Use the Rational Zeros Theorem to write a list of all possible rational zeros of the function.


f(x) = -2x4 + 4x3 + 3x2 + 18

Verified Answer :

\(f(x) = -2x^4 + 4x^3 + 3x^2 + 18\\\\18:\{\pm1;\ \pm2;\ \pm3;\ \pm6;\ \pm9;\ \pm18\}\\2:\{\pm1;\ \pm2\}\\\\Answer:\boxed{\{\pm1;\ \pm2;\ \pm\frac{3}{2};\ \pm3;\ \pm\frac{9}{2};\ \pm6;\ \pm9;\ \pm18}\)

Answer: \(\pm1 , \pm2 , \pm3 , \pm6 , \pm9 , \pm 18, \pm \dfrac{1}{2}, \pm \dfrac{3}{2}, \pm\dfrac{9}{2}\)

Step-by-step explanation:

  • The Rational Zeroes theorem says that if f(x) has integer coefficients and  is a rational zero, then q is the factor of leading coefficient and p is the factor of constant term.

Given polynomial= \(f(x) = -2x^4 + 4x^3 + 3x^2 + 18\)

leading coefficient : q = -2

constant term p = 18

Now, Factors of q   =  \(\pm1, \pm2\)

Factors of p= \(\pm1, \pm2, \pm3 ,\pm 6, \pm9, \pm18\)

Then , by rational root theorem the rational roots are in the form :

\(\dfrac{p}{q}=\dfrac{\pm1}{1}, \dfrac{\pm1}{2} , \dfrac{\pm2}{1}, \dfrac{\pm2}{2}, \dfrac{\pm3}{1}, \dfrac{\pm3}{2}, \dfrac{\pm6}{1}, \dfrac{\pm6}{2}, \dfrac{\pm9}{1}, \dfrac{\pm9}{2}, \dfrac{\pm18}{1}, \dfrac{\pm18}{2}\\\\\Rightarrow\ \dfrac{p}{q}=\pm1 , \pm2 , \pm3 , \pm6 , \pm9 , \pm 18, \pm \dfrac{1}{2}, \pm \dfrac{3}{2}, \pm\dfrac{9}{2}\)

Hence , the list of all possible rational zeros of the function are :

\(\pm1 , \pm2 , \pm3 , \pm6 , \pm9 , \pm 18, \pm \dfrac{1}{2}, \pm \dfrac{3}{2}, \pm\dfrac{9}{2}\)