Chapter 2.5, Problem 39E

a.

To determine

Find the zeros of the polynomial function $f(x)=2x^3-3x^2+8x-12$ .

Answer to Problem 39E

  $\frac{3}{2},2i,-2i$

Explanation of Solution

Given:

Function: $f(x)=2x^3-3x^2+8x-12$

Calculation:

The leading coefficient of given polynomial is 2 and the constant term is -12.

So, possible rational zeros are:

  $\frac{\text{Factors of}-12}{\text{Factors of 2}}=\frac{\pm 1,\pm 2,\pm 3,\pm 4,\pm 6,\pm 12}{\pm 1,\pm 2}=\pm 1,\pm 2,\pm 3,\pm 4,\pm 6,\pm 12,\pm \frac{1}{2},\pm \frac{3}{2}$

Using synthetic division to check whether $x=\frac{3}{2}$ , is a zero or not.

  $\begin{array}{ccccc}\frac{3}{2}& 2& -3& 8& -12\\ & & 3& 0& 12\\ & 2& 0& 8& \boxed{0}\end{array}\begin{array}{c}\\ \\ \hspace{1em}\leftarrow \text{Remainder}\end{array}$

Here, remainder is 0. So, $x=\frac{3}{2}$ is a zero of the given polynomial.

So, $f(x)$ factors as

  $f(x)=\left(x-\frac{3}{2}\right)(2x^2+8)$

Now, consider $2x^2+8.$

The above polynomial function is a second degree polynomial.

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

  $\begin{array}{l}\Rightarrow x=\frac{-(0)\pm \sqrt{{(0)}^2-4(2)(8)}}{2(2)}\\ \Rightarrow x=\frac{0\pm \sqrt{0-64}}{4}\\ \Rightarrow x=\frac{\pm \sqrt{-64}}{4}\\ \Rightarrow x=\frac{\pm 8\sqrt{-1}}{4}\\ \Rightarrow x=\pm 2i\\ \Rightarrow x=2i,-2i\end{array}$

Conclusion:

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

b.

To determine

Write the zeros of polynomial $f(x)=2x^3-3x^2+8x-12$ as a product of linear factors.

Answer to Problem 39E

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

Explanation of Solution

Given:

Function: $f(x)=2x^3-3x^2+8x-12$

Calculation:

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

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

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

c.

To determine

Determine the x-intercepts of polynomial $f(x)=2x^3-3x^2+8x-12$ using factorization and verify whether the real zeros are the only x-intercepts.

Answer to Problem 39E

1

Explanation of Solution

Given:

Function: $f(x)=2x^3-3x^2+8x-12$

Calculation:

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

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

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

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

Calculation for graph:

Consider $f(x)=2x^3-3x^2+8x-12$

Values of x	Values of f (x)
0	-12
1	-5
-1	-25
2	8
-2	-56

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.