Chapter 2.7, Problem 11E

Solution

To find: all zeros of the polynomial function.

The zeros $x=-5,-3,1,2.$

Given information: 

The function $f\left(x\right)=x^4+5x^3-7x^2-29x+30$ .

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, $f\left(x\right)=x^4+5x^3-7x^2-29x+30$ .

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 $a_0=30$ .

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

Then,

  $p=$ factors of $30$

  $=\pm 1,\pm 2,\pm 3,\pm 5,\pm 6\pm 10,\pm 15,\pm 30$

  $q=$ 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,

  $\begin{array}{c}\frac{p}{q}=\pm \frac{1}{1},\pm \frac{2}{1},\pm \frac{3}{1}\pm \frac{5}{1},\pm \frac{6}{1},\pm \frac{10}{1},\pm \frac{15}{1},\pm \frac{30}{1}\\ =\pm 1,\pm 2,\pm 3,\pm 5,\pm 6\pm 10,\pm 15,\pm 30\end{array}$

By synthetic division, the numbers that will result to a remainder of zero are $\text{x=-5,-3,1,2}$ as shown in the computations below.

  $\underset{¯}{\begin{array}{rrrrrr}\hfill x=-5& \hfill 1& \hfill 5& \hfill -7& \hfill -29& \hfill 30\\ \hfill & \hfill & \hfill -5& \hfill 0& \hfill 35& \hfill -30\end{array}}$

  $\begin{array}{l}\begin{array}{rrrrrr}\hfill & \hfill 1& \hfill 0& \hfill -7& \hfill 6& \hfill 0\end{array}\\ \underset{¯}{\begin{array}{rrrrrr}\hfill x=-3& \hfill 1& \hfill 5& \hfill -7& \hfill -29& \hfill 30\\ \hfill & \hfill & \hfill -3& \hfill -6& \hfill 39& \hfill -30\end{array}}\end{array}$

  $\begin{array}{l}\begin{array}{rrrrrr}\hfill & \hfill 1& \hfill 2& \hfill -13& \hfill 10& \hfill 0\end{array}\\ \underset{¯}{\begin{array}{rrrrrr}\hfill x=-3& \hfill 1& \hfill 5& \hfill -7& \hfill -29& \hfill 30\\ \hfill & \hfill & \hfill -3& \hfill -6& \hfill 39& \hfill -30\end{array}}\end{array}$

  $\begin{array}{l}\begin{array}{rrrrrr}\hfill & \hfill 1& \hfill 2& \hfill -13& \hfill 10& \hfill 0\end{array}\\ \underset{¯}{\begin{array}{rrrrrr}\hfill x=1& \hfill 1& \hfill 5& \hfill -7& \hfill -29& \hfill 30\\ \hfill & \hfill & \hfill 1& \hfill 6& \hfill -1& \hfill -30\end{array}}\end{array}$

  $\begin{array}{l}\begin{array}{rrrrrr}\hfill & \hfill 1& \hfill 6& \hfill -1& \hfill -30& \hfill 0\end{array}\\ \underset{¯}{\begin{array}{rrrrrr}\hfill x=2& \hfill 1& \hfill 5& \hfill -7& \hfill -29& \hfill 30\\ \hfill & \hfill & \hfill 2& \hfill 14& \hfill 14& \hfill -30\end{array}}\\ \begin{array}{rrrrrr}\hfill & \hfill 1& \hfill 7& \hfill 7& \hfill -15& \hfill 0\end{array}\end{array}$

Hence, the zeros of $f\left(x\right)=x^4+5x^3-7x^2-29x+30$ are $x=-5,-3,1,2.$