Chapter 6.7, Problem 12CYU

To determine

To find all the zeros for the given function.

Answer to Problem 12CYU

The zeros are $-1,1,3+i,3-i$ .

Explanation of Solution

Given information:

The equation $f\left(x\right)=x^4-6x^3+9x^2+6x-10$ .

Formula used:

Synthetic substitution method.

  $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Where $x$ is the root of the quadratic equation.

Step 1:

The given $f\left(x\right)$ has degree 4, the function has 4 zeros.

Step 2:

Examine the number of sign changes for $f\left(x\right)$ and $f\left(-x\right)$

  $f\left(x\right)=x^4-6x^3+9x^2+6x-10$ Write the function for $f\left(x\right)$

  $f\left(x\right)=+-++-$ Show the signs for each term.

  $f\left(-x\right)=x^4+6x^3+9x^2-6x-10$ Write the function for $f\left(x\right)$

  $f\left(-x\right)=+++--$ Show the signs for each term.

For $f\left(x\right)$ , there is 3 sign changes, so the function has 3 or 1 positive real zero.

For $f\left(-x\right)$ , there is 1 sign change, so the function has 1 negative real zero.

Step 3:

Determine the real zeros by using list of possible values and use synthetic substitution to evaluate $f\left(x\right)$ .

  $\begin{array}{cccccc}x& 1& -6& 9& 6& -10\\ -3& 1& -9& 36& -102& 296\\ -2& 1& -8& 25& -44& 78\\ -1& 1& -7& 16& -10& 0\\ 0& 1& -6& 9& 6& -10\end{array}$

Since the negative real zero is known, use synthetic substitution with the depressed polynomial $x^3-7x^2+16x-10$ to find a second negative zero.

  $\begin{array}{ccccc}x& 1& -7& 16& -10\\ 0& 1& -7& 16& -10\\ 1& 1& -6& 10& 0\\ 2& 1& -5& 6& 2\\ 3& 1& -4& 4& 2\end{array}$

The depressed polynomial $x^2-6x+10$ , can still be simplified, solve using quadratic formula:

  $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

  $\begin{array}{l}a=1,b=-6,c=10\\ \\ x=\frac{-\left(-6\right)\pm \sqrt{{\left(-6\right)}^2-4\left(1\right)\left(10\right)}}{2\left(1\right)}\\ \\ x=\frac{6\pm \sqrt{36-40}}{2}\\ \\ x=\frac{6\pm \sqrt{-4}}{2}\\ \\ x=\frac{6\pm \sqrt{-1\cdot 2}}{2}\end{array}$

  $\begin{array}{l}x=\frac{6\pm 2i}{2}\\ \\ x=\frac{6}{2}+\frac{2i}{2},x=\frac{6}{2}-\frac{2i}{2}\\ \\ x=3+i,x=3-i\end{array}$

The zeros are $-1,1,3+i,3-i$