Chapter 2.3, Problem 82E

To determine

Determine the number of rational and irrational zeroes of the polynomial function

  $f(x)=x^3-2$ .

1. Rational zeros: 0 ; Irrational zeros: 1
2. Rational zeros: 3 ; Irrational zeros: 0
3. Rational zeros: 1 ; Irrational zeros: 2
4. Rational zeros: 1 ; Irrational zeros: 0

Answer to Problem 82E

Option A

Explanation of Solution

Given:

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

Formula used:

  $a³–b³=\left(a–b\right)\left(a²+ab+b²\right)$

Calculation:

  $\begin{array}{l}\Rightarrow f(x)=x^3-2\\ \Rightarrow f(x)=x^3-{\left(\sqrt[3]{2}\right)}^3\\ \Rightarrow f(x)=\left(x-\sqrt[3]{2}\right)(x^2+\sqrt[3]{2}x+2^{\frac{2}{3}})[\because a³–b³=\left(a–b\right)\left(a²+ab+b²\right)]\end{array}$

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

  $\begin{array}{l}\Rightarrow \left(x-\sqrt[3]{2}\right)(x^2+\sqrt[3]{2}x+2^{\frac{2}{3}})=0\\ \begin{array}{cc}\begin{array}{l}\Rightarrow x-\sqrt[3]{2}=0\\ \Rightarrow x=\sqrt[3]{2}\end{array}& \Rightarrow x^2+\sqrt[3]{2}x+2^{\frac{2}{3}}=0\end{array}\end{array}$

Now, consider $x^2+\sqrt[3]{2}x+2^{\frac{2}{3}}.$

The above polynomial is a second degree polynomial.

Discriminant $=b^2-4ac=2^{\frac{2}{3}}-4(1)(2^{\frac{2}{3}})=2^{\frac{2}{3}}-4(2^{\frac{2}{3}})=-3(2^{\frac{2}{3}})<0$

Here, the value of discriminant is less than zero.

So, zeros does not exists for above second degree polynomial.

Hence, zeros of given function is $x=\sqrt[3]{2}.$

So, the number of rational zeros are 0 and number of irrational zeros are 1.

The values mentioned above matches with option A.

Conclusion:

Therefore, option A is correct.