Chapter 2.5, Problem 69E

a.

To determine

To find: The real zeros of given function using graphing utility.

Answer to Problem 69E

The real zero of $h\left(x\right)=8x^3-14x^2+18x-9$ is $x=\frac{3}{4}$ .

Explanation of Solution

Given:

The given function is $h\left(x\right)=8x^3-14x^2+18x-9$ .

Concept used:

If the graph of polynomial function $f\left(x\right)$ intersect x -axis at point $\left(a,0\right)$ , then $x=a$ is the real zero of $f\left(x\right)$ , also $\left(x-a\right)$ is its one facto of $f\left(x\right)$ .

Graph:

The graph of function $h\left(x\right)=8x^3-14x^2+18x-9$ is shown in Figure-1 here.

  

The graph of the function $h\left(x\right)$ intersect x-axis at point $\left(\frac{3}{4},0\right)$ , therefore, $x=\frac{3}{4}$ is the real zero of $h\left(x\right)=8x^3-14x^2+18x-9$ .

Conclusion:

The real zero of $h\left(x\right)=8x^3-14x^2+18x-9$ is $x=\frac{3}{4}$ .

b.

To determine

To find:The exact values of the imaginary zeros, using real zero.

Answer to Problem 69E

The imaginary zeros of $h\left(x\right)=8x^3-14x^2+18x-9$ are $x=\frac{1}{2}-i\frac{\sqrt{5}}{2}$ and $x=\frac{1}{2}+i\frac{\sqrt{5}}{2}$ .

Explanation of Solution

Given:

The real zero of $h\left(x\right)=8x^3-14x^2+18x-9$ is $x=\frac{3}{4}$ calculated in part (a)

Concept used:

The number of zeros of a polynomial equals its degree.

Calculations:

The given polynomial function $h\left(x\right)=8x^3-14x^2+18x-9$ is of degree 3, therefore there are at most 3 zeros of $h\left(x\right)$ . As calculated in part (a) one real zero is $x=\frac{3}{4}$ and other two zeros are imaginary. Also, $\left(x-\frac{3}{4}\right)$ is a factor of $h\left(x\right)=8x^3-14x^2+18x-9$ .

If we divide $8x^3-14x^2+18x-9$ by $\left(x-\frac{3}{4}\right)$ then quotient is $8x^2-8x+12$ that implies

  $h\left(x\right)=\left(x-\frac{3}{4}\right)\left(8x^2-8x+12\right)$    … (1)

Now, we factor $8x^2-8x+12$ by solving equation $8x^2-8x+12=0$ using formula for equation $a{x}^2+bx+c=0$ that is $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$ .

Thus, $x=\frac{8\pm \sqrt{64-4\times 8\times 12}}{2\times 8}=\frac{8\pm \sqrt{-320}}{16}$

  $\Rightarrow x=\frac{8\pm 8\sqrt{-5}}{16}$

  $\Rightarrow x=\frac{1\pm i\sqrt{5}}{2}$

  $\because \sqrt{-5}=i\sqrt{5},i=\sqrt{-1}$

The factors of $8x^2-8x+12=0$ are $\left(x-\frac{1}{2}-i\frac{\sqrt{5}}{2}\right)$ and $\left(x-\frac{1}{2}+i\frac{\sqrt{5}}{2}\right)$ .

Thus we have $h\left(x\right)=\left(x-\frac{3}{4}\right)\left(x-\frac{1}{2}-i\frac{\sqrt{5}}{2}\right)\left(x-\frac{1}{2}+i\frac{\sqrt{5}}{2}\right)$

Thus, the imaginary zeros of $h\left(x\right)=8x^3-14x^2+18x-9$ are $x=\frac{1}{2}-i\frac{\sqrt{5}}{2}$ and $x=\frac{1}{2}+i\frac{\sqrt{5}}{2}$

Conclusion:

The imaginary zeros of $h\left(x\right)=8x^3-14x^2+18x-9$ are $x=\frac{1}{2}-i\frac{\sqrt{5}}{2}$ and $x=\frac{1}{2}+i\frac{\sqrt{5}}{2}$ .