Chapter 2.5, Problem 16E

To determine

Find the zeros of the function $f(x)=x^3-4x^2-4x+16$ and find the relationship between the number of real zeros and the number of x-intercepts of the graph.

  

Answer to Problem 16E

  $-2,2,4$

Explanation of Solution

Given:

Function: $f(x)=x^3-4x^2-4x+16$

  

Calculation:

The leading coefficient of given polynomial is 1 and the constant term is 16.

So, possible rational zeros are: $\frac{\text{Factors of}16}{\text{Factors of 1}}=\frac{\pm 1,\pm 2,\pm 4,\pm 8,\pm 16}{\pm 1}=\pm 1,\pm 2,\pm 4,\pm 8,\pm 16$

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

  $\begin{array}{ccccc}4& 1& -4& -4& 16\\ & & 4& 0& -16\\ & 1& 0& -4& \boxed{0}\end{array}\begin{array}{c}\\ \\ \hspace{1em}\leftarrow \text{Remainder}\end{array}$

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

So, $f(x)$ factors as

  $f(x)=\left(x-4\right)(x^2-4)$

Now, consider $x^2-4.$

The above polynomial is a second degree polynomial.

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

  $\begin{array}{l}\Rightarrow x=\frac{-(0)\pm \sqrt{{(0)}^2-4(1)(-4)}}{2(1)}\\ \Rightarrow x=\frac{\pm \sqrt{16}}{2}\\ \Rightarrow x=\frac{\pm 4}{2}\\ \Rightarrow x=\pm 2\\ \Rightarrow x=2,-2\end{array}$

Interpretation from given graph:

The number of real zeros of the function is equal to 3.

Also, the number of x-intercepts is equal to 3.

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

Conclusion:

Therefore, the zeros of given function are $x=-2,2\text{and}4.$