How do you find the zeros of the function $f\left(x\right)=\frac{{x}^{3}+{x}^{2}-6x}{x-1}$?

Natalya Mayer
2022-09-12
Answered

How do you find the zeros of the function $f\left(x\right)=\frac{{x}^{3}+{x}^{2}-6x}{x-1}$?

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A problem in Dummit and Foote states:

Let k be a field and let k(x) be the field of rational functions in x with coefficients from k. Let$t\u03f5k\left(x\right)$ be the rational function $\frac{P\left(x\right)}{Q\left(x\right)}$ with relatively ' polynomials $P\left(x\right),Q\left(x\right)\u03f5k\left[x\right]$ , with $Q\left(x\right)\ne 0$ . Then k(x) is an extension of k(t) and to compute its degree it is necessary to compute the minimal polynomial with coefficients in k(t) satisfied by x.

By k(t), do they mean k adjoin t, i.e. the set of polynomials$k}_{0}+{k}_{1}\frac{P\left(x\right)}{Q\left(x\right)}+\dots +{k}_{n}{\left(\frac{P\left(x\right)}{Q\left(x\right)}\right)}^{n$ ? Or do they mean the set of rational functions in t, e.g. $\frac{\frac{P\left(x\right)}{Q\left(x\right)}+1}{2{\left(\frac{P\left(x\right)}{Q\left(x\right)}\right)}^{2}+3}$

Thanks for answer!

Let k be a field and let k(x) be the field of rational functions in x with coefficients from k. Let

By k(t), do they mean k adjoin t, i.e. the set of polynomials

Thanks for answer!

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$\int \frac{dx}{1+\sqrt{x+1}}$

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