x^3 - 3x^x2 +12x - 10 = 0
1 is a root
Using synthetic division, we can find the other two roots
1 1 - 3 12 -10
1 -2 10
1 -2 10 0
The remaining polynomial is
x^2 - 2x + 10 this has no real roots
The other two roots are:
1- 3i and 1 + 3i
what would the roots be for the equation x*x*x-3*x*x+12x-10
x3−3⋅x2+12⋅x−10=013−3⋅12−12⋅1−10=0⇒x1=1
If we have x1, then we can calcutate x2 and x3
x2=u+v⋅ix3=u−v⋅i
We can calculate u and v
a⋅x3+b⋅x2+c⋅x+d=0u=−12⋅(x1+ba)v=(√da⋅x1+u2)⋅i
x3−3⋅x2+12⋅x−10=0a⋅x3+b⋅x2+c⋅x+d=0a=1b=−3c=12d=−10x1=1u=−12⋅(1+−31)=−12⋅(1−3)=−12⋅(−2)u=1v=√−101⋅1+12⋅i=√−101+1⋅i=√−10+1⋅i=√−9⋅i=√9⋅i⋅i=3⋅i2i2=−1=3⋅(−1)v=−3
x2=u+v⋅ix2=1−3⋅ix3=u−v⋅ix3=1−(−3)⋅ix3=1+3⋅i