00:01
So for part 8, so we know that our p is negative 3 and our q is 2.
00:13
So x is cubed, negative 1 plus root 2 squared over 4, plus negative 3 cubed over 27.
00:32
Plus, you know, cubed.
00:38
Again, negative 1, negative 2 over 2, something, plus root 2 squared over 4, which is 1, plus negative 3 cubed over 27, which is equal to basically just think about it, it's basically negative 2.
01:06
And so to obtain the roots, we have to do polynomial division, so if we do that, do divided by x plus 2, then we get x squared minus 2, which is x minus 1 squared.
01:34
So the equation can be rewritten as x plus 2 times x minus 1 squared equal to 0.
01:46
Okay? and we can get our zeros.
02:01
But if we look at the original equation, we can see that the sum of its coefficients, the 1 minus 3 plus 2 would be equal to 0.
02:14
So that means that at least one of the roots equals 1, which would have led to a polynomial division, right? so like that, and it would have given us x plus 2 times x times x would have been quicker.
02:41
All right, so that's part a we have a part a, a part b, and a part c.
02:51
So for this one, p, it's negative 27 and q is um, is um, negative 54...