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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?2012-07-20
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    @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.2012-07-20
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    Alright. Another thing. When you say factor out $(x-r)$ you just use synthetic division to get your new quadratic equation?2012-07-20
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    @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.2012-07-20
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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?2012-07-20
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    @Austin: They must divide the leading term, which here is $12$.2012-07-20
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    So all the factors of 12?2012-07-20
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    @AustinBroussard: Exactly. That is the coefficient of the highest power.2012-07-20
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    So I must try: $\dfrac{-1}{1,2,3,4,6,12}$ and $\dfrac{1}{1,2,3,4,6,12}$?2012-07-20
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    @AustinBroussard: That is right, though it is a strange way to list a bunch of fractions.2012-07-20
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    I figured you would understand it. It's a lot easier than typing each individual fraction.2012-07-20
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    I only got two answers: $-\frac{1}{2}$ and $\frac{1}{3}$ So where's the third one?2012-07-20
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    @AustinBroussard: $-\frac 12$ is a double root. Having found these two, you can divide your polynomial by $2x+1$ and by $3x-1$ and see what is left. In this case it will again be $2x+1$.2012-07-20
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    @Austin: As soon as you find **one** root, $r$, **factor out** $x-r$; that will reduce the problem from one with a cubic polynomial to one with a degree $2$ polynomial. It will also help you figure out if the polynomial has **repeated** roots. Here, $$12x^3 +8x^2-x-1 = (x+\frac{1}{2})(12x^2+2x-2) = 2(x-\frac{1}{2})(6x^2+x-1).$$ The roots of $6x^2+x-1$ are $\frac{1}{3}$ and $-\frac{1}{2}$, so $-\frac{1}{2}$ is a **double** root.2012-07-20
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    And since the factors must have leading coefficients which multiply to 12 - the leading coefficient of the given polynomial, and since 12 is greater than the sum of the absolute values of the other coefficients - which rules out $\pm1$ you need to find a factorisation of 12 into 3 factors with absolute value greater than 1, and then sort out the signs.2012-07-20
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    Question, when you take out the $2$, how does that make the first set of parenthesis $(x-\frac{1}{2})$?2012-07-20
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    @AustinBroussard: It should be $x+\frac 12$, the minus sign is a typo. We already know the root is $-\frac 12$2012-07-20
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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.