Chapter 2, Problem 54CR

To determine

To calculate: The rational zeros of the function.

Answer to Problem 54CR

None of the values has rational zeros of $f\left(x\right)$ .

Explanation of Solution

Given information ::

  $f\left(x\right)=2x^5-4x+5$

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 5 and q is the factor of 2.

The factor of 5 is $\pm 1,\pm 5$ and the factor of 2 is $\pm 1,\pm 2$

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

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

  $\begin{array}{l}f\left(1\right)=2\times 1^5-4\times 1+5=2-4+5=3\\ f\left(-1\right)=2\times -1^5-4\times -1+5=-2+4+5=7\\ f\left(5\right)=2\times 5^5-4\times 5+5=6250-20+5=6235\\ f\left(-5\right)=2\times -5^5-4\times -5+5=-6250+20+5=-6225\\ f\left(\frac{1}{2}\right)=2\times {\frac{1}{2}}^5-4\times \frac{1}{2}+5=\frac{1}{16}-2+5=\frac{49}{16}\\ f\left(-\frac{1}{2}\right)=2\times -{\frac{1}{2}}^5-4\times -\frac{1}{2}+5=-\frac{1}{16}+2+5=\frac{111}{16}\\ f\left(-\frac{5}{2}\right)=2\times -{\frac{5}{2^5}}^5-4\times -\frac{5}{2}+5=\frac{3125}{16}+10+5=\frac{3365}{16}\\ f\left(\frac{5}{2}\right)=2\times {\frac{5}{2^5}}^5-4\times \frac{5}{2}+5=\frac{3125}{16}-10+5=\frac{3205}{16}\end{array}$

Hence none of the values has rational zeros of $f\left(x\right)$ .