Question 145754: how do I solve: y^4-3y^3-2y^2+10y-12=0, given root 1+i
Answer by nabla(475)   (Show Source):
You can put this solution on YOUR website! I am going to switch notation from y to x.
Since 1+i is a zero, 1-i will be as well.
Thus we have:
(x-(1+i))(x-(1-i))(ax^2+bx+c)=0
c - c i^2 + b x - 2 c x - b i^2 x + a x^2 - 2 b x^2 + c x^2 -
a i^2 x^2 - 2 a x^3 + b x^3 + a x^4=0
Now, the coefficients of the original 4th degree polynomial are 1, -3, -2, 10,-12.
SO,
2c=-12 implies c=-6.
2b-2c=10 implies b=-1
2a-2b+c=-2 implies a=1.
This gives (x-(1+i))(x-(1-i))(x^2-x-6)=0.
We can solve for the other two zeroes by factoring.
(x-3)(x+2)=0
gives x=3 or x=-2
So, x=3,x=-2,x=1 +i, x=1 -i are the roots.
We can see this in the graph (only the real roots of 3 and -2 will show)...
As with my last answer, you can E-mail me if you would like the problem done in a different way.
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