Chapter 2.3, Problem 81E

To determine

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

  $f(x)=x^3-1$ .

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 81E

Option D

Explanation of Solution

Given:

Function: $f(x)=x^3-1$

Formula used:

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

Calculation:

  $\begin{array}{l}\Rightarrow f(x)=x^3-1\\ \Rightarrow f(x)=x^3-1^3\\ \Rightarrow f(x)=\left(x-1\right)(x^2+x+1)[\because a³–b³=\left(a–b\right)\left(a²+ab+b²\right)]\end{array}$

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

  $\begin{array}{l}\Rightarrow \left(x-1\right)(x^2+x+1)=0\\ \begin{array}{cc}\begin{array}{l}\Rightarrow x-1=0\\ \Rightarrow x=1\end{array}& \Rightarrow x^2+x+1=0\end{array}\end{array}$

Now, consider $x^2+x+1.$

The above polynomial is a second degree polynomial.

Discriminant $=b^2-4ac=1-4(1)(1)=1-4=-3<0$

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 = 1.

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

The values mentioned above matches with option D.

Conclusion:

Therefore, option D is correct.