Chapter 2.6, Problem 11E

Solution

To find: all real zeros of the function.

The real zeros are $3,8$ and $1$ .

Given information: 

The function $f\left(x\right)=x^3-12x^2+35x-24$ .

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:

Consider the function

  $f\left(x\right)=x^3-12x^2+35x-24$

Since the leading coefficient is 1 and the constant term is $-24$ , the possible rational zeros are:

  $\pm \frac{1}{1},\pm \frac{2}{1},\pm \frac{3}{1},\pm \frac{4}{1},\pm \frac{6}{1},\pm \frac{8}{1},\pm \frac{12}{1},\pm \frac{24}{1},\text{ i}\text{.e}\text{.,}\pm 1,\pm 2,\pm 3,\pm 4,\pm 6,\pm 8,\pm 12,\pm 24$

Test these potential zeros by the synthetic division until we have found 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 the 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:

  $\backslash$ Polly Horner scheme $[x=1]\left\{x^3-12x^2+35x-24\right\}$ .

And 1 is indeed a zero.

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

By using the last row of the synthetic division, we have $x^2-11x+24=0$ .

Solve this by factoring:

  $\begin{array}{c}{x}^2-11x+24=0\\ (x-3)(x-8)=0\\ x=3\text{ or }x=8\end{array}$

Therefore, the real zeros are 3,8 and 1 (from the synthetic division).