Chapter 5.7, Problem 36PPS

To determine

To find all the zeros for the given function.

Answer to Problem 36PPS

The zeros are $-3,-3,8i,-8i$ .

Explanation of Solution

Given information:

The equation $f\left(x\right)=x^4+6x^3+73x^2+384x+576$ .

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 $f\left(-x\right)$

  $f\left(x\right)=x^4+6x^3+73x^2+384x+576$ Write the function for $f\left(x\right)$

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

  $f\left(x\right)=x^4-6x^3+73x^2-384x+576$ 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 0 sign change, so the function has 0 positive real zero.

For $f\left(-x\right)$ , there are 4 sign changes, so the function has 4,2 or 0 negative real zeros.

Step 3:

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

  $\begin{array}{cccccc}x& 1& 6& 73& 384& 576\\ -3& 1& 3& 64& 192& 0\\ -2& 1& 4& 65& 254& 68\\ -1& 1& 5& 68& 316& 260\\ 0& 1& 6& 73& 384& 576\\ 1& 1& 7& 80& 464& 1040\end{array}$

The depressed polynomial $x^3+3x^2+64x+192$ , can still be simplified by factoring.

  $x^3+3x^2+64x+192=0$ Show as an equation

  $\left(x+3\right)\left(x^2+64\right)=0$ Factor the Equation

  $\left(x+3\right)=0$ or $\left(x^2+64\right)=0$ Use zero product property

Solve for the result in the zero-product property

  $x+3=0$ Write the first result.

  $x+3-3=0-3$ Subtract 3 from both sides.

  $x=-3$ Simplify

  $x^2+64=0$ Write the second result.

  $x^2+64-64=0-64$ Subtract 64 from both sides.

  $x^2=-64$ Simplify

  $\sqrt{x^2}=\sqrt{-64}$ Use square root to remove power in the variable.

  $x=\pm 8i$ Extract square root and show imaginary unit for the negative sign.

The zeros are $-3,-3,8i,-8i$ .