Chapter 5.7, Problem 35PPS

To determine

To find all the zeros for the given function.

Answer to Problem 35PPS

Therefore, the zeroes are $-4,7,5i,-5i$ .

Explanation of Solution

Given information:

The equation $f\left(x\right)=x^4-3x^3-3x^2-75x-700$ .

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 $f\left(-x\right)$

  $f\left(x\right)=x^4-3x^3-3x^2-75x-700$ Write the function for $f\left(x\right)$

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

  $f\left(x\right)=x^4+3x^3-3x^2+75x-700$ 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 3 sign changes, so the function has 3 or 1 negative real zeros.

Step 3:

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

  $\begin{array}{cccccc}x& 1& -3& -3& -75& -700\\ 1& 1& -2& -5& -80& -780\\ 2& 1& -1& -5& -85& -870\\ 3& 1& 0& -3& -84& -952\\ 4& 1& 1& 1& -71& -984\\ 5& 1& 2& 7& -40& -900\\ 6& 1& 3& 15& 15& -610\\ 7& 1& 4& 25& 100& 0\end{array}$

Since the 1 positive real zero is known, use synthetic substitution with the depressed polynomial $x^3+4x^2+25x+100$ to find a real negative zero.

  $\begin{array}{ccccc}x& 1& 4& 25& 100\\ 0& 1& 4& 25& 100\\ -1& 1& 3& 22& 78\\ -2& 1& 2& 21& 58\\ -3& 1& 1& 22& 34\\ -4& 1& 0& 25& 0\end{array}$

The depressed polynomial $x^2+25$ , can still be simplified, solve this as equation.

  $x^2+25=0$ Show as an equation

  $x^2+25-25=0-25$ Subtract 25 from each side

  $x^2=-25$ Simplify

  $\sqrt{x^2}=\pm \sqrt{-25}$ Use square root to remove power in variable

  $x=\pm 5i$ Extract square root and show negative sign as imaginary root.

Therefore, the zeroes are $-4,7,5i,-5i$ .