Chapter 2.5, Problem 42E

a.

To determine

Find the zeros of the polynomial function $f(x)=2x^3-5x^2+18x-45$ .

Answer to Problem 42E

  $\frac{5}{2},3i,-3i$

Explanation of Solution

Given:

Function: $f(x)=2x^3-5x^2+18x-45$

Calculation:

The leading coefficient of given polynomial is 2 and the constant term is -45.

So, possible rational zeros are: $\frac{\text{Factors of}-45}{\text{Factors of 2}}=\frac{\pm 1,\pm 3,\pm 5,\pm 9,\pm 15,\pm 45}{\pm 1,\pm 2}=\pm 1,\pm 3,\pm 5,\pm 9,\pm 15,\pm 45,$

  $\pm \frac{1}{2},\pm \frac{3}{2},\pm \frac{5}{2},\pm \frac{9}{2},\pm \frac{15}{2},\pm \frac{45}{2}$

Using synthetic division to check whether $x=\frac{5}{2}$ , is a zero or not.

  $\begin{array}{ccccc}\frac{5}{2}& 2& -5& 18& -45\\ & & 5& 0& 45\\ & 2& 0& 18& \boxed{0}\end{array}\begin{array}{c}\\ \\ \hspace{1em}\leftarrow \text{Remainder}\end{array}$

Here, remainder is 0. So, $x=\frac{5}{2}$ is a zero of the given polynomial.

So, $f(x)$ factors as

  $f(x)=\left(x-\frac{5}{2}\right)(2x^2+18)$

Now, consider $2x^2+18.$

The above polynomial function is a second degree polynomial.

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

  $\begin{array}{l}\Rightarrow x=\frac{-(0)\pm \sqrt{{(0)}^2-4(2)(18)}}{2(2)}\\ \Rightarrow x=\frac{0\pm \sqrt{0-144}}{4}\\ \Rightarrow x=\frac{\pm \sqrt{-144}}{4}\\ \Rightarrow x=\frac{\pm 12\sqrt{-1}}{4}\\ \Rightarrow x=\pm 3i\\ \Rightarrow x=3i,-3i\end{array}$

Conclusion:

Therefore, the zeros of given polynomial function are $x=\frac{5}{2},3i\text{and}-3i.$

b.

To determine

Write the zeros of polynomial $f(x)=2x^3-5x^2+18x-45$ as a product of linear factors.

Answer to Problem 42E

  $f(x)=\left(x-\frac{5}{2}\right)\left(x-3i\right)\left(x+3i\right)$

Explanation of Solution

Given:

Function: $f(x)=2x^3-5x^2+18x-45$

Calculation:

From above answer, the zeros of given function are $x=\frac{5}{2},3i\text{and}-3i.$

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

  $\Rightarrow f(x)=\left(x-\frac{5}{2}\right)\left(x-3i\right)\left(x+3i\right)$

c.

To determine

Determine the x-intercepts of polynomial $f(x)=2x^3-5x^2+18x-45$ using factorization and verify whether the real zeros are the only x-intercepts.

Answer to Problem 42E

1

Explanation of Solution

Given:

Function: $f(x)=2x^3-5x^2+18x-45$

Calculation:

Factorization of given function is $f(x)=\left(x-\frac{5}{2}\right)\left(x-3i\right)\left(x+3i\right).$

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

  $\begin{array}{l}\Rightarrow \left(x-\frac{5}{2}\right)\left(x-3i\right)\left(x+3i\right)=0\\ \\ \begin{array}{ccc}\begin{array}{l}\Rightarrow x-\frac{5}{2}=0\\ \Rightarrow x=\frac{5}{2}\end{array}& \begin{array}{l}\Rightarrow x-3i=0\\ \Rightarrow x=3i\end{array}& \begin{array}{l}\Rightarrow x+3i=0\\ \Rightarrow x=-3i\end{array}\end{array}\\ \\ \Rightarrow x=\frac{5}{2},3i,-3i\end{array}$

Hence, the x-intercepts are $x=\frac{5}{2},3i\text{and}-3i.$

Calculation for graph:

Consider $f(x)=2x^3-5x^2+18x-45$

Values of x	Values of f (x)
0	-45
1	-30
-1	-70
2	-13
-2	-117

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 1.

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

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