# Solve for $x; 12x^3+8x^2-x-1=0$

Solve for $x$. $12x^3+8x^2-x-1=0$ all solutions are rational and between $\pm 1$

As mentioned in my previous answers, I'm guessing I have to use the Rational Root Theorem. But I've done my research and I do not understand what to plug in or anything about it at all. Can someone please dumb this theorem down so I can try to solve this equation. I also do not want anyone to solve this problem for me. Thanks!

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The Rational Root Theorem tells you that if the equation has any rational solutions (it need not have any), then when you write them as a reduced fraction $\frac{a}{b}$ (reduced means that $a$ and $b$ have no common factors), then $a$ must divide the constant term of the polynomial, and $b$ must divide the leading term.

Here, the constant term is $1$, so that means that $a$ must be either $\pm 1$. And the leading terms is $12$; the integers that divide $12$ are $\pm 1$, $\pm 2$, $\pm 3$, $\pm 4$, $\pm 6$, and $\pm 12$. That gives twelve possible things to try.

As soon as you find a root $r$, you should stop and factor out $x-r$ from the polynomial, thus reducing the problem to one with smaller degree.

For example, say you wanted to find the roots of $$6x^3 -25x^2+ 10x - 1.$$ The rational root theorem says that any rational root $\frac{a}{b}$ must have $a$ dividing $1$ (so $a=1$ or $a=-1$), and $b$ dividing $6$ (so $b=\pm 1$, $b=\pm 2$, $b=\pm 3$, or $b=\pm 6$). It doesn't tell you all of them are roots, just that if there are any rational roots, then they must be among $$\pm\frac{1}{2},\quad\pm\frac{1}{3},\quad \pm\frac{1}{6}.$$ Now, you can just test them. $\pm\frac{1}{1}$ does not work (plugging in $1$ gives $-10$, plugging in $-1$ gives $-42$). Plugging in $\frac{1}{2}$, $-\frac{1}{2}$, $\frac{1}{3}$, $-\frac{1}{3}$ doesn't work either. Then when you plug in $\frac{1}{6}$, we get $$\frac{6}{6^3} - \frac{25}{6^2} + \frac{10}{6} - 1 = \frac{1}{36}-\frac{25}{36}+\frac{60}{36} - \frac{36}{36} = 0,$$ so $x=\frac{1}{6}$ is a root. We can then factor out $x-\frac{1}{6}$ from the original polynomial, $$6x^3 -25x2 + 10x -1 = \left(x - \frac{1}{6}\right)\left(6x^2-24x+6\right),$$ so we now just need to find the roots of the other factor, $6x^2-24x+6 = 6(x^2-4x+1)$. We can solve this using the quadratic formula, and the roots are $$\frac{4+\sqrt{16-4}}{2}=2 + \sqrt{3},\qquad \text{and}\qquad \frac{4-\sqrt{16-4}}{2} = 2-\sqrt{3}.$$ So the rational root theorem gave us a finite collection of possible roots; we check them, and if we get lucky and find a root among them, we can use it to reduce the degree of the polynomial by $1$ (by factoring out $x-r$) and so exchange the original problem for a simpler one (instead of a polynomial of degree 3, we now have a polynomial of degree 2).

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Okay, so you get the first root and you take it out to get: $(x-r)(ax^2+bx+c)$ and then use the quadratic formula to get the other two roots? –  Austin Broussard Jul 20 '12 at 4:59
@AustinBroussard: Or you can apply the Rational Root Test again, but now to $ax^2+bx+c$ (which is hopefully simpler), or any other methods you may know for finding roots of a quadratic. –  Arturo Magidin Jul 20 '12 at 5:00
Alright. Another thing. When you say factor out $(x-r)$ you just use synthetic division to get your new quadratic equation? –  Austin Broussard Jul 20 '12 at 5:02
@Austin: You use whatever method you are comfortable with to do the division: synthetic, long, etc. So long as you factor it out right, it doesn't matter what method you use. –  Arturo Magidin Jul 20 '12 at 5:04

The rational root theorem tells you that any rational roots of the polynomial have numerators that divide the constant term and denominators that divide the coefficient of the highest power. So here the numerator must be $\pm1$ and the denominator can be any of $1,2,3,4,6,12$. So you have $12$ possibilities for rational roots. You test each to see if it satisfies the polynomial. As there are only three roots of a cubic, you can quit when you have them.

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How do you know what the denominators are though? –  Austin Broussard Jul 20 '12 at 4:43
@Austin: They must divide the leading term, which here is $12$. –  Arturo Magidin Jul 20 '12 at 4:43
So all the factors of 12? –  Austin Broussard Jul 20 '12 at 4:43
@AustinBroussard: Exactly. That is the coefficient of the highest power. –  Ross Millikan Jul 20 '12 at 4:45
@AustinBroussard: It should be $x+\frac 12$, the minus sign is a typo. We already know the root is $-\frac 12$ –  Ross Millikan Jul 20 '12 at 5:10

The reciprocals of the roots are roots of the (negated) reversed polynomial $\rm\:x^3+x^2-8\,x-12.\:$ By the Rational Root Test all its roots are integers. If the roots are $\rm\:a,b,c\:$ then by Vieta's Formulas we have $\rm\:abc = 12,\ a+b+c =-1\:$ so $\rm\:a,b,c = \ldots$

Remark $\$ I chose to work with the reciprocals of the roots because I know they are integers by RRT, and it is more intutive to do arithmetic with integers than with their reciprocals. Notice that this transformation makes the problem so simple that it can easily be solved mentally. Indeed, it took me less than $10$ seconds to do so. With a little practice, anyone can be just as proficient.

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