Chapter 2.6, Problem 27E

Solution

To find: all real zeros of the function.

The real zeros of the function $-\frac{1}{3},-4,3$ .

Given information: 

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

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

  $f\left(x\right)=3x^3+4x^2-35x-12$

Since the leading coefficient is 3 and the constant term is -12, the possible rational zeros are:

  $\pm \frac{1}{1},\pm \frac{2}{1},\pm \frac{3}{1},\pm \frac{4}{1},\pm \frac{6}{1},\pm \frac{12}{1},\pm \frac{1}{3},\pm \frac{2}{3},\pm \frac{3}{3},\pm \frac{4}{3},\pm \frac{6}{3},\pm \frac{12}{3}$ i.e.,

  $\pm 1,\pm 2,\pm 3,\pm 4,\pm 6,\pm 12,\pm \frac{1}{3},\pm \frac{2}{3},\pm \frac{4}{3}$

Test these potential zeros by the synthetic division until it finds a zero. If a test number is a zero, the last number of the last row of the synthetic division will be $0$ . If a test number is not a zero, the last number of the last row of the synthetic division won't be $0$ .

Let's test $1$ first:

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

And $1$ is not a zero.

Let's test $2$ :

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

And $2$ is not a zero.

Let's test $3$ :

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

And $3$ is indeed a zero.

The degree of the polynomial is $3$ and one zero using synthetic division.

By using the last row of the synthetic division, It is $3x^2+13x+4=0$ .

Solve this by factoring:

  $\begin{array}{c}3x^2+13x+4=0\\ \left(3x+1\right)\left(x+4\right)=0\\ x=-\frac{1}{3}\end{array}$

or $x=-4$

Therefore, the real zeros are $-\frac{1}{3},-4$ and 3 (from the synthetic division).