Chapter 2.7, Problem 19E

Solution

To find: all zeros of the polynomial function.

The zeros $-4,-2,-\frac{3}{2},1$ .

Given information: 

The function $h\left(x\right)=2x^4+13x^3+19x^2-10x-24$ .

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, $h\left(x\right)=2x^4+13x^3+19x^2-10x-24$ .

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 $a_n=2$ and the constant is $a_0=-24$ .

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

Then,

  $p=$ factors of $-24$

  $=\pm 1,\pm 2,\pm 3,\pm 4,\pm 6,\pm 8,\pm 12,\pm 24,$

  $q=$ factors of $2$

  $=\pm 1,\pm 2.$

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

That is,

  $\begin{array}{c}\frac{p}{q}=\pm \frac{1}{1},\pm \frac{2}{1},\pm \frac{3}{1},\pm \frac{4}{1},\pm \frac{6}{1},\pm \frac{8}{1},\pm \frac{12}{1},\pm \frac{24}{1}\\ \pm \frac{1}{2},\pm \frac{2}{2},\pm \frac{3}{2},\pm \frac{4}{2},\pm \frac{6}{2},\pm \frac{8}{2},\pm \frac{12}{2},\pm \frac{24}{2}\\ =\pm 1,\pm 2,\pm 3,\pm 4,\pm 6,\pm 8,\pm 12,\pm 24,\pm \frac{1}{2},\pm \frac{3}{2}\end{array}$

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

  $x=-4,-2,1,-\frac{3}{2}$ shown in the computations below.

  $\begin{array}{l}\underset{¯}{\begin{array}{rrrrrr}\hfill x=-4& \hfill 2& \hfill 13& \hfill 19& \hfill -10& \hfill -24\\ \hfill & \hfill & \hfill -8& \hfill -20& \hfill 4& \hfill 24\end{array}}\\ \begin{array}{rrrrrr}\hfill & \hfill 2& \hfill 5& \hfill -1& \hfill -6& \hfill 0\end{array}\\ \underset{¯}{\begin{array}{rrrrr}\hfill x=-2& \hfill 2& \hfill 5& \hfill -1& \hfill -6\\ \hfill & \hfill & \hfill -4& \hfill -2& \hfill 6\end{array}}\\ \begin{array}{rrrrr}\hfill & \hfill 2& \hfill 1& \hfill -3& \hfill 0\end{array}\\ \underset{¯}{\begin{array}{rrrr}\hfill x=1& \hfill 2& \hfill 1& \hfill -3\\ \hfill & \hfill & \hfill 2& \hfill 3\end{array}}\\ \begin{array}{rrrr}\hfill & \hfill 2& \hfill 3& \hfill 0\end{array}\\ \underset{¯}{\begin{array}{rrr}\hfill x=-\frac{3}{2}& \hfill 2& \hfill 3\\ \hfill & \hfill & \hfill -3\end{array}}\\ \begin{array}{rrr}\hfill & \hfill 2& \hfill 0\end{array}\end{array}$

Thus, $-4,-2,-\frac{3}{2},1$ are zeros of $h\left(x\right)$ .

Hence, the zeros of $h\left(x\right)=2x^4+13x^3+19x^2-10x-24$ are $-4,-2,-\frac{3}{2},1$ .