# 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.

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Note that trying to factorize anything as $x(x^2-6x+11)=6$ is almost useless, since $6$ need not factor nicely. The rare instance when such a factorization is useful, is for example $(x+1)(x+2)(x+3)=6$, in which case we can observe that $x=0$ is a root. –  Calvin Lin Jan 14 '13 at 20:21
Do you mean $6x^2$ instead of $6x$? –  Git Gud Jan 14 '13 at 20:22
Yes, sorry forgot to square that one. –  Leonardo Jan 14 '13 at 20:23
You should also know that there are closed formulas to get the roots of polynomials of degree up to 4. –  Git Gud Jan 14 '13 at 20:28

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 $$x^3 - 6x^2 + 11x - 6 =(x-1)(x-2)(x-3)$$

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Luckily this is a nice equation, thank you I have at least one extra tool in my belt and I will need to read up more on the rational root theorem among other things. Your answer is very clear and concise. –  Leonardo Jan 14 '13 at 20:27

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 $a,b,c$.

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I kind of get what you and Will Jagy are saying, but it will take time to absorb. The only thing that is immediately obvious from your last equations is that c = 6. I appreciate your helps. –  Leonardo Jan 14 '13 at 20:22
@Leonardo: Look at Marvis's. I noted some points just for starting. He gave you the answer in detailed. –  Babak S. Jan 14 '13 at 20:25

$x=1$ 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.

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$$x^3 - 6x^2 + 11x - 6 =0$$ $$x^3 -1-( 6x^2 - 11x + 6)+1 =0$$ $$x^3 -1-( 6x^2 - 11x + 5) =0$$ $$x^3 -1-( 6x^2 - 6x-5x + 5) =0$$ $$(x-1)(x^2+x+1)-( 6x(x-1)-5(x-1) =0$$ $$(x-1)(x^2+x+1)-(x-1)(6x-5) =0$$ $$(x-1)(x^2+x+1-6x+5) =0$$ $$(x-1)(x^2-5x+6) =0$$ $$(x-1)(x^2-2x-3x+6=0$$ $$(x-1)(x(x-2)-3(x-2)=0$$ $$(x-1)(x-2)(x-3)=0$$ $$x_1=1,x_2=2,x_3=3$$

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Nice Adi,nice. ;-) –  Babak S. Jan 14 '13 at 20:26
My brain does not work like yours, but I hope it starts to soon. –  Leonardo Jan 14 '13 at 20:29
thank you Babak –  Adi Dani Jan 14 '13 at 23:31

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.

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