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

$$\frac{x^3-2x^2-8}{x^2-3x}=0$$

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 :)

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  • $\begingroup$ 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

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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.

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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.

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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: $x≈2.9311....$

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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
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-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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