Chapter 2.5, Problem 46E

a.

To determine

Find the zeros of the polynomial function $f(x)=x^4-8x^3+17x^2-8x+16$ .

Answer to Problem 46E

  $4,4,-i,i$

Explanation of Solution

Given:

Function: $f(x)=x^4-8x^3+17x^2-8x+16$

Calculation:

  $\begin{array}{l}\Rightarrow f(x)=x^4-8x^3+17x^2-8x+16\\ \Rightarrow f(x)=x^4-8x^3+16x^2+x^2-8x+16\\ \Rightarrow f(x)=x^2(x^2-8x+16)+1(x^2-8x+16)\\ \Rightarrow f(x)=(x^2-8x+16)(x^2+1)\\ \Rightarrow f(x)=(x^2-4x-4x+16)(x^2+1)\\ \Rightarrow f(x)=(x-4)(x-4)(x^2+1)\end{array}$

To find the roots of given function, put $f(x)=0,$

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

Conclusion:

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

b.

To determine

Write the zeros of polynomial $f(x)=x^4-8x^3+17x^2-8x+16$ as a product of linear factors.

Answer to Problem 46E

  $f(x)=\left(x-4\right)\left(x-4\right)\left(x+i\right)\left(x-i\right)$

Explanation of Solution

Given:

Function: $f(x)=x^4-8x^3+17x^2-8x+16$

Calculation:

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

Now, writing the given function as product of linear factors,

  $\Rightarrow f(x)=\left(x-4\right)\left(x-4\right)\left(x+i\right)\left(x-i\right)$

c.

To determine

Determine the x-intercepts of polynomial $f(x)=x^4-8x^3+17x^2-8x+16$ using factorization and verify whether the real zeros are the only x-intercepts.

Answer to Problem 46E

2

Explanation of Solution

Given:

Function: $f(x)=x^4-8x^3+17x^2-8x+16$

Calculation:

Factorization of given function is $f(x)=\left(x-4\right)\left(x-4\right)\left(x+i\right)\left(x-i\right).$

To find x-intercepts, put $f(x)=0.$

  $\begin{array}{l}\Rightarrow \left(x-4\right)\left(x-4\right)\left(x+i\right)\left(x-i\right)=0\\ \\ \begin{array}{cccc}\begin{array}{l}\Rightarrow x-4=0\\ \Rightarrow x=4\end{array}& \begin{array}{l}\Rightarrow x-4=0\\ \Rightarrow x=4\end{array}& \begin{array}{l}\Rightarrow x+i=0\\ \Rightarrow x=-i\end{array}& \begin{array}{l}\Rightarrow x-i=0\\ \Rightarrow x=i\end{array}\end{array}\\ \\ \Rightarrow x=4,4,-i,i\end{array}$

Hence, the x-intercepts are $x=4,4,-i\text{and}i.$

Calculation for graph:

Consider $f(x)=x^4-8x^3+17x^2-8x+16$

Values of x	Values of f (x)
0	16
1	18
-1	50
2	20
-2	180

By taking different values of x, the graph can be plotted.

Graph:

Interpretation:

From graph, it is clear that, the number of x-intercepts is equal to 2.

The number of real zeros of the function is equal to 2.

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