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Find all solutions, other than $2$ for $12x^3-23x^2-3x+2=0$

I started off by taking out an $x$ and got $$x(12x^2-23x-3)+2=0$$
I do not know if this is the correct first step, if it is, then am I able to use the quadratic formula or complete the square to get the answers. Can anyone give me general hints. Please do not solve this for me in anyway. Just give me hints.

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That first step doesn't work very well. It's good if you can find something to factor out of the entire expression so that you end up with something like $AB=0$, because then you can break up the problem into "$A=0$ or $B=0$". But changing it to $AB+C=0$ does not usually simplify the problem. – Arturo Magidin Jul 20 '12 at 3:39
The step is technically not incorrect. But it is not at all useful. – André Nicolas Jul 20 '12 at 5:59
Since $12x^2-23x-3$ at $x=2$ is $-1$, then $x=2$ is a solution of the polynomial. Then divide the original polynomial by $x-2$ to get a quadratic which can be easily solved for the other roots 1/4 and -1/3. – i. m. soloveichik Jul 22 '12 at 14:33
up vote 8 down vote accepted


The great bit here is to note that $2$ is a solution. By the factor theorem, we know that the linear factor $(x - 2)$ divides our polynomial.

So perhaps you should divide out $(x-2)$. You'll be left with a quadratic, which we know how to solve very quickly.

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So, I synthetically divide and then use the quadratic formula for the rest? – Austin Broussard Jul 20 '12 at 3:39
That sounds like a great strategy to me. Good luck! – mixedmath Jul 20 '12 at 3:40
I'll recomment with my answer and hopefully it'll be right! Thanks for the help! – Austin Broussard Jul 20 '12 at 3:40
@Austin: You don't need to ask. You could just plug in those values and check whether they satisfy your equation. – Javier Jul 20 '12 at 3:49
@Austin: Hey that's great! Although, unless my math is off, I happen to get $1/4$ and $-1/3$, as if one of us dropped a negative somewhere. – mixedmath Jul 20 '12 at 3:55

The preceding answers have provided you with sufficient information. However the following might help you with future endeavors:

How would you factor the polynomial if 2 wasn't given as a root?

There exists a useful theorem: the Rational Root Theorem, that helps factor polynomials (indirectly). It gurantees that any root of the polynomial will have a numerator that has $c$, the constant, as a factor and a denominator that has $a_n$, the leading coefficient, as a factor.

Therefore by investigating all possible conbinations of the factors of the constant and the factors of the leading coefficient you can eventually arrive at a valid root. The polynomial can then be further factored into a quadratic through synthetic division which can be factored (as you know) through the quadratic formula.

The interesting thing about this method is its large scope of applicability: it not only works on cubic polynomials but on a polynomial with any degree.

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The reciprocals of the roots $\rm\,r,s,1/2\:$ are roots of the reversed polynomial $\rm\:2\, x^3\! -\! 3\, x^2\! -\! 23\, x\! +\! 12.\: $ Thus by Vieta's Formulas $\rm\:r\!+\!s\!+\!1/2 = 3/2,\ rs/2 = -6,\:$ so $\rm\:r\!+\!s = 1,\ rs = -12,\:$ so $\rm\:r,s = \ldots$

Alternatively, by the Factor Theorem, $\rm\:f(2) = 0\:$ $\Rightarrow$ $\rm\:f(x)\:$ has $\rm\:x\!-\!2\:$ as a factor. Comparing coef's

$$\rm (x-2)(a\, x^2 + b\, x + c)\ =\ 12\, x^3 - 23\,x^2 -3\,x + 2$$

$\rm\qquad x^3\:$ coef $\rm\:\Rightarrow\: a = 12$

$\rm\qquad x^0\:$ coef $\rm\:\Rightarrow\: -2\,c = 2\:\Rightarrow\: c = -1$

$\rm\qquad x^1\:$ coef $\rm\:\Rightarrow\: -3 = c-2b = -1-2b\:\Rightarrow\: b = 1 $

So the quadratic factor is $\rm\: 12\,x^2 + x - 1,\:$ which can be solved by either the Quadratic Formula or Rational Root Test, or $\rm\:(c\,x\!-\!1)\,(d\,x\!+\!1) = 12\,x^2\!+\!x\!-\!1\:$ $\Rightarrow$ $\rm\:cd=12,\ c\!-\!d = 1,\:$ so $\rm\:c,d = \ldots$

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$c=4$ and $d=3$? – Austin Broussard Jul 20 '12 at 4:07
@Austin Indeed, they satisfy the equations. – Bill Dubuque Jul 20 '12 at 4:12

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