1

$$x^3 - 3x^2 - 3x +2 = 0$$

The rational root test does not work; there are no rational roots.

Mhenni Benghorbal
  • 47,431
IRSAgent
  • 317

2 Answers2

4

Here is a step by step procedure.

First shift $x$ so there is no quadratic term. To do this we shift by 1/3 the quadratic term (assuming leading coefficient is 1) This give $$ x=y+1 \Rightarrow y^3-6\,y-3 =0 $$ We now reduce the cubic by the substitution $y = A(z+1/z)$ $$ y = A(z+1/z) \Rightarrow \frac{z^6\,A^3+3\,z^4\,A^3+3\,z^2\,A^3+A^3-6\,z^4\,A-6\,z^2\,A-3\,z^3}{z^3}=0 $$ We now pick $A \neq 0$ so that $Z^4$ term is zero. This requires $$ A(A^2-2)=0$$ So pick $A=\sqrt{2}$ and the equation becomes $$2^{{{3}\over{2}}}\,z^6-3\,z^3+2^{{{3}\over{2}}}=0$$ This is a quadratic in $z^3$. So set $z=u^{1/3}$ to get $$ z=u^{1/3} \Rightarrow 2^{{{3}\over{2}}}\,u^2-3\,u+2^{{{3}\over{2}}}=0$$

Solve this quadratic for $u$. You will have two roots. Use them to find $z$. Recall each number will have 3 cube roots, but you will have only 3 possible values and not 6.

Once you have $z$, calculate $y$ and then finally $x$.

user44197
  • 9,730
  • This is cool. Having problems finding a video explaining the process on youtube. Do you have a link? – Chris Feb 19 '14 at 14:25
  • @Chris Not that I know of. The other methods are Cardono's substitution and Vieta's substitution instead of the substitution $y=A*(z+1/z)$. I prefer this if we have complex roots and we use polar form for complex numbers. Then these become trigonometric formulas. – user44197 Feb 19 '14 at 14:32
  • Thanks I'll look them up. This is so much cleaner and more precise than my graphing method. – Chris Feb 19 '14 at 14:44
  • Thanks a lot!! The answer is very well explained ! thank you! – IRSAgent Feb 20 '14 at 05:31
  • This is way beyond high school right?? Why did they even ask such a question? – IRSAgent Feb 20 '14 at 05:34
  • @user129852 You CAN solve this without this method. Using a graphing calculator will show you the zeros. But that is what I like about the stackexchange the folks here will often show you a simpler or more elegant process that gets you the same answer with less headache. – Chris Feb 20 '14 at 13:50
  • @Chris There is a problem with graphing calculator approach. It will not show the complex roots and in many applications the complex roots are just as important. – user44197 Feb 21 '14 at 02:05
  • Which is why I like your method so much more. – Chris Feb 21 '14 at 02:13
0

There plainly there are no rational roots but the function does have inflexion points at $x=3(1\frac{+}{-} \sqrt2)$ subbing this into the original equation yields a positive and negative y value , respectively. So you may be able to find your root by graphing.

Chris
  • 643