Chapter 2.7, Problem 15E

Solution

To find: all zeros of the polynomial function.

The zeros $\text{x=-2,1,-2 i, 2 i}$ .

Given information: 

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

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

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 $\text{x=-2,1}$ shown in the computations below.

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

Thus, $-2$ and $1$ are zeros of $f\left(x\right)$ .

The last row of the synthetic division above represents the expression $x^2+4$ .

Equating this expression to zero and solving for $x$ result to

  $\begin{array}{ccc}{x}^2+4& =0& \\ x^2& =-4& \\ \sqrt{x^2}& =\sqrt{-4}& \text{ (Take square root of both sides) }\\ x& =\pm \sqrt{-4}& \\ x& =\pm \sqrt{4}\cdot \sqrt{-1}& \\ x& =\pm 2i& (\text{ Use }i=\sqrt{-1})\end{array}$

Thus, $2i$ and $-2i$ are zeros of $f\left(x\right)$ .

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