Chapter 2.6, Problem 32E

Solution

To find: all real zeros of the function.

The real zeros of the function $\text{-3,4}$ .

Given information: 

The function $h\left(x\right)=3x^4-6x^3-32x^2+35x-12$ .

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:

Consider the function

  $h\left(x\right)=3x^4-6x^3-32x^2+35x-12$

The list of probable rational zeros is below, and to the right are some of their synthetic divisions.

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

  $\backslash$ polyhornerscheme $[x=1]\left\{3x^4-6x^3-32x^2+35x-12\right\}$

  $\backslash$ polyhornerscheme $[x=2]\left\{3x^4-6x^3-32x^2+35x-12\right\}$

  $\backslash$ polyhornerscheme $[x=3]\left\{3x^4-6x^3-32x^2+35x-12\right\}$

\polyhornerscheme $[x=4]\left\{3x^4-6x^3-32x^2+35x-12\right\}$

all the positive integer potential zeros up to $4$ , found a rational zero.

Now, use the last row (without the last number $0$ ) of the synthetic division (for $4$ ) to test other numbers (e.g., negative integers).

  $\backslash$ polyhornerscheme $[x=-1]\left\{3x^3+6x^2-8x+3\right\}$

  $\backslash$ polyhornerscheme $[x=-3]\left\{3x^3+6x^2-8x+3\right\}$

all the negative integer potential zeros down to $-3$ , found a rational zero.

The polynomial has a degree of $4$ , and used synthetic division to find two zeros.

By using the last row of the synthetic division, $3x^2-3x+1=0$ .

However, this is not factorable, use the quadratic formula:

  $\begin{array}{c}x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\\ =\frac{3\pm \sqrt{9-12}}{2}\\ =\frac{3\pm i\sqrt{3}}{2}\end{array}$

However, these two zeros are imaginary.

Therefore, the only real zeros are $-3$ and $4$ .