Chapter 6.7, Problem 11CYU

To determine

To find all the zeros for the given function.

Answer to Problem 11CYU

The zeroes are $6,-4,4i,-4i$ .

Explanation of Solution

Given information:

The equation $f\left(x\right)=x^4-2x^3-8x^2-32x-384$ .

Formula used:

Synthetic substitution method.

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 $f\left(-x\right)$

  $f\left(x\right)=x^4-2x^3-8x^2-32x-384$ Write the function for $f\left(x\right)$

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

  $f\left(-x\right)=x^4+2x^3-8x^2+32x-384$ 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 1 sign change, so the function has 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 $f\left(x\right)$ .

  $\begin{array}{cccccc}x& 1& -2& -8& -32& -384\\ 0& 1& -2& -8& -32& -384\\ 1& 1& -1& -9& -41& -425\\ 2& 1& 0& -8& -48& -480\\ 3& 1& 1& -5& -47& -525\\ 4& 1& 2& 0& -32& -512\\ 5& 1& 3& 7& 3& -369\\ 6& 1& 4& 16& 64& 0\end{array}$

Since the 1 positive real zero is known, use synthetic substitution with the depressed polynomial $x^3+4x^2+16x+64$ to find a real negative zero.

  $\begin{array}{ccccc}x& 1& 4& 16& 64\\ -1& 1& 3& 13& 51\\ -2& 1& 2& 12& 40\\ -3& 1& 1& 13& 25\\ -4& 1& 0& 16& 0\end{array}$

The depressed polynomial $x^2+16$ , can still be simplified, solve this as equation.

  $x^2+16=0$ Show as an equation

  $x^2+16-16=0-16$ Subtract 16 from each side

  $x^2=-16$ Simplify

  $\sqrt{x^2}=\pm \sqrt{-16}$ Use square root to remove power in variable

  $x=\pm 4i$ Extract square root and show negative sign as imaginary root.

Therefore, the zeroes are $6,-4,4i,-4i$ .