factorise, $x^3-13x^2+32x+20$ 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 $x=10$
 A: 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
$$
\begin{equation*}
x^{3}-13x^{2}+32x+20=(x-x_1)(x-x_2)(x-x_3).\tag{5}
\end{equation*}
$$
--
$^1$ A deduction of $(3)$ and $(4)$ can be found in this blog post of mine, in Portuguese.
