Chapter 2.5, Problem 43E

a.

To determine

Find the zeros of the polynomial function $f(x)=x^3-11x+150$ .

Answer to Problem 43E

  $-6,3+4i,3-4i$

Explanation of Solution

Given:

Function: $f(x)=x^3-11x+150$

Calculation:

The leading coefficient of given polynomial is 1 and the constant term is 150.

So, possible rational zeros are:

  $\frac{\text{Factors of}150}{\text{Factors of 1}}=\frac{\pm 1,\pm 2,\pm 3,\pm 5,\pm 6,\pm 10,\pm 15,\pm 25,\pm 30,\pm 50,\pm 75,150}{\pm 1}=\pm 1,\pm 2,\pm 3,\pm 5,\pm 6,$

  $\pm 10,\pm 15,\pm 25,\pm 30,\pm 50,\pm 75,150$

Using synthetic division to check whether $x=-6$ , is a zero or not.

  $\begin{array}{ccccc}-6& 1& 0& -11& 150\\ & & -6& 36& -150\\ & 1& -6& 25& \boxed{0}\end{array}\begin{array}{c}\\ \\ \hspace{1em}\leftarrow \text{Remainder}\end{array}$

Here, remainder is 0. So, $x=-6$ is a zero of the given polynomial.

So, $f(x)$ factors as

  $f(x)=\left(x+6\right)(x^2-6x+25)$

Now, consider $x^2-6x+25.$

The above polynomial function is a second degree polynomial.

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

  $\begin{array}{l}\Rightarrow x=\frac{-(-6)\pm \sqrt{{(-6)}^2-4(1)(25)}}{2(1)}\\ \Rightarrow x=\frac{6\pm \sqrt{36-100}}{2}\\ \Rightarrow x=\frac{6\pm \sqrt{-64}}{2}\\ \Rightarrow x=\frac{6\pm 8\sqrt{-1}}{2}\\ \Rightarrow x=\frac{6\pm 8i}{2}\\ \Rightarrow x=3\pm 4i\\ \Rightarrow x=3+4i,3-4i\end{array}$

Conclusion:

Therefore, the zeros of given polynomial function are $x=-6,3+4i\text{and}3-4i.$

b.

To determine

Write the zeros of polynomial $f(x)=x^3-11x+150$ as a product of linear factors.

Answer to Problem 43E

  $f(x)=\left(x+6\right)\left(x-(3+4i)\right)\left(x-(3-4i)\right)$

Explanation of Solution

Given:

Function: $f(x)=x^3-11x+150$

Calculation:

From above answer, the zeros of given function are $x=-6,3+4i\text{and}3-4i.$

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

  $\Rightarrow f(x)=\left(x+6\right)\left(x-(3+4i)\right)\left(x-(3-4i)\right)$

c.

To determine

Determine the x-intercepts of polynomial $f(x)=x^3-11x+150$ using factorization and verify whether the real zeros are the only x-intercepts.

Answer to Problem 43E

1

Explanation of Solution

Given:

Function: $f(x)=x^3-11x+150$

Calculation:

Factorization of given function is $f(x)=\left(x+6\right)\left(x-(3+4i)\right)\left(x-(3-4i)\right).$

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

  $\begin{array}{l}\Rightarrow \left(x+6\right)\left(x-(3+4i)\right)\left(x-(3-4i)\right)=0\\ \\ \begin{array}{ccc}\begin{array}{l}\Rightarrow x+6=0\\ \Rightarrow x=-6\end{array}& \begin{array}{l}\Rightarrow x-(3+4i)=0\\ \Rightarrow x=3+4i\end{array}& \begin{array}{l}\Rightarrow x-(3-4i)=0\\ \Rightarrow x=3-4i\end{array}\end{array}\\ \\ \Rightarrow x=-6,3+4i,3-4i\end{array}$

Hence, the x-intercepts are $x=-6,3+4i\text{and}3-4i.$

Calculation for graph:

Consider $f(x)=x^3-11x+150$

Values of x	Values of f (x)
0	150
1	140
-1	160
2	136
-2	164

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.