1. ## rational roots

f(x)=3x^3-26x^2+52x-24

write down all the solutions of the equation f(x)=0

2. Originally Posted by outlaw
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:

$(3x^3-26x^2+52x-24) \div (x-2)=3x^2-20x+12$

$3x^2-20x+12 = 0$ for x to get the last 2 solutions of the equation

3. Hey outlaw!

Solve: $f (x) = 3x^3 - 26x^2 + 52x - 24$ for $f (0)$.
Looks like you can factor out a 3, but 3 doesn't go into 52 evenly.
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: $\frac{p}{q} = \frac{\pm 1\,,\,\pm2\,,\,\pm 4 \,,\,\pm 6\,,\,\pm8\,,\,\pm12\,,\,\pm24}{\pm 1\,,\,\pm 3}$

So plug all possible values $\frac{p}{q}$ into the equation to see which ones give you zero.

When you do that, I get $f(2), f(6)$ and $f(\frac{2}{3})$ give you zero.