b) > 0 or vice versa. Then, take m := $\frac{a+b}{2}$.

If f(m)=0, you are done. Otherwise choose a or b, such that the signs differ and continue the process.

The newton-method iterates as follows :

$$x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}\ ,$$

with some starting value $x_0$

Regula falsi is similar, but uses

$$\frac{f(b)-f(a)}{b-a}$$

instead of f '(x).

Here a,b are numbers with f(a) < 0 and f(b) > 0 or vice versa.

Regula falsi does not need the derivates, but converges slightly slower than the newton-method.

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  • $\begingroup$ I've recently skimmed through a college algbra and trigonometry book and did not find this concept. Under what level math can I learn about this topic? $\endgroup$ – user121357 Jan 14 '14 at 20:35
  • $\begingroup$ Most handbooks or formula collections should mention those methods. $\endgroup$ – Peter Jan 14 '14 at 20:37
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As you note, there is no rational root. I'm afraid you have to either use a numerical approximation or use the general solution of cubic. There are many methods to find the general solution in closed form. Take a look at http://en.wikipedia.org/wiki/Cubic_function

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To count the number of roots in a given interval, we can use the Sturm Sequence.

The Sturm Sequence for $f(x)=x^3-5x^2+x+8$ is the sequence of polynomals $$ x^3-5x^2+x+8\\ 3x^2-10x+1\\ \frac{44}{9}x-\frac{77}{9}\\ \frac{117}{16} $$ If $N(x)$ is the number of sign changes in the Sturm Sequence at $x$, then for $a<b$, $N(a)-N(b)$ is the number of real roots of $f$ in $(a,b)$.

As $x\to-\infty$, the signs are "$-,+,-,+$"; that is, three sign changes: $N(-\infty)=3$.

At $x=0$, the signs are "$+,+,-,+$"; that is, two sign changes: $N(0)=2$.

As $x\to+\infty$, the signs are "$+,+,+,+$"; that is, no sign changes: $N(+\infty)=0$.

Thus, there are three real roots of $f$, one negatve and two positive.

We can use the Sturm Sequence to find intervals with a single root, then use bisection or Newton to find that root.

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