Chapter 2.3, Problem 60E

To determine

To calculate: The rational zeros of the function.

Answer to Problem 60E

There is no rational zero of the function $g\left(x\right)$ .

Explanation of Solution

Given information:

  $g\left(x\right)=x^3+8x^2+12x+18$

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:

  $g\left(x\right)=x^3+8x^2+12x+18$

Here the constant term is 18, therefore the factors of 18 are $p=\pm 1,\pm 2,\pm 3,\pm 6,\pm 9,\pm 18$

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

  $\frac{p}{q}=\pm 1,\pm 2,\pm 3,\pm 6,\pm 9,\pm 18$

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

  $\begin{array}{l}g\left(1\right)=1^3+8\times 1^2+12\times 1+18=1+8+12+18=39\\ g\left(-1\right)=-1^3+8\times -1^2+12\times -1+18=-1+8-12+18=13\\ g\left(2\right)=2^3+8\times 2^2+12\times 2+18=8+32+24+18=82\\ g\left(-2\right)=-2^3+8\times -2^2+12\times -2+18=-8+32-24+18=18\\ g\left(3\right)=3^3+8\times 3^2+12\times 3+18=27+72+36+18=153\\ g\left(-3\right)=-3^3+8\times -3^2+12\times -3+18=-27+72-36+18=27\\ g\left(6\right)=6^3+8\times 6^2+12\times 6+18=216+288+72+18=594\\ g\left(-6\right)=-6^3+8\times -6^2+12\times -6+18=-216+288-72+18=18\\ g\left(9\right)=9^3+8\times 9^2+12\times 9+18=729+648+108+18=1503\\ g\left(-9\right)=-9^3+8\times -9^2+12\times -9+18=-729+648-108+18=-171\\ g\left(18\right)={18}^3+8\times {18}^2+12\times 18+18=5832+2592+216+18=8658\\ g\left(-18\right)=-{18}^3+8\times -{18}^2+12\times -18+18=-5832+2592-216+18=-3438\end{array}$

From the above result we can say that there is no rational zero of $g\left(x\right)$ .