Chapter 4.5, Problem 36E

To determine

To find: all real zeros of the given function.

Answer to Problem 36E

The real zeros are $1,6$ , and $7$ .

Explanation of Solution

Given information:

Given function,

  $f\left(x\right)=x^3-14x^2+55x-42$

Calculation:

Consider the function.

  $f\left(x\right)=x^3-14x^2+55x-42$

The above function is of degree 3, so it has 3 zeros.

Here, the leading coefficient is $1$ and constant term is $-42$ .

The possible rational zeros are $\pm 1,\pm 2,\pm 3,\pm 6,\pm 7,\pm 14,\pm 21,\text{and}\pm 42$ .

Test possible zeros using synthetic division until a solution is found.

For $x=1$

  $\begin{array}{ccccc}1& 1& -14& 55& -42\\ & & 1& -13& 42\\ & 1& -13& 42& 0\end{array}$

Therefore, $x-1$ is a factor.

Factor completely using the result of above synthetic division.

  $\begin{array}{c}f\left(x\right)=x^3-14x^2+55x-42\\ =\left(x-1\right)\left(x^2-13x+42\right)\\ =\left(x-1\right)\left(x-6\right)\left(x-7\right)\end{array}$

Solve further for $f\left(x\right)=0$ .

  $\begin{array}{c}\left(x-1\right)\left(x-6\right)\left(x-7\right)=0\\ x=1,6,7\end{array}$

Therefore, all real zeros are $1,6$ , and $7$ .