Chapter 2.5, Problem 38E

a.

To determine

Find the zeros of the polynomial function $f(x)=x^2-12x+34$ .

Answer to Problem 38E

  $6+\sqrt{2},6-\sqrt{2}$

Explanation of Solution

Given:

Function: $f(x)=x^2-12x+34$

Calculation:

The above polynomial function is a second degree polynomial.

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

  $\begin{array}{l}\Rightarrow x=\frac{-(-12)\pm \sqrt{{(-12)}^2-4(1)(34)}}{2(1)}\\ \Rightarrow x=\frac{12\pm \sqrt{144-136}}{2}\\ \Rightarrow x=\frac{12\pm \sqrt{8}}{2}\\ \Rightarrow x=\frac{12\pm 2\sqrt{2}}{2}\\ \Rightarrow x=6\pm \sqrt{2}\\ \Rightarrow x=6+\sqrt{2},6-\sqrt{2}\end{array}$

Conclusion:

Therefore, the zeros of given polynomial function are $x=6+\sqrt{2}\text{and}6-\sqrt{2}.$

b.

To determine

Write the zeros of polynomial $f(x)=x^2-12x+34$ as a product of linear factors.

Answer to Problem 38E

  $f(x)=\left(x-(6+\sqrt{2})\right)\left(x-(6-\sqrt{2})\right)$

Explanation of Solution

Given:

Function: $f(x)=x^2-12x+34$

Calculation:

From above answer, the zeros of given function are $x=6+\sqrt{2}\text{and}6-\sqrt{2}.$

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

  $\Rightarrow f(x)=\left(x-(6+\sqrt{2})\right)\left(x-(6-\sqrt{2})\right)$

c.

To determine

Determine the x-intercepts of polynomial $f(x)=x^2-12x+34$ using factorization and verify whether the real zeros are the only x-intercepts.

Answer to Problem 38E

2

Explanation of Solution

Given:

Function: $f(x)=x^2-12x+34$

Calculation:

Factorization of given function is $f(x)=\left(x-(6+\sqrt{2})\right)\left(x-(6-\sqrt{2})\right).$

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

  $\begin{array}{l}\Rightarrow \left(x-(6+\sqrt{2})\right)\left(x-(6-\sqrt{2})\right)=0\\ \\ \begin{array}{cc}\begin{array}{l}\Rightarrow x-(6+\sqrt{2})=0\\ \Rightarrow x=6+\sqrt{2}\end{array}& \begin{array}{l}\Rightarrow x-(6-\sqrt{2})=0\\ \Rightarrow x=6-\sqrt{2}\end{array}\end{array}\\ \\ \Rightarrow x=6+\sqrt{2},6-\sqrt{2}\end{array}$

Hence, the x-intercepts are $x=6+\sqrt{2}\text{and}6-\sqrt{2}.$

Calculation for graph:

Consider $f(x)=x^2-12x+34$

Values of x	Values of f (x)
0	34
1	23
-1	47
2	14
-2	62

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 2.

The number of real zeros of the function is equal to 2.

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