Option C would be the correct answer :)
The trial and error method is used to find an initial factor:
If we let f(x) = x³ - x² - 24x - 36 and all we have to do is sub' in values of x until
f(x) = 0, we can use this to find an initial factor by the factor theorem:
f(1) = (1)³ - (1)² - 24(1) - 36 = -60
f(2) = (2)³ - (2)² - 24(2) - 36 = -80
f(5) = (5)³ - (5)² - 24(5) - 36 = -56
*** f(6) = (6)³ - (6)² - 24(6) - 36 = 0 ***
f(6) = 0 so (x - 6) is a factor of f(x).
This means that: f(x) = x³ - x² - 24x - 36 = (x - 6)(ax² + bx + c).
To find a,b and c, use long division (or inspection) to divide x³ - x² - 24x - 36 by x - 6.
The other 2 factors of f(x) can then be found by factorizing the
ax² + bx + c quadratic the way you would with any other quadratic (i.e. by quadratic formula, CTS or inspection).
6x - 4 < 8
Add 4 to both sides:
6x < 12
Divide both sides by 6:
x < 2
Answer: x < 2
There is no local extremas but here is the objective function graphed
