Chapter 2.5, Problem 15E

To determine

Find the zeros of the function $f(x)=x^3-4x^2+x-4$ and find the relationship between the number of real zeros and the number of x-intercepts of the graph.

  

Answer to Problem 15E

  $4,-i,i$

Explanation of Solution

Given:

Function: $f(x)=x^3-4x^2+x-4$

  

Calculation:

The leading coefficient of given polynomial is 1 and the constant term is -4.

So, possible rational zeros are: $\frac{\text{Factors of}-4}{\text{Factors of 1}}=\frac{\pm 1,\pm 2,\pm 4}{\pm 1}=\pm 1,\pm 2,\pm 4$

Using synthetic division to check whether $x=4$ , is a zero or not.

  $\begin{array}{ccccc}4& 1& -4& 1& -4\\ & & 4& 0& 4\\ & 1& 0& 1& \boxed{0}\end{array}\begin{array}{c}\\ \\ \hspace{1em}\hspace{1em}\leftarrow \text{Remainder}\end{array}$

Here, remainder is 0. So, $x=4$ is a zero of the given polynomial.

So, $f(x)$ factors as

  $f(x)=\left(x-4\right)(x^2+1)$

Now, consider $x^2+1.$

The above polynomial is a second degree polynomial.

So, zeroes of a second degree polynomial can be found using the formula $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}.$

  $\begin{array}{l}\Rightarrow x=\frac{-(0)\pm \sqrt{{(0)}^2-4(1)(1)}}{2(1)}\\ \Rightarrow x=\frac{\pm \sqrt{-4}}{2}\\ \Rightarrow x=\frac{\pm 2\sqrt{-1}}{2}\\ \Rightarrow x=\frac{\pm 2i}{2}\\ \Rightarrow x=\pm i\\ \Rightarrow x=i,-i\end{array}$

Interpretation from given graph:

The number of real zeros of the function is equal to 1.

Also, the number of x-intercepts is equal to 1.

So, the number of real zeros and number of x-intercepts are equal to each other.

Conclusion:

Therefore, the zeros of given function are $x=4,-i\text{and}i.$