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This is the result of determinant evaluation:

$$p(x) = (x-3)((x-1)(x-2)-1)+1$$

How can I factor this polynomial?

N. F. Taussig
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jmiller
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  • Get it into the form $ax^3+bx^3+cx+d$ and look for rational roots of the form $\frac p q$ with $q \vert a$ and $p \vert d$. – Zardo Jul 23 '15 at 10:12
  • The next move would be: $(x-3)(x-1)(x-2)-(x-3)+1 = (x-3)(x-1)(x-2)-x+4$. – jmiller Jul 23 '15 at 10:13
  • when simplified $x^3-6x^2+10x-2$ ,you can try hit and trial from here – Nebo Alex Jul 23 '15 at 10:16
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    There are no rational roots for this equation, and it has only one real root, the method I suggest is to solve the equation using the genral method for solvving an equation of $3$rd degree, If this is your first time to solve an equation od degree $3$ then it's worth to try it ( cualculations) if this is not your first time or you did not want to do so, you can use a software like Mathematica and give you the exact form of the roots, see :https://en.wikipedia.org/wiki/Cubic_function – Elaqqad Jul 23 '15 at 11:09
  • The polynomial is irreducible over $\mathbb Q[x]$. – Lucian Jul 23 '15 at 16:06

1 Answers1

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I walk you through this one. The first step is to expand the products. You get $$f(x)=x^3-6x^2+10x-2.$$ You want fo find the zeros of this. The rational root test tells you to try $x=\pm1$ and $\pm2$. None of those are zeros, so we foresee that this will become ugly.

Next we get rid of the quadratic term. This is a bit like completing square, but no need to worry about that. The coefficient $-6=3\cdot(-2)$, so we make the substitution $x=y-(-2)=y+2$. Plugging that in and expanding (leaving those steps to you) gives $$ 0=f(y+2)=y^3-2y+2. $$ Then comes the bit of Cardano magic. Look it up from a book, wikipedia, or my answer here. Cardano's observation was that if the numbers $u$ and $v$ satisfy the equations (here $p=-2$ is the coefficient of the linear term, and $q=2$ is the constant term) $$ \left\{\begin{array}{rcl}u^3+v^3&=&-q=-2,\\uv&=&-\dfrac p3=\dfrac23, \end{array}\right. $$ then $u+v$ is a solution.

How does that help? We see that $u^3v^3$ should be equal to $8/27$, so $$ g(T):=(T-u^3)(T-v^3)=T^2-(u^3+v^3)T+u^3v^3=T^2+2T+\frac8{27}. $$ So $u^3$ and $v^3$ are zeros of the quadratic (!!!) polynomial $g(T)$. Here the quadratic formula comes in handy, and it gives $$ u^3,v^3=-1\pm \sqrt{1-\frac8{27}}=\frac19(-9\pm\sqrt{57}). $$ Observe that $(-9+\sqrt{57})$ and $(-9-\sqrt{57})$ are both negative real numbers. Therefore one solution of the cubic is $$ y=\root3\of{\frac{-9+\sqrt{57}}9}+\root3\of{\frac{-9-\sqrt{57}}9}, $$ and one of the zeros of your cubic is thus $$ x_1=-2+y=-2+\root3\of{\frac{-9+\sqrt{57}}9}+\root3\of{\frac{-9-\sqrt{57}}9}\approx-0.231. $$ You get the other roots by using complex numbers. We need the cubic root of unity $\omega=(-1+i\sqrt3)/2$, when $\omega^2=\omega^{-1}=(-1-i\sqrt3)/2$. The point is that if $(u_1,v_1)$ works, then so do $(u_2,v_2)=(\omega u_1,\omega^{-1}v_1)$ and $(u_3,v_3)=(\omega^{-1}u_1,\omega v_1)$. This is because multiplying $u$ (or $v$) by a power of $\omega$ does not change the value of $u^3$. But we need to exercise a little bit of care and multiply $u$ and $v$ with opposite powers of $\omega$. For otherwise the value of $uv$ would change!

Your other solutions are thus $$ x_{2,3}=-2+\frac{-1\pm i\sqrt3}2\root3\of{\frac{-9+\sqrt{57}}9}+ \frac{-1\mp i\sqrt3}2\root3\of{\frac{-9-\sqrt{57}}9}\approx2.885\pm i 0.590. $$

Either there was a mistake, your teacher was a bit naughty, or you were not really expected to solve the eigenvalues :-)

Anyway, the factors of $f(x)$ are $x-x_i, i=1,2,3$.

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Jyrki Lahtonen
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