Chapter 2.7, Problem 17E

Solution

To find: all zeros of the polynomial function.

The zeros $x=-1,1,1-\sqrt{3},1+\sqrt{3}$ .

Given information: 

The function $g\left(x\right)=x^4-2x^3-3x^2+2x+2$ .

Formula used:

The Rational Zero Theorem:

If $f\left(x\right)=a_n{x}^n+\cdots +a_{1}x+a_0$ consists of integer coefficients, then the shape of each rational zero of $f$ is as follows:

  $\frac{p}{q}=\frac{\text{ factor of constant term }{a}_0}{\text{ factor of leading coefficient }{a}_n}$

Calculation:

Use the Fundamental Theorem of Algebra and the Rational Zero Theorem to find all zeros of the given function, $g\left(x\right)=x^4-2x^3-3x^2+2x+2$ .

The degree (highest exponent of the variable) of the given polynomial is $4$ .

By the Fundamental Theorem of Algebra, the given polynomial has $4$ zeros.

In the given function, the top correlation is $a_n=1$ and the constant is $a_0=2$ .

Let $p$ be the factors of $a_0$ and $q$ be the factors of $a_n$ .

Then,

  $p=$ factors of $2$

  $=\pm 1,\pm 2$

  $q=$ factors of $1$

  $=\pm 1$

By the Rational Zero Theorem, the possible rational zeros of the given function are given by $\frac{p}{q}$ .

That is,

  $\begin{array}{c}\frac{p}{q}=\pm \frac{1}{1},\pm \frac{2}{1}\\ =\pm 1,\pm 2\end{array}$

By synthetic division, the numbers that will result to a remainder of zero are $\text{x=-1,1}$ shown in the computations below.

  $\begin{array}{l}\underset{¯}{\begin{array}{rrrrrr}\hfill x=-1& \hfill 1& \hfill -2& \hfill -3& \hfill 2& \hfill 2\\ \hfill & \hfill & \hfill -1& \hfill 3& \hfill 0& \hfill -2\end{array}}\\ \begin{array}{rrrrrr}\hfill & \hfill 1& \hfill -3& \hfill 0& \hfill 2& \hfill 0\end{array}\\ \underset{¯}{\begin{array}{rrrrr}\hfill x=1& \hfill 1& \hfill -3& \hfill 0& \hfill 2\\ \hfill & \hfill & \hfill 1& \hfill -2& \hfill -2\end{array}}\\ \begin{array}{rrrrr}\hfill & \hfill 1& \hfill -2& \hfill -2& \hfill 0\end{array}\end{array}$

Thus, $-1$ and $1$ are zeros of $g\left(x\right)$ .

The last row of the synthetic division above represents the expression $x^2-2x-2$ .

Equating this expression to zero results to the quadratic equation

  $x^2-2x-2=0$

In $a{x}^2+bx+c=0$ , the solutions are given by the Quadratic Formula,

  $x=-\frac{-b\pm \sqrt{b^2-4ac}}{2a}$ .

With $a=1$ , $b=-2$ and $c=-2$ , then

  $\begin{array}{c}x=-\frac{-b\pm \sqrt{b^2-4ac}}{2a}\\ x=-\frac{-(-2)\pm \sqrt{{(-2)}^2-4(1)(-2)}}{2(1)}\\ =-\frac{2\pm \sqrt{4+8}}{2}\\ =-\frac{2\pm \sqrt{12}}{2}\\ =-\frac{2\pm \sqrt{4\cdot 3}}{2}\\ =-\frac{2\pm \sqrt{4}\cdot \sqrt{3}}{2}\\ =-\frac{2\pm 2\sqrt{3}}{2}\\ =-\frac{2(1\pm \sqrt{3})}{2}\\ =-\frac{2(1\pm \sqrt{3})}{2}\\ =1\pm \sqrt{3}\end{array}$

(Factor out $2$ )

Thus, $1-\sqrt{3}$ and $1+\sqrt{3}$ are zeros of $g\left(x\right)$

Hence, the zeros of $g\left(x\right)=x^4-2x^3-3x^2+2x+2$ are $x=-1,1,1-\sqrt{3},1+\sqrt{3}$ .