Find the rational zeros and then other zeros of the polynomial function f(x) = x^{3} - 17x^{2} + 55x + 25, that is, solve f(x) = 0 Factor f(x) into linear factors

To find the rational zero and other zeros of the polynomial
$f(x)=x^3-17x^2+55x+25$
Factor $f(x)$ into linear factors
Let us solve the equation.
$f(x)=0$
$x^3-17x^2+55x+25=0$
$\because$ the constant term has factors 1 and 5.
Let us whether these can be zeros.
$f(1)=1-17+55+25=64\ne 0$
$\therefore x=1$ can not be zero of f(x).
$f(5)=5^3-17(5{)}^2+55(5)+25=125-425+275+25=-300+300=0$
$\therefore x=5$ is a rational zero of f(x).
We can write,
$f(x)=x^2(x-5)-12x(x-5)-5(x-5)$
$=(x-5)(x^2-12x-5)$
Now we only need to solve the quadratic equation.
$x^2-12x-5=0$
Then $x=\frac{-(-12)\pm \sqrt{(-12{)}^{2}4(1)(-5)}}{2(1)}$
$=\frac{12\pm \sqrt{144+20}}{2}$
$=\frac{12\pm \sqrt{164}}{2}$
$=\frac{12\pm 2\sqrt{41}}{2}$
$=6\pm \sqrt{41}$
Therefore, there is only one rational zero, 5 and other zeros are
$6+\sqrt{41},6-\sqrt{41}$
The factorization of f(x),
$f(x)=(x-5)(x-(6-\sqrt{41}))(x-(6+\sqrt{41}))$
$=(x-5)(x-6+\sqrt{41})(x-6-\sqrt{41})$