Chapter 2.4, Problem 53E

Solution

To find: All the zeros of the function $f\left(x\right)=x^4-3x^3-6x^2+6x+8$ .

  $-1,4,\sqrt{2},-\sqrt{2}$

Given information:

  $f\left(x\right)=x^4-3x^3-6x^2+6x+8$

Concept Used:

Divide a polynomial function $g\left(x\right)$ by $\left(x-k\right)$ , using synthetic division and the number at last is zero, then $k$ is a real root of the equation $g\left(x\right)=0$ .

Calculation:

Setup the synthetic division for $x=-1$ ,

  $\begin{array}{cccccc}-1& 1& -3& -6& 6& 8\\ & \downarrow & -1& 4& 2& -8\\ & 1& -4& -2& 8& \boxed{0}\end{array}$

Since the last number of the bottom rows is 0, the $x=-1$ is a real root of the function.

The quotient of the polynomial is $x^3-4x^2-2x+8$ .

Setup the synthetic division for $x=4$ ,

  $\begin{array}{ccccc}4& 1& -4& -2& 8\\ & \downarrow & 4& 0& -8\\ & 1& 0& -2& \boxed{0}\end{array}$

Since the last number of the bottom rows is 0, the $x=4$ is a real root of the function.

The quotient of the polynomial is $x^2-2$ .

The other two zeros are,

  $\begin{array}{c}{x}^2-2=0\\ x^2=2\\ x=\pm \sqrt{2}\end{array}$