f(x)=3x^3-26x^2+52x-24
write down all the solutions of the equation f(x)=0
If there exist integer solution they must be factors of 24. Determine all factors of
24 : ±1, ±2, ±3, ±4, ±6, ±8, ±12, ±24
Check if one of these 16 fctors is a solution of the equation. I've found x = 2.
Use synthetic division to cancel out the factor (x - 2). I've got:
$\displaystyle (3x^3-26x^2+52x-24) \div (x-2)=3x^2-20x+12$
Solve the quadratic
$\displaystyle 3x^2-20x+12 = 0$ for x to get the last 2 solutions of the equation
Hey outlaw!
Looks like you can factor out a 3, but 3 doesn't go into 52 evenly.Solve: $\displaystyle f (x) = 3x^3 - 26x^2 + 52x - 24$ for $\displaystyle f (0)$.
And it looks like you can't Factor by grouping.
So use the Rational Roots Theorem or Rational Zeros Theorem.
With that, you'll have: $\displaystyle \frac{p}{q} = \frac{\pm 1\,,\,\pm2\,,\,\pm 4 \,,\,\pm 6\,,\,\pm8\,,\,\pm12\,,\,\pm24}{\pm 1\,,\,\pm 3}$
So plug all possible values $\displaystyle \frac{p}{q}$ into the equation to see which ones give you zero.
When you do that, I get $\displaystyle f(2), f(6)$ and $\displaystyle f(\frac{2}{3})$ give you zero.