How do you use the Rational Zeros theorem to make a list of all possible rational zeros, and use the Descartes rule of signs to list the possible positive/negative zeroes of \[f\left( x \right)={{x}^{3}}-7{{x}^{2}}+x-7\]?

Question

How do you use the Rational Zeros theorem to make a list of all possible rational zeros, and use the Descartes rule of signs to list the possible positive/negative zeroes of \[f\left( x \right)={{x}^{3}}-7{{x}^{2}}+x-7\]?

Vedantu Content Team · Accepted Answer

Hint: These types of problems are pretty straight forward and are very easy to solve. For such problems we first of all need to understand what does the Descartes rule state? It is basically a rule of signs. It states that the number of positive real zeros is equal to the number of sign changes in the given function $$f\left( x \right)$$ or polynomial or minus an even integer. For negative real roots, we need to find the number of sign changes by applying a transformation $$f\left( -x \right)$$ . Thus we can solve the given problem just by observation by calculating the number of sign changes. Complete step-by-step solution:Now, we start off with the solution of the problem, we can write that,In the given function or polynomial there are a total of $$4$$ terms. The first term is positive, the second term is negative, the third term is positive and the fourth term is negative again. Thus from this we can easily see that there are a total of $$3$$ sign changes in the given polynomial i.e. from the first term to the second, from the second to the third and from the third to the fourth. Thus there are $$3$$ positive rational roots or $$3-2=1$$ positive rational roots. We subtracted $$2$$ here, because according to the rule, we have to minus an even integer, and the least even integer which satisfies here is $$2$$ . Now we do,$$\begin{align}  & f\left( -x \right)={{\left( -x \right)}^{3}}-7{{\left( -x \right)}^{2}}+\left( -x \right)-7 \\  & =f\left( -x \right)=-{{x}^{3}}-7{{x}^{2}}-x-7 \\ \end{align}$$ From the above formed equation, we see that all the four terms here are negative, and hence there is no sign change, which means that there are no negative rational roots. Note: For smooth solving of these types of problems, we need to remember the Descartes rule and its application for finding positive and negative real roots. We first need to find the number of sign changes for the positive real roots, then transform the function to $$f\left( -x \right)$$ and find the number of sign changes for negative roots. This problem can also be done graphically, by plotting it on the graph paper and then analysing it.