Chapter 2.7, Problem 10E

Solution

To find: all zeros of the polynomial function.

The zeros $x=-1,1,2,4.$

Given information: 

The function $f\left(x\right)=x^4-6x^3+7x^2+6x-8$ .

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, $f\left(x\right)=x^4-6x^3+7x^2+6x-8$

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 $a_0=-8$ .

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

Then,

  $p=$ factors of $-8$

  $=\pm 1,\pm 2,\pm 4,\pm 8,$

  $q=$ 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,

  $\begin{array}{c}\frac{p}{q}=\pm \frac{1}{1},\pm \frac{2}{1},\pm \frac{4}{1},\pm \frac{8}{1}\\ =\pm 1,\pm 2,\pm 4,\pm 8\end{array}$

By synthetic division, the numbers that will result to a remainder of zero are

  $x=-1,1,2,4$ shown in the computations below.

  $\begin{array}{l}\underset{¯}{\begin{array}{rrrrrr}\hfill x=-1& \hfill 1& \hfill -6& \hfill 7& \hfill 6& \hfill -8\\ \hfill & \hfill & \hfill -1& \hfill 7& \hfill -14& \hfill 8\end{array}}\\ \begin{array}{rrrrrr}\hfill & \hfill 1& \hfill -7& \hfill 14& \hfill -8& \hfill 0\end{array}\\ \underset{¯}{\begin{array}{rrrrrr}\hfill x=1& \hfill 1& \hfill -6& \hfill 7& \hfill 6& \hfill -8\\ \hfill & \hfill & \hfill 1& \hfill -5& \hfill 2& \hfill 8\end{array}}\\ \begin{array}{rrrrrr}\hfill & \hfill 1& \hfill -5& \hfill 2& \hfill 8& \hfill 0\end{array}\\ \underset{¯}{\begin{array}{rrrrrr}\hfill x=2& \hfill 1& \hfill -6& \hfill 7& \hfill 6& \hfill -8\\ \hfill & \hfill & \hfill 1& \hfill -5& \hfill 2& \hfill 8\end{array}}\\ \begin{array}{rrrrrr}\hfill & \hfill 1& \hfill -5& \hfill 2& \hfill 8& \hfill 0\end{array}\\ \underset{¯}{\begin{array}{rrrrrr}\hfill x=4& \hfill 1& \hfill -6& \hfill 7& \hfill 6& \hfill -8\\ \hfill & \hfill & \hfill 2& \hfill -8& \hfill -2& \hfill 8\end{array}}\\ \begin{array}{rrrrrr}\hfill & \hfill 1& \hfill -4& \hfill -1& \hfill 4& \hfill 0\end{array}\end{array}$

Hence, the zeros of $f\left(x\right)=x^4-6x^3+7x^2+6x-8$ are $x=-1,1,2,4.$