Chapter 2, Problem 111RE

a.

To determine

To find the zeros of the function $f(x)=x^3-4x^2+6x-4$ .

Answer to Problem 111RE

  $2,1+i,1-i$

Explanation of Solution

Given:

Function: $f(x)=x^3-4x^2+6x-4$

Calculation:

1. Performing rational zero test:

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

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

2. Finding rational zeros:

Using synthetic division to check whether x = 2, is a zero or not.

  $\begin{array}{ccccc}2& 1& -4& 6& -4\\ & & 2& -4& 4\\ & 1& -2& 2& \boxed{0}\end{array}\begin{array}{c}\\ \\ \leftarrow \text{Remainder}\end{array}$

Here, remainder is 0. So, x = 2 is a zero of the given polynomial.

So, $f(x)$ factors as

  $f(x)=(x-2)(x^2-2x+2)$

Now, consider $x^2-2x+2.$

The above polynomial is a second degree polynomial.

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

  $\begin{array}{l}\Rightarrow x=\frac{-(-2)\pm \sqrt{{(-2)}^2-4(1)(2)}}{2(1)}\\ \Rightarrow x=\frac{2\pm \sqrt{4-8}}{2}\\ \Rightarrow x=\frac{2\pm \sqrt{-4}}{2}\\ \Rightarrow x=\frac{2\pm 2\sqrt{-1}}{2}\\ \Rightarrow x=\frac{2\pm 2i}{2}\\ \Rightarrow x=1\pm i\\ \Rightarrow x=1+i,1-i\end{array}$

Conclusion:

Therefore, the zeros of given polynomial function are $x=2,1+i\text{and}1-i.$

b.

To determine

To write the factors of the function $f(x)=x^3-4x^2+6x-4$ as a product of linear factors

Answer to Problem 111RE

  $f(x)=(x-2)(x-(1+i))(x-(1-i))$

Explanation of Solution

Given:

Function: $f(x)=x^3-4x^2+6x-4$

Calculation:

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

  $f(x)=(x-2)(x-(1+i))(x-(1-i))$ .

Conclusion:

Therefore, $f(x)=(x-2)(x-(1+i))(x-(1-i))$ is complete factorization to given polynomial function.

c.

To determine

To find the x - intercepts of the function $f(x)=x^3-4x^2+6x-4$ .

Answer to Problem 111RE

  $x=2$

Explanation of Solution

Given:

Function: $f(x)=x^3-4x^2+6x-4$

Calculation:

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

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

Calculation for graph:

Consider $f(x)=x^3-4x^2+6x-4$

Values of x	Values of f ( x )
0	-4
1	-1
-1	-15
2	0
-2	-40

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 $x=2.$

Hence, the x - intercept of the function is $x=2.$

Here, $x=2$ is also a real zero of given function.

Conclusion:

Therefore, the x - intercept of given function is $x=2.$