I know about more than half of this, the rest... I had to look up. :/
Don't give me much credit!! LOL
Cubics and up are very difficult to calculate roots, unless they are easily factorable
so ideally we can factor this into something like (x+a)(x+b)(x+c) = 0, then finding the roots will be very easy
Unfortunately we cannot reduce it this way.as our only choices for (a,b,c) will be (1,-1,2), (1,1,-2) or (-1,-1,-2) and none of these will satisfy the equation.
So our only other option is to reduce this to a quadratic aka
(x+a)(x^2+bx+c) and solve for a,b,c
This expands to
x^3+(a+b)x^2 + (c + ab)x +ac
so for this problem we will require that (a+b)= -6, as it is the coefficient for x^2, (c+ab) = 7 and ac = -2.
Starting with ac = -2 we have the following integer potential solutions:
(-1,2)
(1,-2)
Let's now try each one starting with (-1,2).
With these values of (a,c) we can solve for b:
(c+ab) = 7
thus
2 -b = 7
b = -5
Now we check to see if (a,b,c) = (-1,-5,2) works
(a+b) = -6 thus -1 -5 = -6 which works
so therefore we can reduce our cubic to
(x-1)(x^2 -5x +2) =0
This will be true when (x-1) =0 ie x =1 is a solution
or x^2 -5x +2 =0
we must use quadratic theorem to solve this quadratic:
x = (-b+/-sqrt(b^ - 4ac))/2a
ie x = (5+/- sqrt(25-8))/2
sooooo..
x = (5+ sqrt(17))/2
and
x = (5 - sqrt(17))/2
so those are the roots (zeros) of the cubic
