Chapter 6.8, Problem 53SGR

To determine

To calculate; to find all the zeros of given function

Answer to Problem 53SGR

Solution: $-2,-1+\sqrt{2},-1-\sqrt{2}$

Explanation of Solution

In order to find the possible rational zeros of the function, find first the factors of p and q based on the function.

  $\begin{array}{l}p=2andq=1p=constantandq=leadingcoefficient\\ p=\pm 1,\pm 2Writethefactorsof2\\ q=\pm 1Writethefactorsof1\end{array}$

Possible values in simplest form are $\pm 1,\pm 2$

The given function has degree 3, therefore, it has 3 zeros. To find the positive real zeros, count the number of changes in sign for the coefficients of $f\left(x\right)$

  $\begin{array}{l}f\left(x\right)=x^3+4x^2+3x-2functionf\left(x\right)\\ f\left(x\right)=+++-showthesignsforeachterm\end{array}$

There is 1 sign change, so there is 1 positive real zeros

  $\begin{array}{l}f\left(-x\right)=-x^3+4x^2-3x-2functionf\left(x\right)\\ f\left(-x\right)=-+--showthesignsforeachterm\end{array}$

There are 2 sign changes, so there are 2 or 0 negative real zeros.

Make a table and test possible rational zeros.

  $\begin{array}{ccccc}x& 1& 4& 3& -2\\ 1& 1& 5& 8& 6\\ -1& 1& 3& 0& -2\\ 2& 1& 6& 15& 28\\ -2& 1& 2& -1& 0\end{array}$

From the table, it is clear that one zero occurs at $x=-2$

The depressed polynomial is $x^2+2x-1$ can be factored using quadratic formula:

  $\begin{array}{l}x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\\ a=1;b=2;c=-1\\ x=\frac{-2\pm \sqrt{2^2-4\left(1\right)\left(-1\right)}}{2\left(1\right)}\\ x=\frac{-2\pm \sqrt{4+4}}{2}\\ x=\frac{-2\pm \sqrt{8}}{2}\\ x=\frac{-2\pm 2\sqrt{2}}{2}\\ x=\frac{-2+2\sqrt{2}}{2}orx=\frac{-2-2\sqrt{2}}{2}\\ x=-1+\sqrt{2}orx=-1-\sqrt{2}\end{array}$