Chapter 2, Problem 116RE

a.

To determine

To find the zeros of the function $f(x)=x^4+10x^3+26x^2+10x+25$ .

Answer to Problem 116RE

  $2,1+i,1-i$

Explanation of Solution

Given:

Function: $f(x)=x^4+10x^3+26x^2+10x+25$

Calculation:

1. Performing rational zero test:

The leading coefficient of given polynomial is 1 and the constant term is 25.

So, possible rational zeros are: $\frac{\text{Factors of 25}}{\text{Factors of 1}}=\frac{\pm 1,\pm 5,\pm 25}{\pm 1}=\pm 1,\pm 5,\pm 25$

2. Finding rational zeros:

Using synthetic division to check whether x = -5, is a zero or not.

  $\begin{array}{ccccc}-5& 1& 10& 26& 10\text{ 25}\\ & & -5& -25& -5\text{ -25 }\\ & 1& 5& 1\text{ 5}& \boxed{0}\end{array}\begin{array}{c}\\ \\ \leftarrow \text{Remainder}\end{array}$

Here, remainder is 0. So, x = -5 is a zero of the given polynomial.

So, $f(x)$ factors as

  $f(x)=(x+5)(x^3+5x+x+5)$

Now, consider $x^3+5x+x+5$

The above polynomial is a third degree polynomial.

Now follow the same steps as before:

1. Performing rational zero test:

The leading coefficient of given polynomial is 1 and the constant term is 25.

So, possible rational zeros are: $\frac{\text{Factors of 5}}{\text{Factors of 1}}=\frac{\pm 1,\pm 5}{\pm 1}=\pm 1,\pm 5$

2. Finding rational zeros:

Using synthetic division to check whether x = -5, is a zero or not.

  $\begin{array}{ccccc}-5& 1& 5& 1& \text{5 }\\ & & -5& 0& -5\\ & 1& 0& 1& \boxed{0}\end{array}\begin{array}{c}\\ \\ \leftarrow \text{Remainder}\end{array}$

Here, remainder is 0. So, x = -5 is a zero of the given polynomial.

So, $f(x)$ factors as

  $f(x)=(x+5)(x^2+1)$

Therefore, the entire polynomial $f(x)=(x+5)(x^3+5x+x+5)=(x+5)(x+5)(x^2+1)$

Further,

  $\begin{array}{l}\Rightarrow x^2+1=0\\ \Rightarrow (x+i)(x-i)=0\end{array}$

Conclusion:

Therefore, the zeros of given polynomial function are $x=-5,i,-i$

b.

To determine

To write the factors of the function $f(x)=x^4+10x^3+26x^2+10x+25$ as a product of linear factors

Answer to Problem 116RE

  $f(x)=(x+5)(x+5)(x+i)(x-i)$

Explanation of Solution

Given:

Function: $f(x)=x^4+10x^3+26x^2+10x+25$

Calculation:

From above answers, the polynomial as a product of linear factors is

  $f(x)=(x+5)(x+5)(x+i)(x-i)$ .

Conclusion:

Therefore, $f(x)=(x+5)(x+5)(x+i)(x-i)$ is complete factorization to given polynomial function.

c.

To determine

To find the x - intercepts of the function $f(x)=x^4+10x^3+26x^2+10x+25$ .

Answer to Problem 116RE

  $x=-5$

Explanation of Solution

Given:

Function: $f(x)=x^4+10x^3+26x^2+10x+25$

Calculation:

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

  $\begin{array}{l}\Rightarrow (x+5)(x+5)(x+i)(x-i)=0\\ \Rightarrow x=-5,i,-i\end{array}$

Calculation for graph:

Consider $f(x)=x^4+10x^3+26x^2+10x+25$

Values of x	Values of f ( x )
0	25
1	72
-1	32
-5	0

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

Graph:

Interpretation:

By observing graph, it is clear that the curve of the function meets x -axis at $x=-5$

Hence, the x - intercept of the function is $x=-5$

Here, $x=-5$ is also a real zero of given function.

Conclusion:

Therefore, the x -intercept of given function is $x=-5$