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

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

$f(x)=x(x^2-13x+30)+2x+20$

$f(x)=x(x-3)(x-10)+2x+20$

$f(-1)\lt 0$, $f(0)\gt 0$, 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 $x=10$

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  • $\begingroup$ Factorise...over what? $\endgroup$ – DonAntonio Mar 14 '13 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 '13 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 '13 at 2:19
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    $\begingroup$ maybe it is supposed to be $-20$ so that $x=1$ would be a root. $\endgroup$ – Jonathan Mar 14 '13 at 2:42
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    $\begingroup$ @Jonathan The two other roots end up rational that way too, as it turns out. $\endgroup$ – Josephine Moeller Mar 14 '13 at 3:03
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From the elementary theory of polynomials we know that the cubic polynomial $ax^3+bx^2+cx+d$ can be factored as

$$ax^3+bx^2+cx+d=a(x-x_1)(x-x_2)(x-x_3),$$

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

$$ax^3+bx^2+cx+d=0.$$

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

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

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    $\begingroup$ 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 '13 at 21:17
  • $\begingroup$ @egreg Thanks for the link. $\endgroup$ – Américo Tavares May 27 '13 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). $\endgroup$ – Américo Tavares May 28 '13 at 9:28

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