Chapter 6.8, Problem 50SGR

To determine

To calculate; to find the possible number of positive real zeros, negative real zeros, and imaginary zeros of given function

Answer to Problem 50SGR

Solution: $\begin{array}{cccc}\begin{array}{l}no.ofpositive\\ realzeros\end{array}& \begin{array}{l}no.ofnegative\\ realzeros\end{array}& \begin{array}{l}no.ofimaginary\\ zeros\end{array}& \begin{array}{l}totalnumber\\ ofzeros\end{array}\\ 3& 1& 2& 6\\ 1& 1& 4& 6\end{array}$

Explanation of Solution

The given function has degree 6, therefore, it has 6 zeros. To find the positive real zeros, count the number of changes in sign for the coefficients of $f\left(x\right)$

  $\begin{array}{l}f\left(x\right)=x^6-5x^3+x^2+x-6functionf\left(x\right)\\ f\left(x\right)=+-++-showthesignsforeachterm\end{array}$

There are 3 sign changes, so there are 3 or 1 positive real zeros

  $\begin{array}{l}f\left(-x\right)=x^6+5x^3+x^2-x-6functionf\left(x\right)\\ f\left(-x\right)=+++--showthesignsforeachterm\end{array}$

There is 1 sign change, so there is 1 negative real zero.