Chapter 6.7, Problem 9CYU

To determine

To find all the zeros for the given function.

Answer to Problem 9CYU

The real zeroes are $-8,-2,1$ .

Explanation of Solution

Given information:

The equation $f\left(x\right)=x^3+9x^2+6x-16$ .

Formula used:

Synthetic substitution method.

Step 1:

The given $f\left(x\right)$ has degree 3, the function has 3 zeros.

Step 2:

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

  $f\left(x\right)=x^3+9x^2+6x-16$ Write the function for $f\left(x\right)$

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

  $f\left(x\right)=-x^3+9x^2-6x-16$ 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 are 2 sign changes, so the function has 2 or 0 negative real zeros.

Step 3:

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

  $\begin{array}{ccccc}x& 1& 9& 6& -16\\ -3& 1& 6& -12& 20\\ -2& 1& 7& -8& 0\\ -1& 1& 8& -2& -14\\ 0& 1& 9& 6& -16\\ 1& 1& 10& 16& 0\\ 2& 1& 11& 28& 40\\ 3& 1& 12& 42& 110\end{array}$

Since there are 2 positive real zero is known, use synthetic substitution with the depressed polynomial $x^2+7x-8$ to find a second negative zero.

  $\begin{array}{cccc}x& 1& 7& -8\\ -4& 1& 3& -20\\ -5& 1& 2& -18\\ -6& 1& 1& -14\\ -7& 1& 0& -8\\ -8& 1& -1& 0\end{array}$

The real zeroes are $-8,-2,1$ .