factorise, $x^3-13x^2+32x+20$

Let, $f(x)=x^3-13x^2+32x+20$



$f(-1)\lt 0$, $f(0)\gt 0$, which shows there is a root between $x=-1$ and $x=0$

$f(4)\gt 0$, $f(5)\lt 0$, which shows there is a root between $x=4$ and $x=5$

$f(9)\lt 0$, $f(10)\gt 0$, which shows there is a root between $x=9$ and $x=10$

  • $\begingroup$ Factorise...over what? $\endgroup$
    – DonAntonio
    Mar 14, 2013 at 2:11
  • $\begingroup$ @DonAntonio Given the use of the intermediate value theorem, I believe it is over $\mathbb{R}$ or some extension of $\mathbb{R}$. $\endgroup$
    – Julien
    Mar 14, 2013 at 2:17
  • $\begingroup$ Good observations: there are three distinct real roots. So $f$ factors into $(x-x_1)(x-x_2)(x-x_3)$ over $\mathbb{R}$. $\endgroup$
    – Julien
    Mar 14, 2013 at 2:19
  • 8
    $\begingroup$ maybe it is supposed to be $-20$ so that $x=1$ would be a root. $\endgroup$
    – Jonathan
    Mar 14, 2013 at 2:42
  • 1
    $\begingroup$ @Jonathan The two other roots end up rational that way too, as it turns out. $\endgroup$ Mar 14, 2013 at 3:03

1 Answer 1


From the elementary theory of polynomials we know that the cubic polynomial $ax^3+bx^2+cx+d$ can be factored as


where $x_k$ ($k=1,2,3$) are the roots of the general cubic equation


If we use the substitution $x=t+\frac{13}{3}$, the given equation $$ \begin{equation*} x^{3}-13x^{2}+32x+20=0\tag{1} \end{equation*} $$ is transformed into the reduced cubic equation $$ \begin{equation*} t^{3}+pt+q=0,\qquad p=-\frac{73}{3},q=-\frac{110}{27},\tag{2} \end{equation*} $$ a solution of which is$^1$ $$ \begin{eqnarray*} t_{1} &=&\left( \frac{-q+\sqrt{\Delta }}{2}\right) ^{1/3}+\left( \frac{-q- \sqrt{\Delta }}{2}\right) ^{1/3}, \qquad \Delta &=&q^{2}+\frac{4p^{3}}{27}.\tag{3} \end{eqnarray*} $$ When the discriminant $\Delta <0$ the three solutions of the original cubic $(1)$ are real. By using complex numbers we can write them in the form
$$ \begin{equation*} x_{k}=2\sqrt{-\frac{p}{3}}\cos \left( \frac{1}{3}\arccos \left( -\frac{q}{2} \sqrt{-\frac{27}{p^{3}}}\right) +\frac{2\pi (k-1)}{3}\right) -\frac{b}{3} , \end{equation*}\tag{4} $$ where $k=1,2,3$, and $b=-13$ is the coefficient of $x^{2}$ in $(1)$. Since $$\Delta =-\dfrac{57\,184}{27}<0,$$ we have: $$ \begin{eqnarray*} x_{1} &=&\frac{2\sqrt{73}}{3}\cos \left( \frac{1}{3}\arccos \left( \frac{55\sqrt{73}}{5329}\right) \right) +\frac{13}{3} \\ &\approx &9.347\,9, \\ && \\ x_{2} &=&\frac{2\sqrt{73}}{3}\cos \left( \frac{1}{3}\arccos \left( \frac{55\sqrt{73}}{5329}\right) +\frac{2\pi }{3}\right) +\frac{13}{3} \\ &\approx &-0.513\,6, \\ && \\ x_{3} &=&\frac{2\sqrt{73}}{3}\cos \left( \frac{1}{3}\arccos \left( \frac{55\sqrt{73}}{5329}\right) +\frac{4\pi }{3}\right) +\frac{13}{3} \\ &\approx &4.165\,7. \end{eqnarray*} $$

Therefore the factorization of $(1)$ is $$ \begin{equation*} x^{3}-13x^{2}+32x+20=(x-x_1)(x-x_2)(x-x_3).\tag{5} \end{equation*} $$


$^1$ A deduction of $(3)$ and $(4)$ can be found in this blog post of mine, in Portuguese.

  • 1
    $\begingroup$ The found roots are probably expressed by using Viète's method of reducing a cubic with three real roots to an application of the cosine triplication formulas or a similar method. See, for example, capone.mtsu.edu/jhart/cardan.pdf $\endgroup$
    – egreg
    May 27, 2013 at 21:17
  • $\begingroup$ @egreg Thanks for the link. $\endgroup$ May 27, 2013 at 21:38
  • $\begingroup$ @egreg I decided to rewrite the answer using the formulas I derived in a blog post of mine (the link is indicated in the foot note). $\endgroup$ May 28, 2013 at 9:28

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.