solve this polynomial equation: p(x) = x^3 - 6x^2+7x+2

x^3 - 6x^2 + 7x + 2

1. Possible no. of complex roots......either none or two, at most

2. Possible number of real roots.....either one or three, at most

3. Possible rational roots  =  +/- 1 , +/- 2

4.   2 is a root  because  (2)^3 - 6(2)^2 + 7(2) + 2   =  8 - 24 + 14 + 2 =  0

The other two roots are either non-real or non-rational and can be found by synthetic division

2   [ 1     - 6      7      2 ]

                2     -8     -2

      -------------------------

       1     -4      -1      0

So the remaining polynomial is   x^2 - 4x -1

And the remaining roots are  2 + sqrt(5)  and 2 - sqrt(5)

Sounds like a problem I have.. What grade are you in....??

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

Mr. Chris's explanation will be muchhh better than mine!