Is it more practice and intuition or rather algorithmic to solve third degree polynomials of this type? Consider
$x^3 - 6x^2 + 11x - 6 = 0$
I can not reasonably factor this intuitively in any short amount of time with my skill level. Is this the only hope to solving such equations by hand? What tools can I use to help ease the pain or make it more clear to myself?
I try to:
$(x + ?)(x^2 - 6x)$ but it does not seem promising, in my eyes
I cant do anything, that I know about, with:
$x(x^2 - 6x + 11) = 6$
I don't want to give up, but I do anyway and ask my calculator, there is three solutions, alas I have no idea how that helps me (with my knowledge, it does not)!
Should I just practice with smaller polynomials more, and this one might look easier?
Any suggestions are appreciated, I will tag this as homework because it should be.
 A: If you have a polynomial with integer coefficients of the form $$p(x) = x^n + a_{n-1} x^{n-1} + a_{n-2} x^{n-2} + \cdots + a_1 x + a_0$$ and if you have any hope of factorizing this "nicely", you first need to check if the divisors of the constant term, $a_0$, are factors. In your case, I assume the correct polynomial you are interested in is $p(x) = x^3 - 6x^2+11x-6$. The divisors of $-6$ are $\{\pm1,\pm2, \pm3, \pm6\}$. A quick check reveals that $x^*=1,2,3$ satisfy $p(x^*) = 0$. Hence, we have $$x^3 - 6x^2 + 11x - 6 =(x-1)(x-2)(x-3)$$
A: Your equation is $$x^3-6x^2+11x-6=0$$ The coefficients of the equation are $1,-6,+11,-6$ and the summation of them is $$1-6+11-6=0$$ Whenever you face to this fact, you always have $x=1$ as one solution. In fact the equation has a factor as form $x-1$. It means that $$x^3-6x^2+11x-6=(x-1)(ax^2+bx+c)$$ for some proper available constants $a,b,c$.
A: $$x^3 - 6x^2 + 11x - 6 =0$$
$$x^3 -1-( 6x^2 - 11x + 6)+1 =0$$
$$x^3 -1-( 6x^2 - 11x + 5) =0$$
$$x^3 -1-( 6x^2 - 6x-5x + 5) =0$$
$$(x-1)(x^2+x+1)-( 6x(x-1)-5(x-1) =0$$
$$(x-1)(x^2+x+1)-(x-1)(6x-5) =0$$
$$(x-1)(x^2+x+1-6x+5) =0$$
$$(x-1)(x^2-5x+6) =0$$
$$(x-1)(x^2-2x-3x+6=0$$
$$(x-1)(x(x-2)-3(x-2)=0$$
$$(x-1)(x-2)(x-3)=0$$
$$x_1=1,x_2=2,x_3=3$$
A: $x=1$ is an evident solution. Therefore we can write
$$ x^3 - 6 x^2 + 11 x - 6 = (x-1)(a x^2 + b x + c)   $$
and solve for $a,b,c$ carefully. Then the rest is the quadratic formula. 
I suppose the main advice is to check for integer roots and rational roots before factoring, when you have no feeling for what is going on in the problem. There is a Rational Roots Theorem that applies here.
A: You need to know a few things. For example if $p(x)$ is a polynomial of odd degree with rational coefficients it will necessarily have a real root.
Another useful fact is that if $a$ is a root of $p(x)=x^3+px^2+qx+r=0$ then $(x-a)$ is a factor of $p(x)$. This works because:
$x^3-a^3=(x-a)(x^2+ax+a^2)$ and $x^2-a^2=(x-a)(x+a)$ so if $p(a)=0$ we can write:$$p(x)=p(x)-p(a)=[x^3-a^3]+p[x^2-a^2]+q[x-a]=(x-a)([x^2+ax+a^2]+p[x+a]+q)$$
We also note that if $p,q,r,a \in \mathbb Z$, then if $p(a)=a(a^2+pa+q)+r=0$ then $a$ must be a factor of $r$.
[There are various generalisations and extensions of these comments].
So one thing we do when faced with a cubic with integer coefficients to factorise is to try the factors ($\pm$) of the constant term as "easy candidates" - and once we have one factor we can divide through and solve the remaining quadratic.
