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Say for example:


I sort of look at it and don't know where to start, other than just guessing what the first one would be and trying to do from there. Is there a good technique for approaching such polynomials?

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Relevant background. – user2468 Jul 30 '12 at 3:54
The rational root theorem, of course, is a good starting point. – Hurkyl Jul 30 '12 at 3:55
up vote 1 down vote accepted

First, to minimize the lead coefficient, we reverse the polynomial (which $\rm\color{brown}{reciprocates}$ the roots) and then we scale it to make the leading coefficient $1,\,$ i.e. we apply the AC-method, which yields

$$\begin{eqnarray}\rm f(x) \,&=&\,\rm 3\,x^3 - 4\,x^2-17\,x+ 6 \\ \rm \Rightarrow\ \ 9\,f(x)\, &=&\,\rm (3x)^3 - 4\,(3x)^2 - 51\,(3x) + 54 \\ \rm &=&\rm\ \ z^3 - \color{blue}4\, z^2 - 51\,z + \color{#0A0}{54},\ \ \ z = 3x \end{eqnarray}$$

Note $\rm\:z=\color{#C00}1\:$ is a root. By Vieta, other roots have product $\color{#0A0}{-54/\color{#C00}1},\,$ sum $\color{blue}4\!-\!\color{#C00}1 = 3,\:$ so are $\:9,\,-6.\:$

Hence $\rm\:z\, =\, 1,\,9,\,-6\,\:$ so $\rm\:x\, =\, z/3\, =\, 1/3,\, 3,\, -2,\,$ which $\rm\color{brown}{reciprocated}$ yields $\:3,\,1/3,\, -1/2.$

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Oh... would you mind adding one more step between the 1st n second steps? I am getting myself in a muddle. – Magpie Jul 30 '12 at 7:09
yeah it's the 54 I can't see where that came from – Magpie Jul 30 '12 at 7:36
@Magpie To compute $\rm\:9\,f(x)\:$ we multiply all the coefficients of $\rm\:f(x)\:$ by $9,\,$ namely $\rm\:9(3x^3) = 27 x^3 = (3x)^3,\:$ $\rm\:9(4x)^2 = 4(9x^2) = 4(3x)^2,\:$ $\rm\:9(17x) = 17\cdot 3(3x) = 51(3x),\:$ $\,9\cdot 6 = 54.$ – Bill Dubuque Jul 30 '12 at 13:06
This blows my mind. Practice needed. Thank you. It is the best method, I think. Would you mind putting that line into your answer so others can spot it more easily too? I would do it myself but I am still learning the syntax. – Magpie Jul 30 '12 at 16:50

$f(x)= 6x^3 -17x^2 -4x +3$

Always observe the polynomials behavior for different values of x




Here we see that there are two roots of $f(x)=0$ between $-1$ and $1$

Now the constant part of the term $6x^3$ is $6=(1)(2)(3)$

Check for $x=\frac{-1}{3}, \frac{1}{3},\frac{-1}{2}, \frac{1}{2}$


$f(x)$ is not an even function so discard $x=\frac{-1}{3}$


$f(x)$ is not an even function so discard $x=\frac{1}{2}$

we see that two of the roots are $x=\frac{-1}{2}$ and $x=\frac{1}{3}$

Now $f(x)= (x +\frac{1}{2})(6x^2 -20x^2 +6)$

i.e. $f(x)= (x +\frac{1}{2})(x -\frac{1}{3})(6x -18)$

i.e. $f(x)= (6)(x +\frac{1}{2})(x -\frac{1}{3})(x -3)$

i.e. $f(x)= (1)(2)(3)(x +\frac{1}{2})(x -\frac{1}{3})(x -3)$

the solution set for $f(x)=0$ is $(\frac{-1}{2}, \frac{1}{3}, 3)$

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This method isn't the best for me unfortunately, I am terrible at mental arithmetic so I need a fool-proof method! Thanks though. – Magpie Jul 30 '12 at 7:16

You can ask Wolfram Alpha You can use the rational root theorem which works well for class problems (less well for real world problems). You can do numeric solutions-see any numerical analysis text. In this case (you didn't say where it came from) the rational root theorem works very well.

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are you able to explain the rational root theorum in terms of my example? – Magpie Jul 30 '12 at 4:04
Did you look at the Wikipedia page? The rational roots will have numerators that are factors of the constant term, here $\pm1, \pm3$ and denominators that are factors of the highest term, here $1,2,3,6$, so the choices are limited. Try each one in your polynomial and see which ones work. $\frac 13$ is one that works. Having found that one, you can keep on with rational roots (which will find them all this time), or divide your polynomial by $x- \frac 13$ to get a quadratic which you have a formula to solve. – Ross Millikan Jul 30 '12 at 4:08
Yes but I could not see what terms they meant. There are several constant terms in the polynomial and by highest do you mean highest number or final term? – Magpie Jul 30 '12 at 4:31
@Magpie: The constant term is the one that does not multiply any factors of the variable. In your case it is $3$. The highest term is the one that multiplies the highest exponent of the variable. Here it is $6$ because it multiplies $x^3$ – Ross Millikan Jul 30 '12 at 4:33

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