Chapter 2.7, Problem 18E

Solution

To find: all zeros of the polynomial function.

The zeros $x=-\frac{1}{2}\left(\text{multiplicity }2\right),-\sqrt{3},\sqrt{3}$ .

Given information: 

The function $g\left(x\right)=4x^4+4x^3-11x^2-12x-3$ .

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, $g\left(x\right)=4x^4+4x^3-11x^2-12x-3$ .

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 $a_n=4$ and the constant is $a_0=3$ .

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

Then,

  $p=$ factors of $3$ .

  $=\pm 1,\pm 3,$

  $q=$ factors of $4$

  $=\pm 1,\pm 2,\pm 4.$

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

That is,

  $\begin{array}{c}\frac{p}{q}=\pm \frac{1}{1},\pm \frac{1}{2},\pm \frac{1}{4},\pm \frac{3}{1},\pm \frac{3}{2},\pm \frac{3}{4}\\ =\pm 1,\pm \frac{1}{2},\pm \frac{1}{4},\pm 3,\pm \frac{3}{2},\pm \frac{3}{4}\end{array}$

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

  $\begin{array}{l}\underset{¯}{\begin{array}{rrrrrr}\hfill x=-\frac{1}{2}& \hfill 4& \hfill 4& \hfill -11& \hfill -12& \hfill -3\\ \hfill & \hfill & \hfill -2& \hfill -1& \hfill 6& \hfill 3\end{array}}\\ \begin{array}{rrrrrr}\hfill \left(divideby2\right)& \hfill 4& \hfill 2& \hfill -12& \hfill -6& \hfill 0\end{array}\\ \begin{array}{rrrrrr}\hfill & \hfill 2& \hfill 1& \hfill -6& \hfill -3& \hfill 0\end{array}\\ \underset{¯}{\begin{array}{rrrrr}\hfill x=-\frac{1}{2}& \hfill 2& \hfill 1& \hfill -6& \hfill -3\\ \hfill & \hfill & \hfill -1& \hfill 0& \hfill 3\end{array}}\\ \begin{array}{rrrrr}\hfill \left(divideby2\right)& \hfill 2& \hfill 0& \hfill -6& \hfill 0\\ \hfill & \hfill 1& \hfill 0& \hfill -3& \hfill 0\end{array}\end{array}$

Thus, $-\frac{1}{2}$ and $\frac{1}{2}$ are zeros of $g\left(x\right)$ .

The last row of the synthetic division above represents the expression $x^2-3$ .

Equating this expression to zero results to the quadratic equation

  $\begin{array}{c}{x}^2-3=0\\ x^2=3\\ x=\pm \sqrt{3}\end{array}$

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

Hence, the zeros of $g\left(x\right)=4x^4+4x^3-11x^2-12x-3$ are $x=-\frac{1}{2}\left(\text{multiplicity }2\right),-\sqrt{3},\sqrt{3}$ .