Say for example:
$6x^{3}-17x^{2}-4x+3=0$
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?
Say for example:
$6x^{3}-17x^{2}-4x+3=0$
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?
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.
$f(x)= 6x^3 -17x^2 -4x +3$
Always observe the polynomials behavior for different values of x
$f(-1)=-16$
$f(0)=3$
$f(1)=-12$
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 $x=\frac{-1}{3}, \frac{1}{3},\frac{-1}{2}, \frac{1}{2}$
$f(\frac{1}{3})=0$
$f(x)$ is not an even function so discard $x=\frac{-1}{3}$
$f(\frac{-1}{2})=0$
$f(x)$ 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 $(\frac{-1}{2}, \frac{1}{3}, 3)$
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 $\:3,\,1/3,\, -1/2.$