
Andy C. answered 10/15/17
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Applying the rational root theorem, the possible candidates are
{+or- 1, +or- 2, +or- 3, +or- 5, +or- 6, +or- 10, +or- 15, +or- 30}
It turns out that x=5 is a zero.
Synthethic Division:
5 | 1 -5 -6 30
5 0 -30
------------------------
1 0 -6 0
x^2 - 6 = 0
x = +or- sqrt(6)
Using synthetic division on x=7 to check for upper bounds:
7 | 1 -5 -6 30
7 14 56
-------------------------
1 2 8 86
So 7 is an upper bound
Checking lower bounds with x=-3
-3 | 1 -5 -6 30
-3 24 -54
------------------------
1 -8 18 -24
The signs of these coefficients alternate,
so -3 is a lower bound
Since it is a cubic polynomial, there are only 3 zeros.
They are bounded as:
-3 < -sqrt(6) < sqrt(6) < 5 < 7