Chapter 2.7, Problem 14E

Solution

To find: all zeros of the polynomial function.

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

Given information: 

The function $f\left(x\right)=x^4+15x^2-16$ .

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, $f\left(x\right)=x^4+15x^2-16$ .

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 $a_0=-16$ .

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

Then,

  $p=$ factors of $-16$

  $=\pm 1,\pm 2,\pm 4,\pm 8,\pm 16$

  $q=$ 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,

  $\begin{array}{c}\frac{p}{q}=\pm \frac{1}{1},\pm \frac{2}{1},\pm \frac{4}{1}\pm \frac{8}{1},\pm \frac{16}{1}\\ =\pm 1,\pm 2,\pm 4,\pm 8,\pm 16\end{array}$

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

  $\begin{array}{l}\underset{¯}{\begin{array}{rrrrrr}\hfill x=-1& \hfill 1& \hfill 0& \hfill 15& \hfill 0& \hfill -16\\ \hfill & \hfill & \hfill -1& \hfill 1& \hfill -16& \hfill 6\end{array}}\\ \begin{array}{rrrrrr}\hfill & \hfill 1& \hfill -1& \hfill 16& \hfill -16& \hfill 0\end{array}\\ \underset{¯}{\begin{array}{rrrrr}\hfill x=1& \hfill 1& \hfill -1& \hfill 16& \hfill -16\\ \hfill & \hfill & \hfill 1& \hfill 0& \hfill 16\end{array}}\\ \begin{array}{rrrrr}\hfill & \hfill 1& \hfill 0& \hfill 16& \hfill 0\end{array}\end{array}$

The last row of the synthetic division above represents the expression $x^2+16$ .

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

  $\begin{array}{ccc}{x}^2+16& =0& \\ x^2& =-16& \\ \sqrt{x^2}& =\sqrt{-16}& \text{ (Take square root of both sides) }\\ x& =\pm \sqrt{-16}& \\ x& =\pm \sqrt{16}\cdot \sqrt{-1}& \\ x& =\pm 4i& \text{ (Use }i=\sqrt{-1})\end{array}$

Thus, $4i$ and $-4i$ are zeros of $f\left(x\right)$ .

Hence, the zeros of $f\left(x\right)=x^4+15x^2-16$ are $x=-1,1,-4i,4i$ .