Chapter 2.6, Problem 28E

Solution

To find: all real zeros of the function.

The real zeros of the function $-1,-6,\frac{2}{3}$ .

Given information: 

The function $f\left(x\right)=3x^3+19x^2+4x-12$ .

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

  $f\left(x\right)=3x^3+19x^2+4x-12$

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 $0$ .

Also notice that when test 1, the last row of the synthetic division has all nonnegative numbers. In that case, that means no real zeros is higher than $1$ . In general, if a positive number test yields all nonnegative numbers in the last row of the synthetic division, then there are no real zeros higher than that positive number.

Let's test $1$ first:

  $\backslash$ polyhornerscheme $\left[x=1\right]\left\{3x^3+19x^2+4x-12\right\}$

And $1$ is not a zero.

Let's test $\frac{2}{3}$ :

  $\backslash$ polyhornerscheme $[x=2/3]\left\{3x^3+19x^2+4x-12\right\}$

And $\frac{2}{3}$ 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, $3x^2+21x+18=0$ .

Solve this by factoring:

  $\begin{array}{c}3x^2+21x+18=0\\ 3\left(x^2+7x+6\right)=0\\ 3\left(x+1\right)\left(x+6\right)=0\\ x=-1\text{ or }x=-6\end{array}$

Therefore, the real zeros are $-1,-6$ and $\frac{2}{3}$ (from the synthetic division).