Question

find all rational roots of f(x).
f(x) = x³ - 6x² - 39x - 10
write your answer as a list of simplified values separated by commas, if there is more than one value.

Explanation:

Step1: Apply Rational Root Theorem

Possible rational roots: $\pm1, \pm2, \pm5, \pm10$

Step2: Test $x=-1$

$f(-1)=(-1)^3 -6(-1)^2 -39(-1)-10 = -1 -6 +39 -10=22
eq0$

Step3: Test $x=-2$

$f(-2)=(-2)^3 -6(-2)^2 -39(-2)-10 = -8 -24 +78 -10=36
eq0$

Step4: Test $x=-5$

$f(-5)=(-5)^3 -6(-5)^2 -39(-5)-10 = -125 -150 +195 -10=-90
eq0$

Step5: Test $x=10$

$f(10)=(10)^3 -6(10)^2 -39(10)-10 = 1000 -600 -390 -10=0$

Step6: Factor polynomial

Use polynomial division or synthetic division to divide $x^3-6x^2-39x-10$ by $(x-10)$:
$x^3-6x^2-39x-10=(x-10)(x^2+4x+1)$

Step7: Check quadratic roots

The quadratic $x^2+4x+1$ has discriminant $\Delta=4^2-4(1)(1)=12$, roots $-2\pm\sqrt{3}$, which are irrational.

Answer:

$10$