I'm a tutor working with an Algebra II student on rational functions, and this problem is stumping both of us:


She is supposed to find the zero of the function by hand, but we cannot figure out a good way to do so. Google tells me that we could use the Newton-Raphson method, but that seems quite advanced for an Algebra II class. I also read about the cubic equation, but that also seems very tedious and complex.

Does anyone have any thoughts? I'm thinking there might just be a typo in the problem given by her teacher, or maybe we are missing an obvious way to solve it! Any suggestions would be helpful :)

  • $\begingroup$ Have you tried to find roots of the numerator by Ruffini method? Test divisors of the independant term 8 $\endgroup$
    – Ripi2
    Jan 26, 2021 at 23:57
  • $\begingroup$ Welcome to MSE, first try substituting the numerator with $\pm 1, \pm 2, \pm 4, \pm 8$ then use polynomial long division to find the remaining roots. $\endgroup$ Jan 26, 2021 at 23:59
  • $\begingroup$ The only real root is approximately 2.9311. $\endgroup$
    – Andy Walls
    Jan 27, 2021 at 0:06
  • $\begingroup$ Do you have to find it. As $2^3 -2*2^3 -8 =-8 < 0$ and $3^3 -2*3^2 -8 = 1>0$ it is clear there is a root between $x=2$ and $x=3$. Is there any reason we have to find it. $\endgroup$
    – fleablood
    Jan 27, 2021 at 0:31

4 Answers 4


The root in the numerator in the answer from @MPW probably comes from Wolfram alpha. I don't think this can be what was expected of the student.

The graph there shows it's very near $3$. If it were $3$ (which is also a root of the denominator) you could proceed formally.

That suggests a typo. The $-8$ should probably be $-9$.

Even if that's the case there is a problem. The function's value everywhere other than $x=3$ and $x=0$ is $$ \frac{x^2 + x + 3}{x}. $$ The quadratic in the numerator has no real roots, so neither does the original.


The only real solution is

$$x = \frac13\left(2 + \sqrt[3]{116 - 12\sqrt{93}} + 2^{2/3}\sqrt[3]{29 + 3\sqrt{93}}\right)$$

My guess is there is a typo in the problem. It seems unlikely an Algebra II student would be expected to find this.


The simplest solution is to multiply both sides by the denominator and then solve using the cubic formula

$$\frac{x^3-2x^2-8}{x^2-3x}=0\implies x^3-2x^2-8=0\\ \implies \qquad a=1\qquad b=-2\qquad c=0 \qquad d=-8 $$

$$x=\sqrt[\Large{3}]{\bigg(\frac{-b^3}{27a^3 }+\frac{bc}{6a^2}-\frac{d}{2a}\bigg)+\sqrt{\bigg(\frac{-b^3}{27a^3}+\frac{bc}{6a^2}-\frac{d}{2a}\bigg)^2+\bigg(\frac{c}{3a}-\frac{b^2}{9a^2}\bigg)^3}}\\ +\sqrt[\Large{3}]{\bigg(\frac{-b^3}{27a^3 }+\frac{bc}{6a^2}-\frac{d}{2a}\bigg)-\sqrt{\bigg(\frac{-b^3}{27a^3}+\frac{bc}{6a^2}-\frac{d}{2a}\bigg)^2+\bigg(\frac{c}{3a}-\frac{b^2}{9a^2}\bigg)^3}}-\frac{b}{3a}$$

This will give you a real solution and, divided into the original, will give you a second degree equation that can be solved with the quadratic formula.

You can also go to WolframAlpha here to find the answer the easy way: $x≈2.9311....$


Perhaps the teacher is pushing students to imagine a method based on iterations. Newton-Rapson is such a method, but it seems too hard for a novice.

Other, simple, iteration method is just trying values and see when the function changes its sign, a good signal that we are getting close to the solution.

The zeros are those of the numerator. But watch out the denominator: x=0 or x=3 make infinity, so can't use them.

Let's build a table:

x | f(x) numerator
-1| -11
1 | -9
2 | -8
3 | 1 (but forbbiden by denominator)
4 | 24

So, the solution goes between 2 and 3.

Then build a new table with 2.2, 2.4, 2.6, 2.8 and 2.9
The next table may use 2.92, 2.94, 2.96, 2.98

And continue this method until you get some decimals.


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