Chapter 2.6, Problem 34E

Solution

To find: all real zeros of the function.

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

Given information: 

The function $f\left(x\right)=x^5-3x^4-5x^3+15x^2+4x-12$ .

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

  $f\left(x\right)=x^5-3x^4-5x^3+15x^2+4x-12$

Since the leading coefficient is 1 and the constant term is -12, the possible rational zeros are:

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

  $\pm 1,\pm 2,\pm 3,\pm 4,\pm 6,\pm 12$

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 $[x=1]\left\{x^{}-3x^4-5x^3+15x^2+4x-12\right\}$

And $1$ is indeed a zero.

Now, use the last row (without the last number $0$ ) of the synthetic division to test other numbers.

Let's test $2$ :

  $\backslash$ polyhornerscheme $[x=2]\left\{x^4-2x^3-7x^2+8x+12\right\}$

And $2$ is indeed a zero.

Again, use the last row (without the last number $0$ ) of the previous synthetic division to test other numbers.

Let's test $3$

\polyhornerscheme $[x=3]\left\{x^3-7x-6\right\}$

And $3$ is indeed a zero.

The polynomial has a degree of $5$ , and used synthetic division to find three zeros.

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

Solve this by factoring:

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

or $x=-1$

Therefore, the real zeros are $-2,-1,1,2$ and 3.