Chapter 2.3, Problem 59E

To determine

To calculate: The rational zeros of the function..

Answer to Problem 59E

(-1) is the only rational zero of the function.

Explanation of Solution

Given information:

  $C\left(x\right)=2x^3+3x^2-1$

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

Now determine the possible zeros by evaluating $f\left(\frac{p}{q}\right)$ .

Calculation:

  $C\left(x\right)=2x^3+3x^2-1$

Here the constant term is -1, therefore the factors of -1 are $p=\pm 1$

And the leading coefficient is 2 , hence the factors of 2 are $q=\pm 1,\pm 2$

  $\frac{p}{q}=+1,-1,+\frac{1}{2},-\frac{1}{2}$

Now determine the actual possible zeros by substituting the above values for $x$ in $C\left(x\right)$

  $\begin{array}{l}C\left(1\right)=2\times 1^3+3\times 1^2-1=2+3-1=4\\ C\left(-1\right)=2\times -1^3+3\times -1^2-1=-2+3-1=0\\ C\left(2\right)=2\times 2^3+3\times 2^2-1=16+12-1=27\\ C\left(-2\right)=2\times -2^3+3\times -2^2-1=-16+12-1=-5\end{array}$

From the above result we can say that (-1) is the only rational zero of $C\left(x\right)$ .