Question

Use the Rational Root Theorem to list all possible rational roots for the equation.

3x^3+9x-6=0

Answer 1

Q^n + k^(n-1).... + P = 0
Like: 3x^3+9x-6=0

All rational roots will be rational factors of P/Q such that:

Q = 3 Factors: 1,3
P = -6 Factors: [+/-] 1,2,3,6

Possible Rational Roots: [1/1,1/3,2/1,2/3,3/1,3/3,6/1,6/3] = [+/-] 1,1/3,2,2/3,3,6

Now you just test them in the equation itself and where the input makes the function equal 0, you have a root.

For this polynomial, no roots are rational, so when you test it you'll find that it must only contain irrational roots and may only be solved by other means.

Answer 2

`\bf \stackrel{q}{3}x^3+9x-\stackrel{p}{6}~\hspace{5em}\stackrel{\textit{factors of \boxed{p}}}{3,2,1,6}\qquad \stackrel{\textit{factors of \boxed{q}}}{3,1}\qquad \qquad \pm\cfrac{p}{q} \\\\\\ \textit{therefore, using the \underline{rational root test}}`

`\bf \begin{cases} \pm \cfrac{3}{3}\implies &\pm 1\\\\ \pm \cfrac{3}{1}\implies &\pm 3\\\\ \end{cases}\quad \begin{cases} \pm \cfrac{2}{3}\\\\ \pm \cfrac{2}{1}\implies \pm 2 \end{cases}\quad \begin{cases} \pm \cfrac{1}{3}\\\\ \pm \cfrac{1}{1}\implies \pm 1 \end{cases}\quad \begin{cases} \pm \cfrac{6}{3}\implies &\pm 2\\\\ \pm \cfrac{6}{1}\implies &\pm 6 \end{cases}`