Chapter 2, Problem 56CR

To determine

To calculate: The rational zeros of the function.

Answer to Problem 56CR

There is no rational zeros of $f\left(x\right)$ .

Explanation of Solution

Given information ::

  $f\left(x\right)=10x^3-21x^2-x-6$

Formula used :: According to rational zero theorem every rational zero of  f ( x ) has the form $\frac{p}{q}$ where  p  is a factor of the constant term and  q  is a factor of the leading coefficient and when the leading coefficient is 1then the possible rational zeros are the factors of the constant term.

Calculation ::

From rational zero theorem p is a factor of -6 and q is the factor of 10.

The factor of -6 is $\pm 1,\pm 2\pm 3,\pm 6$ and the factor of 10 is $\pm 1,\pm 2\pm 5,\pm 10$

Therefore the possible values of $\frac{p}{q}$ are $\pm 1,\pm 2\pm 3,\pm 6,\pm \frac{1}{2},\pm 1,\pm \frac{3}{2},\pm 3,\pm \frac{1}{5},\pm \frac{2}{5},\pm \frac{3}{5},\pm \frac{6}{5}$

Now substituting these values in f ( x ) to get the possible zeros.

Hence there is no rational zeros of $f\left(x\right)$ .